Moles of Na 2 SO 4 to be dissolved in 12 mole water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is:
What is osmolarity of a 0.20 M KCl solution?
Estimate the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at 100 o C
Study of boiling points is called Ebullioscopy. Ebullioscopic constant depends on
A solution has 1 : 4 mole ratio of pentane to hexane. The vapour pressure of pure hydrocarbons at 20 o C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:
The lubricating action of an oil is more if it possess:
Total vapour pressure of mixture of 1 mol A ( p A 0 = 150 torr ) and 2 mol B ( p B 0 = 240 torr ) is 200 torr. In this case
Benzene ( C 6 H 6 , 78 g/mol) and toluene ( C 7 H 8 , 92 g/mol) form an ideal solution. At 60 o C the vapour pressure of pure benzene and pure toluene are 0.507 atm and 0.184, respectively. The mole fraction of benzene in a solution of these two chemicals that has a vapour pressure of 0.350 atm at 60 o C
The vapour pressure of pure liquid A is 10 torr and at the same temperature when 1 g of B solid is dissolved in 20 g of A, its vapour pressure is reduced to 9.0 torr. If the Molecular mass of A is 200 amu, then the molecular mass of B is
The vapour pressure of pure liquid solvent is 0.50 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is
The freezing point of equimolal aqueous solution will be highest for:
Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 2 : 3. Vapour pressure of pure liquid ‘A’ is 600 mm Hg and pure liquid ‘B’ is 400 mm Hg. Mole fraction of ‘B’ in vapour phase will be
4 litres of 2.5 M Hydrochloric acid completely neutralizes 20 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.25 M Nitric acid completely neutralizes 300 ml of 0.75 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.25 M Nitric acid completely neutralizes 300 ml of 0.75 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Nitric acid completely neutralizes 2 litres of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Nitric acid completely neutralizes ‘X’ litres of 0.1 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M Nitric acid completely neutralizes 1 litre of 0.5 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.2 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.33 M Phosphoric acid completely neutralizes 300 ml of 1 M Magnesium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Phosphoric acid completely neutralizes 6 litres of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Phosphoric acid completely neutralizes 20 litres of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Phosphoric acid completely neutralizes 80 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
‘X’ litres of 0.1 M tribasic acid completely neutralizes 2 litres of 0.4 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
4 litres of 2.5 M tribasic acid completely neutralizes 20 litres of ‘X’ M diacidic base. The value of ‘X’ is
2 litres of 1.5 M tribasic acid completely neutralizes 12 litres of ‘X’ M diacidic base. The value of ‘X’ is
80 ml of 5 M tribasic acid completely neutralizes 80 ml of ‘X’ M diacidic base. The value of ‘X’ is
80 ml of 5 M tribasic acid completely neutralizes 80 ml of ‘X’ M diacidic base. The value of ‘X’ is
80 ml of 5 M tribasic acid completely neutralizes 80 ml of ‘X’ M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 600 ml of 2 M diacidic base. The value of ‘X’ is
500 ml of 0.4 M dibasic acid completely neutralizes ‘X’ ml of 0.1 M triacidic base. The value of ‘X’ is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 65% dissociated Sodium sulphate solution is
10% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
Raoult’s law is obeyed by each constituent of a binary liquid solution when:
Two liquids A and B have P A ∘ and P B ∘ in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phase are 1 : 3 then mole fraction of ? in vapour phase in equilibrium with the solution is equal to:
A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour pressure at 25 o C. Which of the following operation results in a decrease in the equilibrium vapour pressure at 25 o C?
The degree of dissociation of an electrolyte is α and its van’t Hoff factor is i. The number of ions .obtained by complete dissociation of 1 molecule of the electrolyte is :
A very diluted saturated solution of a sparingly soluble salt X 3 Y 4 has a vapour pressure of 20 mm Hg at temperature I while pure water exerts a pressure of 20.0126 mm Hg at the same temperature. Calculate molality (m) at temperature I:
The boiling point elevation constant for toluene is 3.32 K kg mol -1 . The normal boiling point of toluene is 110.7 o C. The enthalpy of vaporisation of toluene would be nearly:
A compound has the empirical formula C 10 H 8 Fe. A solution of 0.26 g of the compound in 11.2 g of benzene (C 6 H 6 ) boils at 80.26 o C. The boiling point of benzene is 80.10 o C; the K b is 2.53’C/mo1al. What is the molecular formula of the compound?
Which condition is not satisfied by an ideal solution?
Vapour pressure of two pure liquids A and B are 450 mm and 700 mm Hg respectively at 623 0 C . If the total vapour pressure of the liquid mixture is 600 mm then the mole fraction of ‘A’ in vapour phase is equal to
Which one of the following electrolytes has the same value of van’t Hoff factor (i) as that of Al 2 SO 4 3 (if all are 100 % ionised)?
The freezing point depression constant ( K f ) of benzene is 5.12 K kg mol – 1 . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places):
What is the concentration of O 2 in a fresh water stream in equilibrium with air at 30 o C and 2.0 bar? Given, K H (Henry’s law constant) of O 2 = 2 .0 × 10 – 3 mol/kg bar at 30 o C .
If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500 ml then new concentration of the solution is
The vapour pressure of a liquid in a closed container depends upon
An ideal solution has two components A and B. If A is more volatile than B and also P A o > P T , then the correct relation between mole fraction of A in liquid (X) and vapour (Y) phase is:
Which of the following is correct for a solution showing positive deviations from Raoult’s law?
Water and ethanol form non-ideal solution with positive deviation from Raoult’s law. This solution will have vapour pressure
The vapour pressure of the solution of two liquids A ( p o = 80 mm ) and B ( p o = 120 mm ) is found to be 100 mm when X A = 0 .4 . The result shows that
Mixture of volatile components A and B has total vapour pressure (in torr) p = 254 — 119 X A where X A is mole fraction of A in mixture. Hence p A o and p B o are (in torr)
The freezing point of a solution containing 50 cm 3 of ethylene glycol in 50 g of water is found to be − 34 o C . Assuming ideal behaviour, calculate the density of ethylene glycol ( K f for water = 1.86 K kg mol – 1 )
Which statement is false?
The process of getting fresh water from sea water is known as
x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27 o C . Calculate the value of ‘x’. Assume complete dissociation of NaCl and ideal behaviour of this solution.
A 5.25% solution of a substance is isotonic with a 1.5% solution of rea (molar mass = 60 g mol – 1 ) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm – 3 , molar mass of the substance will be :
The solution containing 4.0 gm of a polyvinyl chloride polymer in 1 litre of dioxane was found to have an osmotic pressure 6 .0 × 10 − 4 atmosphere at 300 K, the value of R used is 0.082 litre atmosphere mole − 1 K − 1 . The molecular mass of the polymer was found to be
The freezing point of 0.2 molal K 2 SO 4 is − 1.1 o C . Calculate percentage degree of dissociation of K 2 SO 4 . K f for water is 1 . 86 o
If a 6.84% (wt./vol.) solution of cane-sugar (mol. Wt. = 342) is isotonic with 1.52% (wt./vol.) solution of thiocarbamide, then the molecular weight of thiocarbamide is
For 0.1 M solution, the colligative property will follow the order
The van’t Hoff factor i for an infinitely dilute solution of N a H S O 4 is:
1 mol each of following solutes are taken in 5 mol water, (A) NaCl (B) K 2 SO 4 (C) Na 3 PO 4 (D) glucose Assuming 100% ionisation of the electrolyte, relative decrease in vapour pressure will be in order:
Aluminium phosphate is 100% ionised in 0.01 molal aqueous solution. Hence, ΔT b / K b is:
A solution of crab haemocyanin, a pigmented protein extracted from crabs, was prepared by dissolving 0.750 g in 125 c m 3 of an aqueous medium. At 4 o C an osmotic pressure rise of 2.6 mm of the solution was observed. The solution has a density of 1.00 g / c m 3 . Determine the molecular weight of the protein.
1.0 molal aqueous solution of an electrolyte X 3 Y 2 is 25% ionized. The boiling point of the solution is ( K b for H 2 O = 0 .52 K kg/mol )
Which of the following aqueous solutions has osmotic pressure nearest to that of an equimolar solution of K 4 [ Fe ( CN ) 6 ]
For an ideal solution containing a non-volatile solute, which of the following expression is correctly represented? Where m is the molality of the solution and K b is molal elevation constant.
Statement A: The boiling and melting points of amides are higher than corresponding acids. Statement B: It is due to strong intermolecular hydrogen bonding in their molecules.
The units of Molarity are
To change the molal concentration to one half, one of the following should be adopted
P A and P B are the vapour pressure of pure liquid components A and B respectively of an ideal binary solution .If X A represents the mole fraction of component A, the total vapour pressure solution will be
At 300 K, osmotic pressure of a deci molar solution of Sodium sulphate was observed to be 3.695 atmospheres. Degree of ionization of Sodium sulphate is
The molality of a urea solution in which 0.0100 g of urea,[(NH 2 ) 2 CO] is added to 0.3000 dm 3 of water at STP is
Which of the following is true about the liquid solution?
If two bottles A and B contain I M and 1 m aqueous solution of sulphuric acid respectively,
Which of the following statements is true about Henry’s law?
The solubility of N 2 in water at 300 K and 500 torr partial pressure, is 0.01g L -1 .The solubility in (g L -1 ) at 750 torr partial pressure is
Which of the following statements is not true?
Anoxia is a condition, generally seen in climbers because of
100 ml of 2 M acidified potassium permanganete completely oxidizes 100 ml of potassium iodide in faint alkaline medium. Molarity of potassium iodide solution is
Among the following, the azeotropic mixture is
When concentration of a salt solution is increased,
A 5% solution (by mass) of cane sugar in water has freezing point = 271 K and freezing point of pure water rs 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is
Osmosis is the process of movement of solvent particles
Solutions A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl 2 and 0.1 M AlCl 3 . Which one of the following pairs is isotonic?
K f for water is 1.86 K kg mol -1 . If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C 2 H 6 O 2 ) must you add to get the freezing point of the solution lowered to -2.8 o C ?
Which of the following will have same value of van’t Hoff factor as that of Al 2 (SO 4 ) 3 ?
During the depression of freezing point experiment an equilibrium is established between the molecules of
The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular weight = 3429 mol -1 ) in 100 g of water is 105.0 o C. If K b and K 6 of water are 1.86 and 0.51 K kg mol -1 respectively, the weight of sucrose in the solution is about
6 g of a compound exerts the same osmotic pressure as that of 0.05 M glucose solution. Find out the molecular formula of the compound if empirical formula of non-electrolyte is CH 2 O.
The van’t Hoff factor (r) for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
The elevation in boiling point of a solution is 13.44 g of CuCl 2 in 1 kg of water using the following information will be (molecular weight of CuCl 2 =134.4 and K b = 0.52 Km -1 ).
The molar mass of the solute, sodium hydroxide obtained from the measurement of the osmotic pressure of its aqueous solution at 27 o C is 25 g mol -1 . Therefore, its ionisation percentage in this solution is
Find out the osmotic pressure of 0.1 M monobasic acid, if pH = 2.0 at 25 o C.
Select the incorrect statements about solubility.
Statement 1: Freezing point of solvent is more than that of solution. Statement 2: When non-volatile solid is added to the solvent, its vapour pressure increases and become equal to solid solvent at the lower temperature.
Match the items of Column I with Column II. Column I (Concentrations term) Column II (Factor on concentration term depends) A. Molarity p. Dependent upon temperature B. Molality q. Not depend upon volume C. Mass per cent r. Depend upon volume of solution D. Volume per cent s. Depend upon moles of solute
Select the correct statements.
Statement 1: The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different. Statement 2: Water is polar and benzene is non-polar.
Two beakers of capacity 500 mL were taken. One of these beakers, labelled as A, was filled with 400 mL water, whereas the other beaker is labelled as ‘B’ was filled with 400 mL of 2 M solution of NaCl. At the same temperature both the beakers were placed in closed containers of same material and same capacity as shown in figure. At a given temperature, which of the following statement is correct about the vapour pressure of pure water and that of NaCl solution?
Which one of them is more volatile component?
Statement 1: When methyl alcohol is added to water, boiling point of water increases. Statement 2: When a volatile solute is added to a volatile solvent, elevation in boiling point is observed.
Inter molecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?
Colligative properties depend on
Which is the correct statement for positive deviation of solution from Raoult’s law?
Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 2 : 3. Vapour pressure of pure liquid ‘A’ is 600 mm Hg and pure liquid ‘B’ is 400 mm Hg. Mole fraction of ‘B’ in vapour phase will be
At 300 K, 0.3 M sucrose solution is isotonic with 0.15 M KCl solution. Degree of dissociation of KCl in the solution is
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
4.5 litres of 1.5 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
4.5 litres of 1.5 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
4 litres of 1 M KOH is diluted to 0.05 M solution. Final volume of the solution is ——- litres
4 litres of 1 M KOH is diluted to 0.05 M solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
1.5 litres of 2 N H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
4 litre of 0.125 M Barium hydroxide is diluted to 8 litre solution. Normality of the resultant solution is —- N
1 litre of 1 M Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
2 litres of 1.5 M Sulphuric acid completely neutralizes 12 litres of ‘X’ M NaOH solution. The value of ‘X’ is
1 litre of 1.5 M Sulphuric acid completely neutralizes 6 litres of ‘X’ M NaOH solution. The value of ‘X’ is
2 litres of 1.5 M Sulphuric acid completely neutralizes 12 litres of ‘X’ M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Hydrochloric acid completely neutralizes 200 ml of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 3 M Hydrochloric acid completely neutralizes 300 ml of 1.5 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
2 litres of 1.5 M Hydrochloric acid completely neutralizes 12 litres of ‘X’ M KOH solution. The value of ‘X’ is
75 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 12 M NaOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Hydrochloric acid completely neutralizes 6 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Nitric acid completely neutralizes ‘X’ litres of 0.2 M Calcium hydroxide solution. The value of ‘X’ is
500 ml of 0.2 M Nitric acid completely neutralizes ‘X’ ml of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Nitric acid completely neutralizes 20 litres of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Nitric acid completely neutralizes 80 ml of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Nitric acid completely neutralizes 80 ml of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.2 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Sulphuric acid completely neutralizes 20 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Sulphuric acid completely neutralizes 6 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Sulphuric acid completely neutralizes 6 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Phosphorous acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
1 litre of 0.2 M Phosphoric acid completely neutralizes ‘X’ litres of 0.1 M Magnesium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 10 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Phosphorous acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
4 litres of 0.2 M Phosphoric acid completely neutralizes ‘X’ litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M tribasic acid completely neutralizes 1 litre of 0.5 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
4 litres of 2.5 M tribasic acid completely neutralizes 20 litres of ‘X’ M diacidic base. The value of ‘X’ is
400 ml of 1 M tribasic acid completely neutralizes ‘X’ ml of 1 M monoacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
‘X’ litres of 2 M dibasic acid completely neutralizes 1 litre of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 300 ml of 0.75 M triacidic base. The value of ‘X’ is
1 litre of 1.5 M dibasic acid completely neutralizes 6 litres of ‘X’ M triacidic base. The value of ‘X’ is
4 litres of 2.5 M dibasic acid completely neutralizes 20 litres of ‘X’ M triacidic base. The value of ‘X’ is
40 ml of 5 M dibasic acid completely neutralizes 80 ml of ‘X’ M triacidic base. The value of ‘X’ is
40 ml of 5 M dibasic acid completely neutralizes 80 ml of ‘X’ M triacidic base. The value of ‘X’ is
100 ml of ‘X’ M dibasic acid completely neutralizes 300 ml of 4 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
1 litre each of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
800 ml of 1 M HCl and 200 ml of 2 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated CH 3 COOH solution is
Van’t Hoff factor of 10% dissociated CH 3 COOH solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 15% dissociated Potassium phosphate solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
For the dissociation of PCl 5 s into PCI 3 and CI 2 , in gaseous phase reaction, if d is the observed vapour density and D the theoretical vapour density with α as degree of dissociation. Variation of D/d with ‘ α is given by which graph?
For an ideal binary liquid solution with P A ∘ > P B ∘ , which relation between X A (mole fraction of A in liquid phase) and Y A (mo1e fraction of A in vapour phase) is correct ?
The solubility of a specific non-volatile salt is 4 g in 100 g of water at 25 o C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25 o C, in system X, y and Z. The vapour pressure would be in the order:
Two liquids A and B have vapour pressure in the ratio P A ∘ : P B ∘ = 1 : 3 at a certain temperature. Assume A and B form an ideal solution and the ratio of mole fractions of A to B in the vapour phase is 4 : 3. Then the mole fraction of B in the solution at the same temperature is :
Which is correct statement?
When a liquid that is immiscible with water was steam distilled at 95.2.C at a total pressure of 748 tort the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 toff at 95.2 o C, what is the molar mass of liquid?
Water and chlorobenzene are immiscible liquids. Their mixture boils at 89 o C under a reduced pressure of 7.7 x 10 4 Pa. The vapour pressure of pure water at 89 o C is 7 x 10 4 pa. Weight per cent of chlorobenzene in the distillate is :
A solution of 0.640 g of azulene in 100.0 g of benzene boils at 80.23 o C. The boiling point of benzene is 80.10’C; the K b is 2.53 o C/molal. what is the molecular weight of azulene?
Molality of 110 grams of aqueous solution containing 10 grams of caustic soda is
0.1 moles of Barium hydroxide is dissolved in a 400 ml solution. Normality of the solution is—–N
Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
All the following liquid mixtures show negative deviations from Raoult’s law except
Which of them is not equal to zero for an ideal solution?
Henry’s constant of Argon gas at 298K temperature is 40kbar. Calculate the weight of Argon dissolved in 1000 grams of water when packed at a pressure of 7.2 bar and 298 K .
Statement I : Solubility of Helium in blood is more than that of Nitrogen Statement II : Henry’s law constant increases with increase in temperature.
Consider the solutions given below. The solution with least freezing point is
Molarity of 0.5N Potassium ferrocyanide solution is
Vapour pressure of water at certain temperature is 90mm Hg. 6 grams of a non–volatile solute (molar mass – 180g) is dissolved in 180 grams of water. Vapour pressure of solution at that temperature would be
Freezing point of 0.1 M aqueous solution of Calcium nitrate cannot be
If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g mol – 1 ) of the solute is [Given that molar mass of n-octane is 114 g mol – 1 ]
The major products C and D formed in the following reaction respectively are H 3 C − CH 2 − CH 2 − O − C CH 3 3 excessHl Δ C + D
Mole fraction of water vapour in moist air is 0.05. If the total pressure of moist air is 1.8 atm then the partial pressure of dry air will be
Weight of Na 2 CO 3 which can neutralize 500ml of 0.1M H 2 SO 4 solution is
A) Molality B) Mole fraction C) Lowering of Vapour pressure (LVP) D) Relative lowering of vapour pressure (RLVP) From the above list identify the property which is influenced by temperature
Molality of a solution is 1m. If the density of the solution is 0.8g/cc, the molarity of the solution will be (molar mass of solute = 80g)
Two volatile liquids A and B are mixed in the mole ratio 2:3 . Calculate the vapour pressure of pure A if the vapour pressure of pure B is 200mm Hg and the total vapour pressure is 300mm Hg
A solution of Al 2 ( SO 4 ) 3 [ d = 1 .253 gm / ml ] contain 22% salt by weight. The molarity, normality and molality of the solution is
Pressure cooker reduces cooking time for food because
Following data has been given for CO 2 for the concentration in H 2 O . Temperature Henry’s constant Pressure 273 K 600 atm 0.30 atm 333 K 3400 atm p 2 If solution of CO 2 in H 2 O is heated from 273 to 333 K, pressure ( p 2 ) needed to keep CO 2 in the solution is
A solid dissolves in water if
The boiling points of C 6 H 6 , CH 3 OH , C 6 H 5 NH 2 and C 6 H 5 NO 2 are 80 o C , 65 o C , 184 o C and 212 o C respectively. Which of the following will have highest vapour pressure at room temperature?
According to William Henry; the solubility of a gas in liquid depends on the pressure of the gas. If ‘m’ is the molality of the gas and ‘P’ is its pressure then which of the following plot is in accordance with the law:
Given P-x curve for a non-ideal liquid mixture (Fig.). Identify the correct T-x curve for the same mixture.
Which of the following plots does not represent the behaviour of an ideal binary liquid solution?
Which of the following is less than zero for ideal solutions?
If vapour pressures of pure liquids ‘A’ & ‘B’ are 300 and 800 torr respectively at 25 o C . When these two liquids are mixed at this temperature to form a solution in which mole percentage of ‘B’ is 92, then the total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution.
If liquid A and B form ideal solution, then:
The mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.500 is (vapour pressure of pure benzene and pure toluene are 119 torn and 37.0 torn respectively at the same temperature).
At 334 K the vapour pressure of benzene ( C 6 H 6 ) is 0.526 atm and that of toluene ( C 7 H 8 ) is 0.188 atm. In a solution containing 0.500 mole of benzene and 0.500 mole of toluene, what is the vapour pressure of toluene above the solution at 334 K?
The mass of a non-volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% will be
Colligative properties depend on the number of solute particles present in the solution. Osmotic pressure of 60% ionized 0.1M BaCl 2 solution at 27 0 c is
100ml of decimolar sulphuric acid is diluted to one litre. Proton concentration of the solution is
18 g of glucose is dissolved in 180 grams of water. Vapour pressure of solution is 178.2 mm . Vapour pressure of water at the same temperature will be
In the following solutions, the solution with highest vapour pressure is
The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is (molal depression constant = 1.86)
When a solution containing w g of urea in 1 kg of water is cooled to − 0 .372 o C , 200 g of ice is separated. If K f for water is 1.86 K kg mol – 1 , w is
The osmotic pressure of blood is 7.65 atm. at 310 K. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)
Which has maximum osmotic pressure at temperature T K?
A 5.8% (wt./vol.) NaCl solution will exert an osmotic pressure closest to which one of the following:
0.5 normal sugar solution is isotonic with
The order of osmotic pressure of equimolar solutions of BaCl 2 , NaCl and glucose will be :
The degree of dissociation ( α ) of weak electrolyte A x B y is related to van’t Hoff factor (i) by the expression
Which has the highest boiling point?
A complex of iron and cyanide ions is 100% ionised at 1m (molal). If its elevation in b.p. is 2.08. Then the complex is ( K b = 0 .52 o mol − 1 kg ) :
Aqueous solution of barium phosphate which is 100% ionised has ΔT f / K f as 0.05 . Hence, given solution is
The fraction of phenol dimerised in benzene if 20 g of phenol in 1 kg benzene exhibits a freezing point depression of 0.69 K. (K f benzene = 5.12 ) , (MW phenol = 94). What is the value of Van’t Hoff for (i) in this reaction?
The ratio of the value of any colligative property for K 4 [ Fe ( CN ) 6 ] solution to that of Fe 4 [ Fe ( CN ) 6 ] 3 (prussian blue), solution is nearly
To halve the molarity of a solution the following should be adopted
Which of the following acid has the same molecular weight and equivalent weight
0.1 gram mole of urea is dissolved in 100g. of water. The molality of the solution is
Example of solid foam is
A molecule ‘M’ associates in a given solvent according to the equation M ⇌ ( M ) n for a certain concentration of ‘M’ , the van’t hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2, the value of n is
At 298 K, 6 grams of urea is dissolved in 90 grams of water. Relative lowering of vapour pressure of the solution is
Boiling point is highest for (assume 100% ionization)
Molefraction of solute in 0.1 molal aqueous solution will be
Respiratory kit of Scuba divers contain X% of Nitrogen, Y% of Oxygen and Z% of Helium. Correct relation between X, Y and Z is
At 298 K, 34.2 grams of sucrose is dissolved in 180 grams of water. Relative lowering of vapour pressure of solution will be
Which of the following is the correct example of solid solution in which the solute is in gas phase?
If at certain temperature, the ;pour pressure of pure water is 25 mm of Hg and that of a very dilute aqueous urea solution is 24.5 mm of Hg, the molality of solution is
At 300 K two pure liquids A and B have 150 mm Hg and 100 mmHg vapour pressures, respectively. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is
1.26 g of the protein is present in the aqueous solution of 200 cm 3 .Calculate the molar mass of the protein,if the osmotic pressure of such solution is 2.57 x 10 -3 bar at 300 K.
The correct statement about semipermeable membrane is
Select the correct statement.
The van’t Hoff factor of BaCl 2 at 0.01 M concentration is 1.98. The percentage of dissociation of BaCl 2 at this concentration is
Arrange the following aqueous solutions in the order of their increasing boiling points. I. 10 – 2 MNaCl II. 10 – 3 M MgCl 2 III. 10 – 4 M Urea IV. 10 – 4 M NaCl
Which of the following has least freezing point?
lnterpret the correct statement for the following figure.
A 6% solution of urea is isotonic with
What is the freezing point of a solution containing 8.1 g HBr in 100 g of water. Assuming the acid to be 90% ionised? (K f for water = 1.86 K kg mol -1 )
Arrange the following in the increasing order of their solubility in n-octane based on solute-solvent interactions.
The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass = 3429 mol -1 ) to 1000 g of water, if the vapour pressure of water at 25 o C is 23.8 mm of Hg is
Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with 8 solution of benzene and toluene having a mole fraction of toluene 0.50, will be
Here A, B C can be respectively,
Solvent is the component of a solution
The order of boiling points of four equimolar aqueous solutions is C < B < A < l). The correct order of their freezing points is
In the graph given below, what does the slope of the line represent ?
In case of which type of solution concentration of solute shows its solubility?
A 5.2 molal aqueous solution of methyl alcohol, CH 3 OH is supplied. What is the mole fraction of methyl alcohol in the solution ?
For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point a 1 atm pressure is 2 o C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b = 0.76K kg mol -1 )
Select the incorrect statement.
one component of a solution follows Raoult’s law over the entire range 0 ≤ x 1 ≤ 1 . The second component must follow Raoult’s law in the range when x 2 is
The total pressure (p total ) over the solution phase in the container will be the sum of the partial pressures of the components of the solution. This statement to belongs
Identify the phase of solute and solvent among the options are given below, for a solution as amalgam of mercury with sodium.
0.01 M solution of KCI and BaCl 2 are prepared in water. The freezing point of KCl is found to be -2 o C. What is the freezing point of BaCl 2 to be completely ionised?
Match the items of Column I with Column II. Column I (Example of solution) Column II (Type of solution) A. Sucrose solution p. Either solute or solvent is liquid B. Air q. Solid solution C. Brass r. Homogeneous mixture D. Amalgam s. Gaseous solution
Which of the following statements are true about osmotic pressure? I. Flow of solvent from solution across a semipermeable membrane can be stopped, if some extra pressure is applied on solution. II. It is the pressure that stops the flow of solvent towards solution. III. It is the pressure that follow flows of solvent across the semipermeable membrane.
Statement 1: 1 . 575 g H 2 C 2 O 4 2 H 2 O in 250 mL solution makes it 0.1 N. Statement 2: H 2 C 2 O 4 . 2 H 2 O is a dihydrate organic acid.
Match the terms given in Column I with expression given in Column II. Select an appropriate answer from the codes given below. Column I Column II A. Isotonic solution 1. Salt concentration is less than 0.9% (m/V), water will flow into cell and it will swell. B. Hypertonic solution 2. Salt concentration is more than 0.9% (m/V) NaCl, water will flow out of cell and the cell will shrink. C. Hypotonic solution 3. No osmosis occurs, if solution is separated by semipermeable membrane.
The air is a mixture of a number of gases. The major components of air are oxygen and nitrogen with approximate proportion of 20% to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, if the Henry’s law constants for oxygen and nitrogen are 330 x 10 7 mm and 6.51 x 10 7 mm respectively, calculate the composition of these gases in water.
Compartments A and B have the following combinations of solution. Answer the following questions on this basis. The solutions in which compartment B is hypertonic are
Henry’s law constant for the solubility of methane in benzene at 298 K is 4.27 x 10 5 mm Hg. Calculate the solubility of methane in benzene at 298K under 760 mmHg.
The diagram given below represents the vapour pressure and mole fraction of an ideal solution of component 1 and 2. Answer the following questions. What does lines I, II and III indicate?
Determine the osmotic pressure of the solution prepared by dissolving 25 mg of K 2 SO 4 in 2L of water at 25 o C. (Assuming it is to be completely dissociated)
If two liquids A and B form minimum boiling azeotrope at some specific composition then
In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl 2 solution is
4L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molarity of the resultant solution is
Which of the following binary mixtures will have same composition in liquid and vapour phase?
For a binary ideal liquid solution, the variation in total vapour pressure versus composition of solution is given by which of the curves?
Soft drinks and soda water bottles are sealed under high pressure
van’t Hoff factor for benzoic acid in benzene undergoing 80% dimerization is
Find the value of mole fraction of the solute when vapour pressure of glucose is 750 mm Hg at 373 K in dilute solution.
If A and B are two components of an ideal solution. Y is molefraction in vapour phase. Total vapour pressure of an ideal solution is represented as P s = 150 + 50 X A mm Hg at 25°C. at If X B = 0.4 then select the incorrect statement.
Boiling point of 0.2m urea solution will be ( K b of water is 0.52 K Kg mole – 1 )
Boiling point of 1 m sucrose solution will be ( K b of water is 0.52 K Kg mole – 1 )
1 litre of 0.5 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
10 litre of 1 M HCl is diluted to 40 litre solution. Normality of the resultant solution is —- N
10 litre of 0.5 M HCl is diluted to 20 litre solution. Normality of the resultant solution is —- N
1 litre of 0.5 M HCl is diluted to 5 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
4.5 litres of 1.5 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 1 M H 2 SO 4 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 1.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
1 litre of 0.5 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
1 litre of 0.5 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
2 litres of 1.5 M Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M NaOH solution. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ litres of 2 M Sulphuric acid completely neutralizes 1 litre of 0.5 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
1 litre of 1.5 M Sulphuric acid completely neutralizes 6 litres of ‘X’ M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Hydrochloric acid completely neutralizes 200 ml of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Hydrochloric acid completely neutralizes 200 ml of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 3 M Hydrochloric acid completely neutralizes 300 ml of 1.5 M KOH solution. The value of ‘X’ is
2 litres of 1.5 M Hydrochloric acid completely neutralizes 12 litres of ‘X’ M KOH solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
80 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
1 litre of 1.5 M Hydrochloric acid completely neutralizes 6 litres of ‘X’ M KOH solution. The value of ‘X’ is
1 litre of 1.5 M Hydrochloric acid completely neutralizes 6 litres of ‘X’ M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
75 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 12 M NaOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M Hydrochloric acid completely neutralizes 1 litre of 0.5 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Hydrochloric acid completely neutralizes 20 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Hydrochloric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
500 ml of 0.2 M Nitric acid completely neutralizes ‘X’ ml of 0.1 M Calcium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Hydrochloric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Hydrochloric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Nitric acid completely neutralizes ‘X’ litres of 0.2 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Nitric acid completely neutralizes 2 litres of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 3 M Nitric acid completely neutralizes 300 ml of 4.5 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 3 M Nitric acid completely neutralizes 300 ml of 4.5 M Calcium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Nitric acid completely neutralizes 80 ml of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Nitric acid completely neutralizes 80 ml of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Nitric acid completely neutralizes 80 ml of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 600 ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Nitric acid completely neutralizes 20 litres of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.2 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M Sulphuric acid completely neutralizes 1 litre of 0.5 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
60 ml of 1 M Sulphuric acid completely neutralizes 180 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Sulphuric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Sulphuric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Sulphuric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphorous acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M Phosphorous acid completely neutralizes 1 litre of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphoric acid completely neutralizes ‘X’ litres of 0.1 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphoric acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphoric acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 1.5 M Phosphoric acid completely neutralizes 12 litres of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 600 ml of 2 M Magnesium hydroxide solution. The value of ‘X’ is
125 ml of ‘X’ M Phosphoric acid completely neutralizes 50 ml of 5 M Magnesium hydroxide solution. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
100 ml of 2 M tribasic acid completely neutralizes ‘X’ ml of 0.5 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
1 litre of 2.5 M tribasic acid completely neutralizes 5 litres of ‘X’ M diacidic base. The value of ‘X’ is
1 litre of 1.5 M tribasic acid completely neutralizes 6 litres of ‘X’ M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 300 ml of 4 M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 300 ml of 4 M diacidic base. The value of ‘X’ is
75 ml of ‘X’ M tribasic acid completely neutralizes 25 ml of 12 M diacidic base. The value of ‘X’ is
75 ml of ‘X’ M tribasic acid completely neutralizes 25 ml of 12 M diacidic base. The value of ‘X’ is
75 ml of ‘X’ M tribasic acid completely neutralizes 25 ml of 12 M diacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
‘X’ litres of 1 M dibasic acid completely neutralizes 2 litres of 0.5 M triacidic base. The value of ‘X’ is
‘X’ litres of 0.2 M dibasic acid completely neutralizes 2 litres of 0.4 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.1 M dibasic acid completely neutralizes 200 ml of 0.4 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 300 ml of 0.75 M triacidic base. The value of ‘X’ is
150 ml of 1 M dibasic acid completely neutralizes 450 ml of ‘X’ M triacidic base. The value of ‘X’ is
150 ml of 1 M dibasic acid completely neutralizes 450 ml of ‘X’ M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
1 litre of 0.6 M NaOH and 3 litres 0.4 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.6 M NaOH and 3 litres 0.4 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.3 M NaOH and 3 litres 0.1 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.3 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
0.3 litres each of 1 M HCl and 2 M H 2 SO 4 are mixed with each other. The solution is diluted to one litre. Proton concentration of the resultant mixture is —– M
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
0.3 litres each of 1 M HCl and 2 M H 2 SO 4 are mixed with each other. The solution is diluted to one litre. Proton concentration of the resultant mixture is —– M
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
1 litre each of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
6 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
4 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
800 ml of 1 M HCl and 200 ml of 2 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 10% dissociated CH 3 COOH solution is
Van’t Hoff factor of 10% dissociated CH 3 COOH solution is
Van’t Hoff factor of 65% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 40% dissociated Potassium sulphate solution is
Van’t Hoff factor of 75% dissociated Barium nitrate solution is
Van’t Hoff factor of 5% dissociated Aluminium fluoride solution is
Van’t Hoff factor of 25% dissociated AlF 3 solution is
20% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
85% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
85% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
85% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
90% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
An ideal solution is formed by mixing two volatile liquids A and B. X A and X B are the mole fractions of A and B respectively in the solution and Yo and Y, are the mole fractions of A and B respectively in the vapour phase. A plot of 1 Y A along y-axis against, along x-axis gives a 1 X A straight line. What is the slope of the straight line ?
For a dilute solution, Raoult’s law states that:
An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same temperature?
The vapour pressure of an aqueous solution of sucrose at 373 K is found to be 750 mm Hg. The molality of the solution at the same temperature will be :
At 25 o C, t]le vapour pressure of pure liquid A (mol. wt. = 40) is 100 torr while that of pure liquid B is 40 torr, (mol. wt. – B0). The vapour pressure at 25 o C of a solution containing 20 g of each A and B is:
If two liquids A ( P A o =100 torr) and B ( P B o =200 torr) are completely immiscible with each other (each one will behave independently of the other) are present in a closed vessel. The total vapour pressure of the system will be:
Total vapour pressure of mixture of 1 mole of volatile component A ( P B o =100 mm Hg) and 3 mole of volatile component B ( P B o =80 mm Hg) is 90 mm HS For such case:
Which solution has the highest vapour pressure?
Four solutions of K 2 SO, with the concentrations 0.1 m, O.O7 m,0.001 m and 0.0001 m are available. The maximum value of colligative property corresponds to :
The van’t Hoff factor i for an electrolyte which undergoes dissociation and association in solvent are respectively:
When 1 mole of a solute is dissolved in 1 kg of H 2 O, boiling point of solution was found to be 100.5 o C. K b for H 2 O is:
Calculate the percentage degree of dissociation of an electrolyte XY 2 (Normal molar mass = 164) in water if the observed molar mass by measuring elevation in boiling point is 65.6:
Calculate the molecular weight of a substance whose 7.0% by mass solution in water freezes at -0.93 o C. The cryoscopic constant of water is 1.86 o C kg mol -1 :
Which of the following method of expressing concentration is independent of temperature and has units.
Mole fraction of water in 46% (w/w) aqueous solution of ethanol is
Which of the following method of expressing concentration is independent of temperature and has units.
Mole fraction of water in 46% (w/w) aqueous solution of ethanol is
24.5 grams of Sulphuric acid is dissolved in a one litre solution. Normality of the solution is…………N
Molality of 110 grams of aqueous solution containing 10 grams of caustic soda is
The exact composition of bronze is
How much percentage of fluoride ions in water prevents tooth decay?
which of the fallowing compound is used in rat poison?
The composition of brass is
which of the fallowing one determine the physical state of the solution?
Binary solution have
Solution have
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Intravenous injections are always dissolved in water containing salts at particular ionic concentrations because of
which of the fallowing component in large excess component in the solution?
Which of the fallowing concentration term is independent on temperature?
Mass %, ppm, mole fraction and molality are independent of temperature because
Which of the following statements about the composition of the vapour over an ideal 1: 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25 0 C . (Given, vapour pressure data at 25 0 C , benzene =12.8kPa, toluene =3.85 kPa
For an ideal solution, the correct option is
At 100 0 C the vapour pressure of a solution of 6 . 5 g of a solute in 100 g water is 732 mm . If k b = 0 . 52 the boiling point of this solution will be
The mixture that forms maximum boiling azeotrope is
What is the mole fraction of the solute in a 1.00 m aqueous solution?
The boiling point of 0.2 mol k g – 1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
Which one of the following is incorrect for ideal solution?
Henry’s law gives a relation between pressure of the gas and its solubility in a particular solvent at constant temperature. At a given temperature, Henry’s law constant is minimum for the following gas when dissolved in water
Normal molar mass and observed molar mass of a solute dissolved in a solvent are 60g and 80g respectively. True statement about degree of association of the solute in the solution is
1 mole of a non–volatile solute is dissolved in 360grams of water. Relative lowering of vapour pressure of the solution is
Vapour pressure of liquid depends on temperature.Which one of the following graph gives correct variation of vapour pressure with temperature (P = vapour pressure of a liquid, T = Absolute temperature)
Froth flotation is useful in concentrating (A) Galena (B) Sphalerite (C) Copper pyrites (D) Pitch blend
5.3 grams of Na 2 CO 3 is dissolved in 200ml of a solution. It is diluted by adding 800 ml of water. 100ml of the resulting solution requires ‘V’ ml of 0.02N H 2 SO 4 solution for complete neutralization. ‘V’ is
Which of the following is not a correct match
Which one of the following is not a true condition for an ideal solution
Molality and molarity of solution are 2.5m and 2M respectively. Weight of water present in 900 ml solution will be
One molal solution of ‘X’ showed a boiling point of 374.3 K. Vanthoff factor of ‘X’ in the solution is ( Molal elevation constant of water is 0 . 52 K Kg mole – 1 )
Osmotic pressure of 50% ionized 0.1 M KCl solution at 127 0 c will be
Most suitable substance for the preparation of semi permeable membrane used in reverse osmosis is
The following graph is plotted by Freundlich for physical adsorption. The correct relation is
The mixture which shows positive deviation from Raoult’s law is:
Isotonic solutions have same
If molality of the dilute solution is doubled, the value of molal depression constant ( K f ) will be
Which of the following is dependent on temperature?
Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene 0.50. If vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature, mole fraction of toluene in vapour phase will be :
Regarding ideal and non ideal solutions, incorrect combination is
Osmotic pressure of 0.1 M monoprotic acid solution at 300K is 4.926 atm . Percentage of ionization of the acid is
Vapour pressure of liquid ‘A’ is 400mm Hg and liquid ‘B’ is 600mm Hg. If the V.P of the mixture containing 1:1 mole ratio of ‘A’ and ‘B’ is 504 mm Hg then the incorrect statement is
A & B form ideal solution. A graph is plotted between Total pressure(y-axis) and Mole fraction(x-axis).From the given data, vapour pressure of pure liquid ‘A’ will be
Compartment A contains decimolar NaCl at 298 K. Compartment B contains decimolar urea at 298K . Both these solutions are separated by a semi permeable membrane. Then the correct statement is
Cryoscopic constant of benzene is 5 .12 KKgmole − 1 . Depression of freezing point for 0.0976m solution containing a non–electrolyte solute in benzene is
Vapour pressure is influenced by many factors. Raoult’s law relates vapour pressure of a liquid with its mole fraction in the solution. False statement among the following is
Mole fraction of solvent in 5 molal aqueous solution is nearly
Vapour pressure of pure liquids A & B are 180 mm and 540 mm respectively. Equimolar mixture of A & B is prepared which behaves like an ideal solution. Molefraction of ‘B’ in vapour phase will be
In the solutions given below, the solution with highest osmotic pressure at 300K will be
NaCl is dissolved in water . Its molecular weight calculated from elevation of boiling point experiment cannot be
FeSO 4 + 2 NaOH Fe OH 2 + Na 2 SO 4 . Equivalent weight of FeSO 4 in the given reaction is M = molar mass of FeSO 4
At certain temperature vapour pressures of four liquids A, B, C, D are 360 mm, 270 mm, 450 mm and 630 mm respectively. Most non-volatile liquid is
Calculate the masses of cane sugar and water required to prepare 250 grams of 25% cane sugar solution-
6.02 × 10 20 molecules of urea are present in 200 mL of its solution. The concentration of urea solution is ( N 0 = 6.02 × 10 23 m o l – 1 )
Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml. solution-
Hardness of a water sample is 100 ppm CaCO 3 . Thus, molarity of CaCO 3 is
An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is
5 ml of N HCl, 20 ml of N/2 H 2 SO 4 and 30 ml of N/3 HNO 3 are mixed together and volume made to one litre. The normality of the resulting solution is
Which of the following aqueous solutions has the highest concentration of Na + ?
At 25 o C , the density of 15 M H 2 SO 4 is 1.8 g cm – 3 . Thus, mass percentage of H 2 SO 4 in aqueous solution is
In what ratio should a 6.5 N HNO 3 be diluted with water to get 3.5 N HNO 3 ?
The molality of a urea solution in which 0.0100 g of urea, [ ( NH 2 ) 2 CO ] is added to 0.3000 dm 3 of water at STP is
15 grams of methyl alcohol is dissolved in 35 grams of water. What is the mass percentage of methyl alcohol in solution?
A 3.5 molal aqueous solution of methyl alcohol ( CH 3 OH ) is supplied. What is the mole fraction of methyl alcohol in the solution?
In which mode of expression of concentration of a solution remains independent of temperature?
The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity and normality of the solution
Calculate the molarity of pure water ( d = 1 g / ml )
Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml 0.1M solution.
Find the molality of H 2 SO 4 solution whose specific gravity is 1.98 g ml -1 and 95% by volume H 2 SO 4
20 ml of 0.02 M KMnO 4 was required to completely oxidise 10 ml of oxalic acid solution. What is the molarity of the oxalic acid solution?
Calculate molality of 1 litre solution of 93% (w/v) H 2 SO 4 (The density of solution is 1.84 g/ml)
Suppose 5 g of CH 3 COOH is dissolved in one litre of Ethanol. Assume no reaction between them. Calculate molality of resulting solution if density of ethanol is 0.789 g/ml.
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol. wt. of urea = 60).
A solution has 25% of water, 25% ethanol and 50% acetic acid by mass. The mole fraction of water,ethanol and acetic acid respectively will be
Calculate normality of the mixture obtained by mixing 100 ml of 0.1 N HCl and 50 ml of 0.25 N NaOH solution.
300 ml 0.1 M HCl and 200 ml of 0.03 M H 2 SO 4 are mixed. Calculate the normality of the resulting mixture
250 mL of a Na 2 CO 3 solution contains 2.65 g of Na 2 CO 3 . 10 mL of this solution is added to X mL of water to obtain 0.001 M Na 2 CO 3 solution. The value of X is:(Molecular mass of Na 2 CO 3 = 106amu)
A solution contains 1.2046 × 10 24 hydrochloric acid molecules in one dm 3 of the solution. The strength of the solution is
What weight of oxalic acid ( H 2 C 2 O 4 . 2 H 2 O ) is required to prepare 1,000 mL of N/10 solution?
Variation of K H (Henry’s law constant) with temperature T is shown by following graphs I — IV. Correct representation is
Which of the following units is useful in relating concentration of solution with its vapour pressure?
An unopened soda has an aqueous concentration of CO 2 at 25 o C equal to 0.0408 molal. Thus, pressure of CO 2 gas in the can is ( K H = 0.034 mole/ Kg bar)
Relation between the volume of gas (2) that dissolves in a fixed volume of solvent (1) and the partial pressure of gas (2) is ( n t = total moles, K 1 and K 2 are Henry’s constants)
CO(g) is dissolved in H 2 O at 30 o C and 0.020 atm. Henry’s law constant for this system is 6.20 × 10 4 atm. Thus, mole fraction of CO(g) is
H 2 S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol Kg -1 . Thus, Henry’s law constant ( in atm molal – 1 ) for H 2 S is
Observe the P-T phase diagrams for a given substance A. Then melting point of A(s), boiling point of A(l), critical point of A and triple point of A (at their respective pressure) are respectively –
During the evaporation of liquid
At higher altitudes the boiling point of water lowers because
The vapour pressure of two liquid P and Q are 80 torr and 60 torr respectively. The total vapour pressure obtained by mixing 3 mole of P and 2 mole of Q would be
Consider a binary mixture of volatile liquids. If at X A = 0.4 the vapour pressure of solution is 580 torr then the mixture could be ( p A o = 300 torr, p B o = 800 torr ):
If liquids A and B form an ideal solution
Mole fraction of toluene in the vapour which is in equilibrium with a solution containing benzene and toluene having 2 moles each is Given: Saturation vapour pressure of benzene = 120 torr, Saturation vapour pressure of toluene = 80 torr
For an ideal binary liquid solution with p A o > p B o . x A and y A represent the mole fraction of A in liquid phase and vapour phase respectively whereas x B and y B represents the mole fraction of B in liquid phase and vapour phase respectively. Therefore which of the following relation is correct?
Which of the following shows negative deviation from Raoult’s law?
A liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface, then the vapour pressure of the liquid in the vessel is:
Given at 350 K p A o = 300 torr and p B o = 800 torr, the composition of the mixture having a normal boiling point of 350 K is:
Which liquid pair shows a positive deviation from Raoult’s law?
In mixture A and B, components show negative deviations as:
Benzene and toluene form nearly ideal solutions. At 20 o C , the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 o C for a solution containing 78 g benzene and 46 g toluene in torr is:
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be:
At 80 o C , the vapour pressure of pure liquid ‘A’ is 520 mmHg and that of pure liquid ‘B’ is 1000 mmHg. If a mixture solution of ‘A’ and ‘ B’ boils at 80 o C and 1 atm pressure, the amount of ‘A’ in the mixture is: (1 atm = 760 mmHg):
The vapour pressure of water at 20 o C is 17.5 mmHg. If 18 g of glucose ( C 6 H 12 O 6 ) is added to 178.2 g of water at 20 o C , the vapour pressure of the resulting solution will be:
Which combination of following terms is matched correctly? (I) Inter molecular forces (II) Latent heat of vaporisation (III) Vapour pressure
A binary liquid solution of n-heptane and ethyl alcohol is prepared. Which of the following statements correctly represents the behaviour of this liquid solution?
For A and B to form an ideal solution which of the following conditions should be satisfied?
Two liquids are mixed together to form a mixture which boils at same temperature and their boiling point is higher than the boiling point of either of them so they shows.
At 35 o C the vapour of pure chloroform is 0.359 atm and that of pure acetone is 0.453 atm. A solution containing 1 mole of chloroform and 4 mole of acetone has a vapour pressure of (in atm)
The vapour pressure of hexane ( C 6 H 14 ) and heptane ( C 7 H 16 ) at 50 o C are 408 Torr and 141 Torr, respectively. The composition of the vapour above a binary solution composition containing a mole fraction of 0.300 hexane is ( Y 6 = mol fraction of hexane and Y 7 = mol fraction of heptane).
When a solution of CHCl 3 is mixed with a solution of acetone, ΔV mix is
At 25 o C the vapour pressure of benzene, C 6 H 6 (78 g/mole), is 93.2 Torr and that of toluene, C 7 H 8 (92 g/mol), is 28.2 torr. A solution of 1.0 mole of C 6 H 6 and 1.0 mol of C 7 H 8 is prepared. Calculate the mole fraction of C 6 H 6 in the vapour above this solution (assume the solution is ideal).
Vapour pressure of CCl 4 at 25 o C is 143 mmHg ,0.05 g of a non-volatile solute (mol. wt. = 65) is dissolved in 100 ml CCl 4 . Find the vapour pressure of the solution (Density of CCl 4 = 1 .58 g / cm 3 )
The vapour pressure of water depends upon:
The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm. Mole fraction of the component B in the solution is:
Among the following substances, the lowest vapour pressure is exerted by
18 g of glucose ( C 6 H 12 O 6 ) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100 o C is:
Two liquids X and Y form an ideal solution at 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively
An ideal solution has equal mol-fractions of two volatile components A and B. In the vapour above the solution, the mol-fractions of A and B:
The vapour pressure of pure water at 75 o C is 296 torr the vapour pressure lowering due to 0.1 m solute is
A sample of 20.0 g of a compound ( molecular weight 120) which is a non-electrolyte is dissolved in 10.0 grams of ethanol ( C 2 H 5 OH ) . If the vapour pressure of pure ethanol at the temperature used is 0.250 atm, what is the vapour pressure of ethanol above the solution?
At 120 o C , the vapour pressure of pure chlorobenzene ( C 6 H 5 Cl ) is 0.736 atm. What is the vapour pressure of a solution of 5.00 g of naphthalene ( C 10 H 8 ) in 50.0 g of chlorobenzene? (Assume that naphthalene is not volatile)
Henry’s constant values for four gases dissolved in water at 298K are given below. The gas with least solubility of the four will have a value of……..kPa
Vanthoff factor (i) is maximum for
Volume of decimolar sulphuric acid which can exactly neutralize 500ml of semimolar caustic soda solution is
The liquid mixture which shows positive deviations from Raoult’s law is
Molal depression constant of water is 1 . 86 K Kg / mole . Freezing point of urea solution containing ‘X’ grams of urea in 100 grams of water is 269.28K. Value of ‘X’ is
When 3 g of a nonvolatile solute is dissolved in 50 g of water, the relative lowering of vapour pressure observed is 0.018 Nm − 2 . Molecular weight of the substance is
Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:
The vapour pressure of solvent is 20 torr, while that of its dilute solution is 17 torr, the mole-fraction of the solvent is
Relative decrease in vapour pressure of an aqueous solution containing 2 moles [ C u ( N H 3 ) 3 C l ] C l in 3 mol H 2 O is 1 2 . When the given solution reacts with excess of A g N O 3 solution the number of moles of AgCl produced is.
The degree of dissociation of Ca ( NO 3 ) 2 in a dilute aqueous solution containing 7.0 g of salt per 100 g of water at 100 o C is 70%. If the vapour pressure of water at 100 o C is 760 mmHg, the vapour pressure of the solution is
Normal boiling point ( T N ) is defined as the temperature when vapour pressure of liquid becomes equal to 1 atm and standard boiling point ( T S ) is defined as the temperature when vapour pressure of liquid becomes equal to 1 bar. Which one is not correct if water is considered?
A solution containing 28 g of phosphorus in 315 g CS 2 ( b . p . 46 .3 o C ) boils at 47.98 o C . If K b for CS 2 is 2.34 K kg mol − 1 . The formula of phosphorus is (at. mass of P = 31).
Y g of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If K b is molal elevation constant, the value of Δ T is given by:
Elevation in boiling point of a molar (1 M) glucose solution ( d = 1.2 g m L − 1 ) is
0.15 g of a substance dissolved in 15 g of solvent boiled at a temperature higher by 0.216 o C than that of the pure solvent. Calculate the molecular weight of the substance. Molal elevation constant for the solvent is 2.16 o C .
A solution of 0.450 g of urea (mol. wt. 60) in 22.5 g of water showed 0 .170 o C of elevation in boiling point. Calculate the molal elevation constant of water.
Given, H 2 O ( l ) ⇌ H 2 O ( g ) at 373 K , ΔH o = 8 .31 kcal mol − 1 Thus, boiling point of 0.1 molal sucrose solution is
At higher altitudes, water boils at temperature < 100 o C because
The elevation in boiling point of a solution of 13.44 g of C u C l 2 (molecular weight = 134.4, K b = 0 .52 K molality − 1 ) in 1 kg water using the following information will be:
Which aqueous solution exhibits highest boiling point?
Calculate the normal boiling point of a sample of sea water found to contain 3.5% of NaCl and 0.13% of MgCl 2 by mass. The normal boiling point of water is 100 o C and K b (water) = 0.51 K kg mol – 1 . Assume that both the salts are completely ionised
Which will have largest ΔT b ?
An aqueous solution of glucose boils at 100.01 o C . The molal elevation constant for water is 0.5 K mol – 1 kg. The number of molecules of glucose in the solution containing 100 g of water is
The latent heat of vaporisation of water is 9700 cal/mole and if the boiling point is 100 o C , ebullioscopic constant of water is
Molal Elevation Constant depends on
A mixture can be homogeneous or heterogeneous, Amalgam is an example of
If for a sucrose solution elevation in boiling point is 0.1 o C then what will be the boiling point of NaCl solution for same molal concentration
0.01 (M) solution of an acid HA freezes at − 0.0205 o C . If K f for water is 1.86 K kg m o l − 1 , the ionisation constant of the conjugate base of the acid will be (consider molarity ≃ molality)
The boiling point of 0.1 m K 4 [ Fe ( CN ) 6 ] is expected to be ( K b for water = 0.52 K kg m o l – 1 )
Equimolal solutions of A and B show depression in freezing point in the ratio 2 : 1. A remains in its normal state in solution. B will be ………… state
The freezing point of 0.05 m solutions of a non-electrolyte in water is
The value of K f for water is 1 . 86 o , calculated from glucose solution. The value of K f for water calculated for NaCl solution will be:
A solution of x moles of sucrose in 100 grams of water freezes at − 0 .2 o C . As ice separates the freezing point goes down to 0 .25 o C . How many grams of ice would have separated?
The freezing point of a dilute solution of acetamide in glacial acetic acid is 298K. This is the value when crystals of
A solution of a non-volatile solute in water freezes at − 0 .40 o C . The vapour pressure of pure water at 298 K is 23.51 torr. For water, K f = 1 .86 K mol − 1 kg . Thus, vapour pressure of the solution (in torr)
In which case depression in freezing point is equal to cryoscopic constant for water:
What freezing point depression would be predicted for 0.2 molal solution of benzoic acid in benzene if latent heat of fusion is 40.00 cal g − 1 at 280 K (freezing point) for benzene? (assume no change in molecular state)
How much ethyl alcohol ( C 2 H 5 OH ) must be added to 1 L of water so that solution will not freeze at – 6 o F ? ( K f of H 2 O = 1 .86 o mol − 1 kg )
The amount of urea to be dissolved in 500 cc of water ( K f = 1 .86 ) to produce a depression of 0.186 o C in the freezing point is:
Freezing point of an aqueous solution is − 0.186 o C . Elevation of boiling point of the same solution is if K b = 0 .512 K molality − 1 and K f = 1 .86 K molality − 1 :
A certain substance ‘A’ tetramerises in water to the extent of 80%. A solution of 2.5g of A in 100 g of water lowers the freezing point by 0 . 3 o C . The molar mass of A is
Cryoscopic constant of a liquid
Addition of 0.643 g of a compound to 50 ml of benzene (density = 0.879 g mol – 1 ) lowers the freezing point from 50.51 o C to 50.03 o C . If K f for benzene is 5.12, the molecular mass of the compound is
Which of the following solution will have highest freezing point?
What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised ( K f for water = 1.86 K molality – 1 )
Which of the following aqueous molal solution have highest freezing point
Increasing amount of HgI 2 is added to 1 litre of an aqueous solution containing 0.1 mole of KI. Which of the following graphs does represent the variation of depression in freezing point of the resulting solution with the amount of HgI 2 added?
3.24 g Hg ( NO 3 ) 2 (Molecular mass 324) dissolved in 1000 g of water constitutes a solution having a freezing point − 0 .0558 o C , while 21.68 g of HgCl 2 (Molecular mass 271) in 2000 g of water constitutes a solution with a freezing point of − 0.0744 o C . The K f of water is 1.86 K Kg mol – 1 . About the state of ionisation of the two solids in water can be inferred that.
π 1 , π 2 , π 3 and π 4 are the osmotic pressures of 5% ( W V ) solutions of urea, fructose, sucrose and KCl respectively at certain temperatures. The correct order of their magnitude is
A living cell contains a solution which is isotonic with 0.3 (M) sugar solution. What osmotic pressure developed when the cell is placed in 0.1 (M) KCl solution at body temperature?
Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionized)
Two solutions (A) containing FeCl 3 (aq) and (B) containing K 4 [ Fe ( CN ) 6 ] are separated by semipermeable membrane as shown below. If FeCl 3 on reaction with K 4 [ Fe ( CN ) 6 ] , produces blue colour of Fe 4 [ Fe ( CN ) 6 ] , the blue colour will be noticed in:
Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be
Osmosis is involved in one or more processes. (I) Interchange of nutrients and waste products between tissue cells and their surroundings. (II) Reverse osmosis (III) Excretion of urine (IV) Evaporation Select the correct processes.
If ‘A’ contains 2% NaCl and is separated by a semipermeable membrane from ‘B’ which contains 10% NaCl, which event will occur?
Molar mass of acetic acid measured by osmotic pressure experiments is 80 grams. Degree of dimerization of acetic acid dissolved in benzene is
Which one of the following statement is incorrect regarding minimum boiling azeotrope
A solution containing 0.5 g of a non-volatile solute in 0.2 d m 3 of the solution exerts an osmotic pressure of 44.44 kPa at 300 K. Thus, molar mass of the solute is
Osmotic pressure of blood is 7.40 atm at 27 o C . Number of mol of glucose to be used per L for an intravenous injection that is to have the same osmotic pressure as blood is:
A solution of glucose ( C 6 H 12 O 6 ) is isotonic with 4 g of urea ( NH 2 − CO − NH 2 ) per litre of solution. The concentration of glucose is:
A 5% solution of can sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is
Consider the separate solution of 0 .500 M C 2 H 5 OH ( aq ) , 0 .100 M Mg 3 ( PO 4 ) 2 ( aq ) , 0 .250 M KBr ( aq ) and 0.125 M Na 3 PO 4 ( aq ) at 25 o C . Which statement is true about these solutions, assuming all salts to be strong electrolytes?
Two solutions of KNO 3 and CH 3 COOH are prepared separately. Molarity of both is 0.1 M and osmotic pressures are P 1 and P 2 respectively. The correct relationship between the osmotic pressures is :
The weight of urea dissolved in 100 ml solution which produce an osmotic pressure of 20.4 atm, will be
In the phenomenon of osmosis, the membrane allow passage of.
Blood plasma has the following composition (milliequivalents per litre). Calculate its osmotic pressure at 37 o C . Na + = 138 , Ca 2 + = 5 .2 , K + = 4 .5 , Mg 2 + = 2 .0 , Cl − = 105 , HCO 3 – = 25 , PO 4 3 − = 2 .2 , SO 4 2 − = 0 .5 , Proteins = 16 , Others = 1 . 0
Osmotic pressure of a sugar solution at 24 o C is 2.5 atmospheres. Determine the concentration of the solution in gram mole per litre.
Sea water is 3.5% by mass of a salt and has a density 1.04 g c m – 3 at 293 K. Assuming the salt to be sodium chloride, calculate the osmotic pressure of sea water. Assume complete ionisation of the salt –
Insulin ( C 2 H 10 O 5 ) n is dissolved in a suitable solvent and the osmotic pressure ( π ) of solution of various concentrations (g/cc) C is measured at 20 o C . The slope of the plot of π against ‘C’ is found to be 4.65 × 10 − 3 . The molecular weight of insulin is
After removing the hard shell of an egg by dissolving in dilute HCl a semipermeable membrane can be visible. If such an egg is kept in a saturated solution of common salt, the size of egg will
A solution containing 8.6 g urea in one litre was found to be isotonic with a 5% (wt./vol.) solution of an organic non-volatile solute. The molecular weight of latter is :
Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P, then 0.05 M solution of A at the same temperature will produce an osmotic pressure equal to
The relationship between osmotic pressure at 273 K when 10 g of glucose ( P 1 ) , 10 g urea ( P 2 ) and 10 g sucrose ( P 3 ) are dissolved in 250 ml of water is
Osmotic pressure of a urea solution at 10 o C is 500 mm. Osmotic pressure of the solution become 105.3 mm. When itis diluted and temperature raised to 25 o C . The extent of dilution is
p H of 0.1 (M) BOH (weak base) is found to be 12. The solution at temperature T K will display an osmotic pressure equal to
A 2% solution of cane sugar is isotonic with 0.5% x solution. The molecular weight of substance ‘x’ is [Assume that x does not undergo association or dissociation]
A compound MX 2 has observed and normal molar masses 65.6 and 164 respectively. Calculate the apparent degree of ionization of MX 2 :
PtCl 4 . 6 H 2 O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of 3 . 72 o . Assume 100% ionization and K f ( H 2 O ) = 1 .86 o mol − 1 kg , then complex is
When 30 g of this acid C 11 H 8 O 2 is dissolved in 60 g of benzene, a freezing point depression of 2K is observed. [ K f ( benzene ) = 1 .72 K mol − 1 kg ] . The van’t Hoff factor (i) is
When only a little quantity of HgCl 2 ( s ) is added to excess KI (aq) to obtain a clear soltion, which of the following is true for this solution? (no volume change on mixing)
Which of the following has been arranged in order of decreasing freezing point?
The freezing point depression of 0.001 m K x [ Fe ( CN ) 6 ] is 7.10 × 10 − 3 K . Determine the value of x. Given, K f = 1 .86 K kg mol − 1 for water
Moles of K 2 SO 4 to be dissolved in 12 mol water to lower its vapour pressure by 10 mmHg at a temperature at which vapour pressure of pure water is 50 mm is:
If α is the degree of dissociation of N a 2 S O 4 , the van’t Hoff factor (i) used for calculating the molecular mass is
pH of a 0.1 M monobasic acid is found to be 2. Hence its osmotic pressure at a given temperatue T K is
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water ( ΔT f ) , when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is ( K f = 1 .86 K kg mol − 1 )
Which salt may show the same value of vant Hoff factor (i) as that of K 4 Fe ( CN ) 6 in very dilute solution state?
20 g of a binary electrolyte (mol. wt. = 100) are dissolved in 500 g of water. The freezing point of the solution is − 0 .74 o C , K f = 1 .86 K molality − 1 . The degree of ionization of the electrolyte is
When the pure solvent diffuses out of the solution through the semi-permeable membrane then the process is called
If air is taken as a binary solution, the solvent is
The characterstic property of solution is a) Formation of solution is physical change b) Solute and solvent in the solution can be separated by filtration c) Solute and solvent in the solution can be separated by decantation d) Solution can be represented with a chemical formula
A mixture of salt and water can be separated by
The volume of decamolar aquous solutions of hydrochloric acid is required to prepare 2dm 3 of 5M HCl solution is
Molarity of the liquid HCl if density of the solution is 1.17 g/cc is
The units of Normality are
A student has 100mL of 0.1 M KCl solution. To make it 0.2 M, he has to
Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL .The molality of the solution is
5 ml of 1 N HCl, 20 ml of N/2 H 2 SO 4 and 30 ml of N/3 HNO 3 are mixed together and the volume made to one litre. The normality of the resulting solution is
The volumes of 4N HCl and 10N HCl required to make 1L of 6 N HCl are
The number of millimoles of H 2 SO 4 present in 5 litres of 0.2N H 2 SO 4 solution is
The correct relationship between molarity (M) and molality (m) is (d = density of the solution, in Kg L -1 , M 2 = molar mass of the solute in kg mol -1 )
A one molal solution is one that contains
0.1 mol of NaCl is dissolved in 100g of water. The mole fraction of NaCl is
At 25 0 C for a given solution M = m, then at 50 0 C the correct relationship is
6g. of Urea is dissolved in 90g. of water. The mole fraction of solute is
The molarity of a 9.8% H 2 SO 4 of density 1.1 g/cc is
M = molarity of the solution m = molality of the solution d = density of the solution (in g. ml -1 ) M 1 = gram molecular weight of solute Which of the following relations is correct
The units of molality are
In which mode of expression, the concentration of solution remains independent of temperature?
Incorrect statement is (K H = Henry’s constant)
If two substances A and B have = 1 : 2and have mole fraction in solution 1 : 2, then mole fraction of A in vapours is
Which of the following statement is correct about Henry’s law
Among the following gases which gas has the highest Henry’s law constant, K H value in water at the same temperature ?
In a nearly saturated solution, the solubility increases with increase in temperature, then ∆ H solution is
In a mixture of A and B, components show negative deviation when
The vapour pressure is least for
If 0.05 mole of gas are dissolved in 500 grams of water under 1 atm. pressure, 0.1 moles will be dissolved if the pressure is 2atm. It illustrates
If liquid A and B form ideal solutions
Lowering of vapour pressure is maximum in
Which of the following liquid pairs shows a positive deviation from Raoults Law ?
The density of 20% (w/w) aqueous NaOH solution is 1.20 g.mL –1 . What is the mole fraction of water ?(molar mass of NaOH = 40g.mol –1 )
A solution that obeys Raolult's law is called
Boiling point is least for
A non-volatile solute (A) is dissolved in a volatile solvent (B). The vapour pressure of resultant solution is Ps. The vapour pressure of pure solvent is P 0 B . If "X" is mole fraction, which of the following is correct?
Concentration of sodium stearate is increased from 10 -5 M to 0.0015M then its Osmotic pressure (T is constant)
Dissolution of gases in solvents is governed by
Normality of 0.1 M potassium permanganate solution when it acts like oxidizing agent in neutal medium is
Molar mass of NaCl dissolved in water is determined from depression of freezing point experiment. Molar mass of NaCl observed can never be
Six grams of urea dissolved in 250 grams of water showed a depression of freezing point of ‘X’. 36 grams of glucose dissolved in 500 grams of water showed a depression of freezing point of ‘Y’. Correct relation between ‘X’ and ‘Y’ is
9 ml of liquid ‘A’ is mixed with 9 ml of liquid ‘B’. The volume of the mixture formed is 18 ml. The two liquids are most likely to be
Which one of the following solution has least vapour pressure
Molar mass of a diacidic base is 40g. Density of its aqueous solution is 1 g/ cc. More concentrated solution among the following is
‘A’ and ‘B’ are two volatile liquids.A mixture of ‘A’ and’B’ is prepared.A graph is plotted between vapour pressure of liquid mixture and mole fraction. The following plot is obtained From the graphical represent identify the liquid mixture
Vanthoff factor for 50% ionized calomel in aqueous solution is
Consider the reaction, C 6 H 5 NH 2 pyridine Z C 6 H 5 NH – COCH 3 ; Here ‘Z’ cannot be
Solubility of a gas in water at STP is 0.2m. Henry’s constant of the gas is equal to
Regarding osmosis, correct statement is
Vapour pressure of a liquid depends on many factors. Correct statement of the following is
Molal elevation constant of water is 0 . 52 K . Kg / mole . If the boiling point of 1M Sodium chloride solution is 374.04 K then the degree of ionization of sodium chloride will be
For the liquid mixture of benzene and toluene, correct statement is
Water movement from soil into plant roots and subsequently into upper portion of the plant is partly due to ?
Consider the solution of ethanol. In it
Which of the following is not an example of a solution?
Which of the following statements is not correct about homogeneous mixtures?
18 g of sucrose is dissolved in 162 g of water. Calculate the mass percentage of solution.
25.3 g of sodium carbonate,Na 2 CO 3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na + and carbonate ions, CO 3 2 – respectively are (Molar mass of Na 2 CO 3 = 106 mol -1 )
How many gram of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO 3 ? The concentrated acid rs 70% HNO 3
29.2% (w/W) HCI stock solution has density of 1.25 g mL -1 . The molecular weight of HCI is 36.5 mol -1 . The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCI is
Mole fraction of a solute in benzene is 0.2. The molality of solute is
184 g ethyl alcohol is mixed with 72 g of water. The ratio of mole fraction of alcohol to water is
What is the mole fraction of the solute in a 1 m aqueous solution
What is the molarity of K + in aqueous solution that contains 17.4 ppm of K 2 SO 4 (molar mass = 174 g mo -1 ) ?
The molarity of H 2 SO 4 solution, which has a density 1.84 g/cc at 35 o C and contains 98% by weight is
6.02x 10 20 molecules of urea are present in 100 mL of its solution. The concentration of solution is
The molarity of a solution obtained by mixing 750 mL of 0.5 M HCI with 250 mL of 2M HCI will be
If mole fraction of a solute in 1 kg benzene is 0.2 then molality of solute is
Concentration terms which are independent of temperature is/are
What happens to the solubility of substance with rise in temperature, if the dissolution process is endothermic?
Which of the following is the incorrect statement about solubility?
15 g of sucrose is dissolved in 50 mL of water and some pressure is applied on the surface of solution formed. It will result in
The value of Henry’s law constant for argon (Ar), carbon dioxide (CO 2 ), methane (CH 4 ) and formaldehyde (HCHO) are respectively 40.3 K bar,1.67 K bar, 0.413 K bar and 1.83 x 10 -5 K bar at 298 K. The correct order of their solubility is
Calculate the concentration of nitrogen present in the water. Assuming that the temperature is 25 o C, total pressure is 1 atm and mole fraction of nitrogen is 0 .78. K H fornitrogen = 8 .42 × 10 − 7 M / mmHg
Ratio of O 2 and N 2 in the air is 1:4. Find out the ratio of their solubilities in terms of mole fractions of O 2 and N 2 dissolved in water at atmospheric pressure and room temperature. K H O 2 = 3 .30 × 10 7 torr K H N 2 = 6 .60 × 10 7 torr
Which of the following statements is true about bends?
What is the composition of gases in the tanks used by the scuba divers?
Which of the following is not the characteristic of solutions of liquids and solids in liquid?
When a binary solution of two volatile liquids is taken in a closed vessel, then
p A and p B are the vapour pressure of pure liquid components A and B respectively of an ideal binary solution.If x A represents the mole fraction of component A, the total pressure of the solution will be
What does A point signifies in the figure given below?
Vapour pressure of pure A is 70 mm of Hg at 25 o C.It forms an ideal solution with ‘B’ in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25 o C the vapour pressure of pure B at 25 o C is
At 40 o C, the vapour pressure of pure liquids,benzene and toluene, are 160 mmHg and 60 mmHg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be
18 g of glucose (C 6 H 12 O 6 ) is added to 178.2 g water. The vapour pressure (in torr) for this aqueous solution is
If two substances A and B have p ∘ A : p ∘ B = 1 : 2 and mole fraction in solution = 1 : 2 then mole fraction of A in vapours is
Vapour pressure of chloroform (CHCI 3 ) and dichloromethane (CH 2 Cl 2 ) at 25 o C are 200 mm of Hg and 41.5 mm of Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl 3 and 40 g of CH 2 Cl 2 at the same temperature will be (Molecular mass of CHCI 3 = 119.5 u and molecular mass of CH 2 Cl 2 – 85 u)
Raoult’s law becomes a special case of Henry’s law when
Decrease in the vapour pressure of the solvent depends on
Two volatile liquids A and B are mixed in the mole ratio 3 : 2 to form an ideal solution. Vapour pressure of pure liquids A and B is 600 mm Hg and 300 mm Hg respectively. Mole fraction of B in vapour phase is
Which one of the following is not correct far an ideal solution
Solution of bromoethane and chloroethane
True statement about ideal solutions is
Which of the following statements is/are true for an ideal solution?
The solution formed by adding carbon disulphide to acetone, shows positive deviation from Raoult’s law. It is because
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct ? Assume that the temperature is constant at 25 o C. (Given, vapour pressure data at 25 o C, benzene = 12.8 kPa and toluene = 3.85 kPa)
Select the incorrect statement.
What does the following figure represent?
The solution which show large positive deviation from Raoult’s law form
Which of the following statements is/are true for the diagram?
Which of the following azeotropic solutions has the boiling point less than the boiling point of its constituents molecules?
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase
If p o and p s are vapour pressures of the solvent and solution respectively, n 1 and n 2 are the mole fractions of solvent and solute respectively, then
Solution of azeotropic nitric acid mixture contain
The solution that forms maximum boiling azeotrope is
The vapour pressure of acetone at 20 o C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 o C, its vapour pressure was 183 torr. The molar mass (g mol -1 ) of the substance is
Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 M aqueous solution at 100 o C is
What does point A and B represent in the following diagram?
If the elevation in boiling point of a solution of 10 g of solute (molecular weight = 100) in 100 g of water is ∆ T b , the ebullioscopic constant of water is
To observe an elevation of boiling point is 0.05 o C, the amount of a solute (molecular weight = 100) to be added to 100 g of water (K b – 0.5) is
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at -6 o C will be (K f for water = 1.86 K kg mol -1 and molar mass of ethylene glycol =62 g mol -1 )
How many grams of methyl alcohol should be added to 10 L tank of water to prevent its freezing at 268 K?
What happens to freezing point of benzene when naphthalene is added?
Dissolution of 1.5 g of a non-volatile solute (molecular weight = 60) in 250 g of a solvent reduces its freezing point by 0.01 o C. Find at the molal depression constant of the solvent.
At a given temperature, osmotic pressure of the concentrated solution of a substance
0.1 M NaCl and 0.005 M BaCl 2 solutions are separated by a semipermeable membrane in a container. For this system, choose the collect answer.
The temperature at which 10% aqueous solution (w/V) of glucose exhibits the osmotic pressure of 16.4 atm, is (R = 0.082 dm 3 atm K -1 mol -1 )
The osmotic pressure of 0.2 molar solution of urea at 27 o C (R =0.082 L atm mol -1 K -1 ) is.
Osmotic pressure present in the fluid inside the blood cell is equivalent to
Osmotic pressure method is used to determine molar mass of protein, macromolecules like polymers, biomolecules etc., because
Reverse osmosis is a process,
A molecule M associates in a given solvent according to the equation M ⇌ ( M ) n .For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is
Which of the following would exert maximum osmotic pressure?
58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (bp) of the resulting solution.
Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
Compound A undergoes tetramerisation in the given solvent. The van’t Hoff factor is
The degree of dissociation ( α ) if a weak electrolyte, AB is related to van’t Hoff factor (r) by the expression
KBr is 80% dissociated in aqueous solution of 0.5 m concentration. (Given, K f for water = 1.86 K kg mol -1 ). The solution freezes at
The freezing point depression constant for water is 1.86 o C molal -1 . If 5.00 g Na 2 SO 4 is dissolved in 45.0 g of H 2 O the freezing point is changed by -3.82 o C. Calculate the van’t Hoff factor for Na 2 SO 4 .
A solution of 1.25 g of P in 50 g of water lowers the freezing point of water by 0.3 o C. Molar mass of P is 94 . K f water = 1 . 86 K kg mol – 1 .’ The degree of association of P in water is
On adding I g arsenic to 80 g benzene, the freezing point of benzene is lowered by 0.19 o C. The formula of arsenic is K f = 50 . 8 K kg mol – 1 ) .
Pure benzene freezes at 5.3 o C. A solution of 0.223 g of phenylacetic acid C 6 H 2 CH 2 COOH in 4.4 g of benzene K f = 5 . 12 K kg mol – 1 freezes at 4.47 o C.From this observation, one can conclude that
1.2 % NaCl solution is isotonic with 7.2% glucose solution. What will be the van’t Hoff factor ‘i’ for NaCl ?
1 × 10 – 3 m solution of Pt NH 3 2 Cl 4 in H 2 O shows depression in freezing point by 0.0054 o C. The ionisable Cl – ions will be (Given, K f ( H 2 O ) = 1 . 860 K kg mno – 1 )
Which of the following statements are correct?
10% (m/m) aqueous potassium iodide has a density of 1.202 g mL -1 . The true statements about this solution are
The correct statement about the mixture of phenol and aniline are
The examples of minimum boiling azeotropes are
The correct relations showing Raoult’s law are
The following is a graph plotted between the vapour pressure of two volatile liquids against their respective mole fractions. Which of the following options are correct?
Consider the following aqueous solutions and / assume 100% ionisation in electrolytes I. 0.1 M urea II. 0.04 M Al 2 (SO 4 ) 3 III. 0.05 M CaCl 2 IV. 0.005 M NaCI The incorrect statements regarding the above solution are
Mark the correct options for the freezing point of a substance.
Which of the following statements are false?
Which of the following statements are true? I. In a binary mixture, mole fraction of, A is, χ A = n A n A + n B II.For solution containing (i) number of components, χ i = n i Σn i III.Sum of mole fractions of all the components of a solution is one. Select the correct option with true statements.
Which of the following statements is/are true? I. Different gases have different K s values at the same temperature. II. K H is a function of the nature of the gas. Choose the correct option
Following conclusions can be derived from the above equation. I. Total vapour pressure of the solution is related to the mole fraction of any one component. II. Total vapour pressure of the solution varies exponentially with the mole fraction of component 2. III.Depending on the vapour pressure of pure components 1 and 2, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1. Select the correct conclusions derived from the given equation.
Mark the incorrect information derived from the diagram.
Which of the following statements is/are correct for K f ? I. K f depends upon the nature of solvent. Il. K f is known as freezing point depression constant / molal depression constant. lll. K f is known as cryoscopic constant. Select the correct conclusion derived from the equation.
Information I. Semipermeable membrane contains network of submacroscopic holes or pores. II. Semipermeable membrane appears to be continuous sheets or films. III. Solvent molecules cannot pass through the holes of semipermeable membrane but solute molecules can pass. On the basis of the information given above select the correct option.
Consider the following statements about osmotic pressure method of molar mass determination. l. Molarity of the solution is used instead of molality. Il. Magnitude of osmotic pressure is very large even for dilute solutions. III. Molar mass of bio molecules can be determined as they are not stable at higher temperatures. IV. Determine the molar mass of polymer as they have poor solubility. Which of the above statements are responsible for advantage of osmotic pressure for determination of molar mass over the other colligative properties?
Statement 1: 1M aqueous solution of glucose contains 180 g of glucose in 1 kg water. Statement 2: Solution containing one mole of solute in 1000 g of solvent is called one molal solution.
Statement 1: Pressure does not have any significant effect on solubility of solids in liquids. Statement 2: Solids and liquids are highly incompressible and remain unaffected by change in pressure.
Statement 1: Polar solute dissolves in polar solvents and non-polar solute dissolves in non-polar solvents. Statement 2: Like dissolves like.
Statement 1: When scuba divers come towards surface, their capillaries get blocked which is painful and dangerous to life. Statement 2: There occurred release of dissolved gases as the pressure decreases and leads to the formation of bubbles of nitrogen in the blood.
Statement 1: Ethanol and acetone show positive deviation from Raoult’s law. Statement 2: Pure ethanol molecule show hydrogen bond and on adding acetone hydrogen bond between ethanol molecules breaks.
Statement 1: When non-volatile solute is added to solvent, the vapour pressure of the solution decreases. Statement 2: As number of solvent molecules escaping from the surface is reduced, the vapour pressure of the solution is also reduced.
Statement 1: Experimentally determined molar mass is always lower than the true value. Statement 2: Lower molar mass is due to dissociation of solute into ions.
Statement 1: The water pouch of instant cold pack for treating athletic injuries breaks when squeezed and NH 4 NO 3 dissolves in water to lower the temperature. Statement 2: Addition of non-volatile solute into solvent results into depression of freezing point of the solvent.
Statement 1: People taking a lot of salty food experience the puffiness or swelling, called edema. Statement 2: There is water retention in tissue cells and inter cellular spaces because of osmosis.
Statement 1: Melting point of a substance is used for testing the purity of the substance. Statement 2: There is no other method to determine the purity of substance.
Match the following columns and choose the correct option from the codes given below. Column I (Solution example) Column II (Type of solution) A. Chloroform mixed with nitrogen gas 1. Gaseous solution B. Ethanol dissolved in water 2. Solid solution C. Amalgam of mercury with sodium 3. Liquid solution
Statement 1: High blood pressure patients are advised to take the minimum quantity of salt. Statement 2: From salt Na + and Cl – ion concentration increases in the body fluid which may rupture the blood cells.
Match the following Henry’s law constant values for gases in water at 298 K. Column I Column II A. Argon 1. 1.83 x 10 -5 B. CO 2 2. 0.413 C. formaldehyde 3. 40.3 D. Methane 4. 0.611 E. Vinyl chloride 5. 1.67
Match the following Column I which represent concentration term to the Column II representing their corresponding formulae. Column I Column II A. Mass percentage 1. Volume of the component Total volume of solution × 100 B. Volume percentage 2. Mass of the component in the solution Total mass of the solution × 100 C. Molality 3. Moles of solute Volume of solution ( in L ) D. Molarity 4. Moles of solute Mass of solvent ( in kg )
Match the following terms given in Column I with the Column II. Column I Column II (van’t Hoff factor appox.) A. KCI 1. 0.5 B. Ethanoic acid 2. 2 C. K 2 SO 4 3. 3
Match the items of Column I with Column II. Column I (Inter molecular interaction) Column II (Example) A. A − B > A − A or B − B Interaction p. Hexane and heptane B. ΔV mix = 0 q. Chloroform and acetone C. ΔV mix < 0 r. Chlorobenzene and bromobenzene D. Follows Raoult’s law in all conditions of T and p. s. Water and nitric acid
Match the items of Column I with Column Il. Column I Column II A. Vapour pressure p. Colligative properties B. Osmotic pressure q. Decreases in the presence of solute. C. Freezing point r. Varies inversely with molecular mass. D. Elevation in boiling point s. Dependent on ebullioscopic constant.
X 1 = 1 signifies that
The diagram given below represents the vapour pressure and mole fraction of an ideal solution of component 1 and 2. Answer the following questions. Which of the following statements is true about the diagram?
What is the vapour pressure of solution prepared by mixing 25 . 5 g of CHCl 3 and 40 g of CH 2 Cl 2 at 298 K ?
Calculate the mole fractions of each component in vapour phase.
Compartments A and B have the following combinations of solution. A B 1 0.1 M KCI 0.2 M KCI 2 0.1% (m/v) NaCl 10% (m/V) NaCl 3 18 gL -1 glucose 34.2 gL -1 sucrose 4 20% (m/V) glucose 10% (m/V) glucose Answer the following questions on this basis. Which of the above solutions is isotonic?
Compartments A and B have the following combinations of solution. Answer the following questions on this basis. Indicate the solution(s) in which compartment B will show an increase in volume
Calculate the mass of urea (NH 2 CONH 2 ) required in making 2.5 kg of 0.25 molal aqueous solution.
Calculate the mole fraction of ethylene glycol (C 2 H 6 O 2 ) in a solution containing 20% of C 2 H 6 O 2 by mass.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the nitric acid, if the density of the solution is 1.504 g mL -1 ?
On dissolving sugar in water at room temperature, solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
A beaker contains a solution of substance.A.Precipitation of substance A takes place when small amount of A is added to the solution. The solution is
Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon
Benzene and naphthalene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.
1.00 g of a non-electrolyte solute is dissolved in 50 g of benzene which lowers the freezing point of benzene by 0.40 K. The freezing point of depression constant of benzene is 5.12 K kg mol -1 . Find the molar mass of the solute.
The boiling point of benzene is 3 53.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (K b for benzene is 2.53 K kg mol -1 )
Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?
On the basis of information given below mark the correct option. I. In bromo ethane and chloro ethane mixture,inter molecular interactions of A – A and B – B type are nearly same as A-B type interactions. II. In ethanol and acetone mixture A-A or B-B type inter molecular interactions are stronger than A-B type interactions. III. In chloroform and acetone mixture A-A or B-B type inter molecular interactions are weaker than A-B type interactions.
Which of the following aqueous solutions should have the highest boiling point?
An unripe mango placed in a concentrated salt solution to prepare pickle shrivels because
Which of the following units is useful in relating concentration of solution with its vapour pressure?
Which of the following statements is false?
Consider the figure and mark the correct option.
The values of van’t Hoff factors for KCl, NaCl and K 2 SO 4 respectively are
Which of the following statements is false?
We have three aqueous solutions of NaCl labelled as A, B and C with concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor for these solutions will be in the order:
Which of the following factors affect the solubility of a gaseous solute in the fixed volume of liquid solvent? (i) Nature of solute (ii) Temperature (iii) Pressure
Colligative properties are observed when
Relative lowering of vapour pressure is a colligative property because
Statement 1: Molarity of a solution in liquid state changes with temperature. Statement 2: The volume of a solution changes with change in temperature.
Statement 1: When NaCl is added to water, a depression in free zing point is observed. Statement 2: The lowering of vapour pressure of a solution causes depression in the freezing point.
Statement 1: When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Statement 2: Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.
Match the items given in Column I and Column II. Column I Column II A. Saturated solution 1. Solution having same osmotic pressure at a given temperature as that of given solution. B. Binary solution 2. A solution whose osmotic pressure is less than that of another. C. Isotonic solution 3. Solution with two components. D. Hypotonic solution 4. A solution which contains maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature. E. Solid solution 5. A solution whose osmotic pressure is more than that of another. F. Hypertonic solution 6. A solution in solid phase.
Match the items given in Column I with the type of solutions given in Column II. Column I Column II A. Soda water 1. A solution of gas in solid B. Sugar solution 2. A solution of gas in gas C. German silver 3. A solution of solid in liquid D. Air 4. A solution of solid in solid E. Hydrogen gas in palladium 5. A solution of gas in liquid 6. A solution of liquid in solid
Match the laws given in Column I with expressions given in Column II. Column I Column II A. Raoult’s law 1. ΔT f = K f m B. Henry’s law 2. π = CRT C. Elevation in boiling point 3. p = x 1 p 1 ∘ + x 2 p 2 ∘ D. Depression in freezing point 4. ΔT b = K b m E. Osmotic pressure 5. p = K H ⋅ x
The mass per cent of different elements present in sodium sulphate, Na 2 SO 4 respectively are
statement 1 : Molality of a solution does not change with temperature statement 2 : Mass is affected with temperature.
which solution will show greater relative lowering of vapour pressure when equal mass of each of the following non electrolyte is separately dissolved in equal quantity of same solvent?
Which of the following gas is least soluble in water at same temperature (K H =Henry’s constant value)
Identify correct statement. Different solutions
When no more solute can be dissolved in solution at a given temperature the solution is known as
A solution will freeze when
Consider a binary mixture of volatile liquids. If at X a = 0.2, the vapour pressure of solution is 580 torr, then the mixture could be(P° A =200 torr, P° B = 600 torr)
Among the given salts, aqueous solution of which salt will show maximum freezing point?
Aqueous solution of which has highest boiling point?
The osmotic pressure of decimolar solution of NaCl at 27°C is, assuming that it is completely dissociated. (R = 0.083 L bar K -1 mol -1 )
If 6.2 g of ethylene glycol (C 2 H 6 O 2 ) is mixed with 500 g of water. The freezing point of the solution obtained will be (K f for water = 1 .86 K kg mol -1 )
200 mmHg and 400 mmHg are the vapour pressures of pure liquid components, CHCl 3 and CH 2 Cl 2 , respectively which forms an ideal binary solution. If 0.75 is the mole fraction of CHCl 3 , then the total vapour pressure of the solution will be
Solubility of a solid substance least likely to depend on
A mixture of phenol and aniline
Which of the following gas is most soluble in water?
The mass of a non-volatile solute (molar mass = 40 g mol -1 ) dissolved in 114 g octane to reduce its vapour pressure to 80% is
Raoult’s law is applicable to A) Volatile liquid mixture B) Solution of non volatile solute C) Aqueous solutions only D) Non-aqueous solutions only
Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 2 : 3. Vapour pressure of pure liquid ‘A’ is 600 mm Hg and pure liquid ‘B’ is 400 mm Hg. Mole fraction of ‘B’ in vapour phase will be
Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 2 : 3. Vapour pressure of pure liquid ‘A’ is 600 mm Hg and pure liquid ‘B’ is 400 mm Hg. Mole fraction of ‘B’ in vapour phase will be
Which of the following method of expressing concentration is dependent on temperature?
At 300 K, 0.3 M sucrose solution is isotonic with 0.15 M KCl solution. Degree of dissociation of KCl in the solution is
Molarity of a solution of density 0.6 g/ml is 1 M. Molality of the same solution under similar conditions will be (Molar mass of solute – 150 g)
Molarity of a solution of density 0.6 g/ml is 1 M. Molality of the same solution under similar conditions will be (Molar mass of solute – 150 g)
Molarity of a solution of density 0.6 g/ml is 1 M. Molality of the same solution under similar conditions will be (Molar mass of solute – 150 g)
Boiling point of 0.5 m NaCl solution will be ( K b of water is 0.52 K Kg mole – 1 )
Boiling point of 1 m NaCl solution will be ( K b of water is 0.52 K Kg mole – 1 )
Colligative properties depend on the number of solute particles present in the solution. Osmotic pressure of 90% ionized 0.1M BaCl 2 solution at 27 0 c is
Molality of 10 grams of aqueous solution containing 2 grams of caustic soda is
Which one of the following solution has least freezing point ?
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
2 litre of 0.25 M Sulphuric acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
10 litre of 1 M HCl is diluted to 40 litre solution. Normality of the resultant solution is —- N
10 litre of 1 M HCl is diluted to 40 litre solution. Normality of the resultant solution is —- N
5 litres of 4 M HCl is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 4 M HCl is diluted to 20 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M H 2 SO 4 is diluted to 10 litre solution. Normality of the resultant solution is —- N
4.5 litres of 4.5 M H 2 SO 4 is diluted to 1.5 M solution. Final volume of the solution is ——- litres
4.5 litres of 1.5 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 2 M H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
10 litre of 4 M KOH is diluted to 3 M solution. Final volume of the solution is ——- litres
5 litre of 4 M KOH is diluted to 1 M solution. Final volume of the solution is ——- litres
5 litre of 4 M KOH is diluted to 1 M solution. Final volume of the solution is ——- litres
5 litre of 4 M KOH is diluted to 1 M solution. Final volume of the solution is ——- litres
5 litre of 4 M KOH is diluted to 1 M solution. Final volume of the solution is ——- litres
3 litre of 4 M NaOH is diluted to 0.125 M solution. Final volume of the solution is ——- litres
2 litres of 2 M NaOH is diluted to 0.05 M solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
6 litres of 3 M H 2 SO 4 is diluted to 1 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M H 2 SO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
6 litres of 2 N H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 2 N H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 2 N H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 2 N H 2 SO 4 is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 2 N H 2 SO 4 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 0.4 N H 2 SO 4 is diluted to 0.025 M solution. Final volume of the solution is ——- litres
1 litre of 0.4 N H 2 SO 4 is diluted to 0.025 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
20 litre of 10 N H 2 SO 4 is diluted to 2 M solution. Final volume of the solution is ——- litres
2 litre of 0.25 M Barium hydroxide is diluted to 5 litre solution. Normality of the resultant solution is —- N
4 litre of 0.125 M Barium hydroxide is diluted to 8 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litres of 1 M Barium hydroxide is diluted to 10 litre solution. Normality of the resultant solution is —- N
4.5 litres of 4.5 M Barium hydroxide is diluted to 1.5 M solution. Final volume of the solution is ——- litres
4.5 litres of 4.5 M Barium hydroxide is diluted to 1.5 M solution. Final volume of the solution is ——- litres
2 litres of 1.5 M Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
2 litres of 1.5 M Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 5 M Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 5 M Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 5 M Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Barium hydroxide is diluted to 1.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Barium hydroxide is diluted to 0.5 N solution. Final volume of the solution is ——- litres
2 litres of 4 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Barium hydroxide is diluted to 1 N solution. Final volume of the solution is ——- litres
1.5 litres of 2 N Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 2 N Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 2 N Barium hydroxide is diluted to 0.5 M solution. Final volume of the solution is ——- litres
2 litres of 1 N Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
2 litres of 1 N Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 2 N Barium hydroxide is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
1 litre of 0.2 N Barium hydroxide is diluted to 0.0125 M solution. Final volume of the solution is ——- litres
100ml of 2.5 M Barium hydroxide is diluted to 0.5 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Barium hydroxide is diluted to 0.25 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 2 M Sulphuric acid is diluted to 0.4 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 1 M Barium hydroxide is diluted to 0.2 N solution. Volume of water added is —— mL
200 ml of 2 M Barium hydroxide is diluted to 0.4 N solution. Volume of water added is —— mL
200 ml of 2 M Barium hydroxide is diluted to 0.4 N solution. Volume of water added is —— mL
200 ml of 2 M Barium hydroxide is diluted to 0.4 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
200 ml of 2 M Barium hydroxide is diluted to 0.4 N solution. Volume of water added is —— mL
100ml of 2.5 M Sulphuric acid is diluted to 0.25 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
10 litre of 0.5 M HNO 3 is diluted to 20 litre solution. Normality of the resultant solution is —- N
10 litre of 0.5 M HNO 3 is diluted to 20 litre solution. Normality of the resultant solution is —- N
10 litre of 0.5 M HNO 3 is diluted to 20 litre solution. Normality of the resultant solution is —- N
1 litre of 0.5 M HNO 3 is diluted to 5 litre solution. Normality of the resultant solution is —- N
10 litre of 0.5 M HNO 3 is diluted to 20 litre solution. Normality of the resultant solution is —- N
5 litres of 4 M HNO 3 is diluted to 20 litre solution. Normality of the resultant solution is —- N
5 litres of 4 M HNO 3 is diluted to 10 litre solution. Normality of the resultant solution is —- N
5 litre of 4 M HNO 3 is diluted to 1 M solution. Final volume of the solution is ——- litres
8 litre of 0.5 M HNO 3 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
8 litre of 0.5 M HNO 3 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
10 litre of 4 M HNO 3 is diluted to 3 M solution. Final volume of the solution is ——- litres
2 litres of 2 M HNO 3 is diluted to 0.05 M solution. Final volume of the solution is ——- litres
5 litre of 4 M HNO 3 is diluted to 1 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M HNO 3 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
8 litre of 0.5 M HNO 3 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
8 litre of 0.5 M H 3 PO 4 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
2 litres of 2 M HNO 3 is diluted to 0.05 M solution. Final volume of the solution is ——- litres
2 litres of 2 M H 3 PO 3 is diluted to 0.05 M solution. Final volume of the solution is ——- litres
2 litres of 2 M H 3 PO 3 is diluted to 0.05 M solution. Final volume of the solution is ——- litres
6 litre of 2 M H 3 PO 3 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
6 litre of 2 M H 3 PO 3 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M H 3 PO 3 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M H 3 PO 3 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 0.5 N H 3 PO 4 is diluted to 5 litre solution. Normality of the resultant solution is —- N
3 litre of 4 M H 3 PO 4 is diluted to 0.125 M solution. Final volume of the solution is ——- litres
4 litres of 1 M H 3 PO 3 is diluted to 0.05 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M H 3 PO 4 is diluted to 0.25 M solution. Final volume of the solution is ——- litres
0.5 moles of Phosphorous acid is dissolved in a ten litre solution. Molarity of the solution is
0.02 moles of Phosphorous acid is dissolved in 400 ml solution. Molarity of the solution is
0.1 moles of Phosphorous acid is dissolved in a two litre solution. Molarity of the solution is
0.05 moles of Phosphorous acid is dissolved in 800 ml solution. Molarity of the solution is
0.05 moles of Phosphorous acid is dissolved in 800 ml solution. Molarity of the solution is
0.05 moles of Phosphorous acid is dissolved in 800 ml solution. Molarity of the solution is
2 moles of Phosphorous acid is dissolved in 40 litre solution. Molarity of the solution is
0.01 gram equivalents of Phosphoric acid is dissolved in a 100ml solution. Normality of the solution is…………N
Two gram equivalents of Phosphoric acid is dissolved in a twenty litre solution. Normality of the solution is…………N
10 milli equivalents of Phosphoric acid is dissolved in a 100ml solution. Normality of the solution is…………N
3 litre of 4 M H 3 PO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
2 litre of 4 M H 3 PO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
2 litre of 4 M H 3 PO 4 is diluted to 0.25 N solution. Final volume of the solution is ——- litres
5 litres of 4 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
5 litres of 1 M H 3 PO 4 is diluted to 0.5 N solution. Final volume of the solution is ——- litres
400 ml of 1 M Phosphoric acid is diluted to 0.4 N solution. Volume of water added is —— mL
600 ml of 1 M Phosphoric acid is diluted to 0.3 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
100 ml of 0.3 M Phosphoric acid is diluted to 0.15 N solution. Volume of water added is —— mL
1 litre of 0.5 M Phosphorous acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
1 litre of 0.5 M Phosphorous acid is diluted to 5 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
0.5 litre of 0.125 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
1000 ml of 0.4 M Sulphuric acid is diluted to 0.1 N solution. Volume of water added is —— mL
5 litres of 1 M Phosphorous acid is diluted to 10 litre solution. Normality of the resultant solution is —- N
4 litres of 2.5 M Phosphorous acid is diluted to 20 litre solution. Normality of the resultant solution is —- N
4 litres of 2.5 M Phosphorous acid is diluted to 20 litre solution. Normality of the resultant solution is —- N
4 litres of 2.5 M Phosphorous acid is diluted to 20 litre solution. Normality of the resultant solution is —- N
4 litres of 2.5 M Phosphorous acid is diluted to 20 litre solution. Normality of the resultant solution is —- N
4 litres of 2.5 M Phosphorous acid is diluted to 20 litre solution. Normality of the resultant solution is —- N
4.5 litres of 4.5 M Phosphorous acid is diluted to 1.5 M solution. Final volume of the solution is ——- litres
2 litres of 1.5 M Phosphorous acid is diluted to 0.5 M solution. Final volume of the solution is ——- litres
4.5 litres of 1.5 M Phosphorous acid is diluted to 0.5 M solution. Final volume of the solution is ——- litres
2 litres of 1.5 M Phosphorous acid is diluted to 0.5 M solution. Final volume of the solution is ——- litres
2 litres of 1.5 M Phosphorous acid is diluted to 0.5 M solution. Final volume of the solution is ——- litres
1 litre of 1 M Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 5 M Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 5 M Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 5 M Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
3 litres of 1.5 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Phosphorous acid is diluted to 1.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
1 litre of 2 M Phosphorous acid is diluted to 0.5 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Phosphorous acid is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Phosphorous acid is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Phosphorous acid is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Phosphorous acid is diluted to 1 N solution. Final volume of the solution is ——- litres
9 litres of 2 M Phosphorous acid is diluted to 1 N solution. Final volume of the solution is ——- litres
1.5 litres of 0.2 M Phosphorous acid is diluted to 0.05 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M Phosphorous acid is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M Phosphorous acid is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M Phosphorous acid is diluted to 0.25 N solution. Final volume of the solution is ——- litres
0.15 litres of 2 M Phosphorous acid is diluted to 0.25 N solution. Final volume of the solution is ——- litres
6 litres of 2 N Phosphorous acid is diluted to 0.5 M solution. Final volume of the solution is ——- litres
6 litres of 1 N Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
6 litres of 1 N Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
6 litres of 1 N Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 2 N Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 2 N Phosphorous acid is diluted to 0.25 M solution. Final volume of the solution is ——- litres
1 litre of 0.4 N Phosphorous acid is diluted to 0.025 M solution. Final volume of the solution is ——- litres
1 litre of 0.4 N Phosphorous acid is diluted to 0.025 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
10 litre of 5 N Phosphorous acid is diluted to 1 M solution. Final volume of the solution is ——- litres
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M NaOH solution. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M NaOH solution. The value of ‘X’ is
1 litre of 0.2 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ litres of 2 M Sulphuric acid completely neutralizes 2 litres of 0.5 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Sulphuric acid completely neutralizes 200 ml of 0.4 M NaOH solution. The value of ‘X’ is
2 litres of 1.5 M Sulphuric acid completely neutralizes 12 litres of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
15 ml of ‘X’ M Sulphuric acid completely neutralizes 7.5 ml of 2 M NaOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Phosphorous acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
250 ml of 0.6 M Phosphorous acid completely neutralizes ‘X’ ml of 0.15 M NaOH solution. The value of ‘X’ is
1 litre of 0.2 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphorous acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphorous acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphorous acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Phosphorous acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ litres of 2 M Phosphorous acid completely neutralizes 1 litre of 0.5 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.25 M Phosphorous acid completely neutralizes 300 ml of 0.75 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.25 M Phosphorous acid completely neutralizes 300 ml of 0.75 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.25 M Phosphorous acid completely neutralizes 300 ml of 0.75 M NaOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphorous acid completely neutralizes 200 ml of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Phosphorous acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
2 litres of 1.5 M Phosphorous acid completely neutralizes 12 litres of ‘X’ M KOH solution. The value of ‘X’ is
1 litre of 1.5 M Phosphorous acid completely neutralizes 6 litres of ‘X’ M KOH solution. The value of ‘X’ is
1 litre of 2.5 M Phosphorous acid completely neutralizes 5 litres of ‘X’ M NaOH solution. The value of ‘X’ is
4 litres of 2.5 M Phosphorous acid completely neutralizes 20 litres of ‘X’ M KOH solution. The value of ‘X’ is
4 litres of 2.5 M Phosphorous acid completely neutralizes 20 litres of ‘X’ M KOH solution. The value of ‘X’ is
40 ml of 5 M Phosphorous acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
40 ml of 5 M Phosphorous acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
60 ml of 1 M Phosphorous acid completely neutralizes 180 ml of ‘X’ M KOH solution. The value of ‘X’ is
60 ml of 1 M Phosphorous acid completely neutralizes 180 ml of ‘X’ M KOH solution. The value of ‘X’ is
60 ml of 1 M Phosphorous acid completely neutralizes 180 ml of ‘X’ M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphorous acid completely neutralizes 300 ml of 4 M KOH solution. The value of ‘X’ is
250 ml of ‘X’ M Phosphorous acid completely neutralizes 25 ml of 20 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M KOH solution. The value of ‘X’ is
1 litre of 0.2 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M NaOH solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M KOH solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M KOH solution. The value of ‘X’ is
‘X’ litres of 2 M Hydrochloric acid completely neutralizes 1 litre of 0.5 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.1 M Hydrochloric acid completely neutralizes 200 ml of 0.4 M KOH solution. The value of ‘X’ is
‘X’ ml of 3 M Hydrochloric acid completely neutralizes 300 ml of 1.5 M KOH solution. The value of ‘X’ is
‘X’ ml of 3 M Hydrochloric acid completely neutralizes 300 ml of 1.5 M KOH solution. The value of ‘X’ is
‘X’ ml of 3 M Hydrochloric acid completely neutralizes 300 ml of 1.5 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M KOH solution. The value of ‘X’ is
2 litres of 1.5 M Hydrochloric acid completely neutralizes 12 litres of ‘X’ M KOH solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
80 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
80 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
80 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M NaOH solution. The value of ‘X’ is
75 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 12 M NaOH solution. The value of ‘X’ is
75 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 12 M NaOH solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
400 ml of 1 M Hydrochloric acid completely neutralizes ‘X’ ml of 1 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
1 litre of 0.2 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 1.5 M Hydrochloric acid completely neutralizes 12 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Hydrochloric acid completely neutralizes 6 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Hydrochloric acid completely neutralizes 600 ml of 2 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Hydrochloric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Hydrochloric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Nitric acid completely neutralizes ‘X’ ml of 2 M Calcium hydroxide solution. The value of ‘X’ is
500 ml of 0.2 M Nitric acid completely neutralizes ‘X’ ml of 0.1 M Calcium hydroxide solution. The value of ‘X’ is
1 litre of 0.2 M Nitric acid completely neutralizes ‘X’ litres of 0.1 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Nitric acid completely neutralizes 2 litres of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Nitric acid completely neutralizes 2 litres of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Nitric acid completely neutralizes ‘X’ litres of 0.2 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Nitric acid completely neutralizes 200 ml of 0.4 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.25 M Nitric acid completely neutralizes 300 ml of 0.75 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.25 M Nitric acid completely neutralizes 300 ml of 0.75 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.25 M Nitric acid completely neutralizes 300 ml of 0.75 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 3 M Nitric acid completely neutralizes 300 ml of 4.5 M Calcium hydroxide solution. The value of ‘X’ is
‘X’ ml of 3 M Nitric acid completely neutralizes 300 ml of 4.5 M Calcium hydroxide solution. The value of ‘X’ is
2 litres of 1.5 M Nitric acid completely neutralizes 12 litres of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
1 litre of 2.5 M Nitric acid completely neutralizes 5 litres of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
1 litre of 1.5 M Nitric acid completely neutralizes 6 litres of ‘X’ M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Nitric acid completely neutralizes 300 ml of 4 M Calcium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
200 ml of 2 M Sulphuric acid completely neutralizes ‘X’ ml of 2 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
500 ml of 0.4 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.2 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.2 M Barium hydroxide solution. The value of ‘X’ is
1 litre of 0.2 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 1 M Sulphuric acid completely neutralizes 2 litres of 0.5 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 1 M Sulphuric acid completely neutralizes 2 litres of 0.5 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
2 litres of 1.5 M Sulphuric acid completely neutralizes 12 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.3 M Sulphuric acid completely neutralizes 600 ml of 0.6 M Barium hydroxide solution. The value of ‘X’ is
60 ml of 1 M Sulphuric acid completely neutralizes 180 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Sulphuric acid completely neutralizes 20 litres of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of 1 M Sulphuric acid completely neutralizes 180 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of 1 M Sulphuric acid completely neutralizes 180 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of 1 M Sulphuric acid completely neutralizes 180 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Sulphuric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
60 ml of ‘X’ M Sulphuric acid completely neutralizes 15 ml of 4 M Barium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of 2 M Phosphorous acid completely neutralizes ‘X’ ml of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
1 litre of 0.2 M Phosphorous acid completely neutralizes ‘X’ litres of 0.1 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.2 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphorous acid completely neutralizes ‘X’ litres of 0.2 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Phosphorous acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
150 ml of 1 M Phosphorous acid completely neutralizes 450 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.1 M Phosphoric acid completely neutralizes ‘X’ litres of 0.1 M Magnesium hydroxide solution. The value of ‘X’ is
2 litres of 0.2 M Phosphoric acid completely neutralizes ‘X’ litres of 0.2 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Phosphoric acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.1 M Phosphoric acid completely neutralizes 2 litres of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ litres of 2 M Phosphoric acid completely neutralizes 1 litre of 0.5 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.1 M Phosphoric acid completely neutralizes 200 ml of 0.4 M Magnesium hydroxide solution. The value of ‘X’ is
‘X’ ml of 0.666 M Phosphoric acid completely neutralizes 200 ml of 2 M Magnesium hydroxide solution. The value of ‘X’ is
4 litres of 2.5 M Phosphoric acid completely neutralizes 20 litres of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
1 litre of 2.5 M Phosphoric acid completely neutralizes 5 litres of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
40 ml of 5 M Phosphoric acid completely neutralizes 80 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Phosphoric acid completely neutralizes 80 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Phosphoric acid completely neutralizes 80 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
80 ml of 5 M Phosphoric acid completely neutralizes 80 ml of ‘X’ M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 600 ml of 2 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
100 ml of ‘X’ M Phosphoric acid completely neutralizes 300 ml of 4 M Magnesium hydroxide solution. The value of ‘X’ is
200 ml of 2 M tribasic acid completely neutralizes ‘X’ ml of 15 M diacidic base. The value of ‘X’ is
500 ml of 0.2 M tribasic acid completely neutralizes ‘X’ ml of 0.4 M diacidic base. The value of ‘X’ is
1 litre of 0.2 M tribasic acid completely neutralizes ‘X’ litres of 0.1 M diacidic base. The value of ‘X’ is
2 litres of 0.1 M tribasic acid completely neutralizes ‘X’ litres of 0.1 M diacidic base. The value of ‘X’ is
2 litres of 0.1 M tribasic acid completely neutralizes ‘X’ litres of 0.1 M diacidic base. The value of ‘X’ is
2 litres of 0.1 M tribasic acid completely neutralizes ‘X’ litres of 0.1 M diacidic base. The value of ‘X’ is
‘X’ litres of 0.1 M tribasic acid completely neutralizes 2 litres of 0.4 M diacidic base. The value of ‘X’ is
‘X’ litres of 0.2 M tribasic acid completely neutralizes 2 litres of 0.4 M diacidic base. The value of ‘X’ is
4 litres of 0.2 M tribasic acid completely neutralizes ‘X’ litres of 0.4 M diacidic base. The value of ‘X’ is
‘X’ litres of 0.1 M tribasic acid completely neutralizes 2 litres of 0.4 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.1 M tribasic acid completely neutralizes 200 ml of 0.4 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M tribasic acid completely neutralizes 300 ml of 0.75 M diacidic base. The value of ‘X’ is
40 ml of 5 M tribasic acid completely neutralizes 80 ml of ‘X’ M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 300 ml of 4 M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 300 ml of 4 M diacidic base. The value of ‘X’ is
100 ml of ‘X’ M tribasic acid completely neutralizes 300 ml of 4 M diacidic base. The value of ‘X’ is
75 ml of ‘X’ M tribasic acid completely neutralizes 25 ml of 12 M diacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
400 ml of 1 M dibasic acid completely neutralizes ‘X’ ml of 1 M triacidic base. The value of ‘X’ is
1 litre of 0.2 M dibasic acid completely neutralizes ‘X’ litres of 0.1 M triacidic base. The value of ‘X’ is
2 litres of 0.1 M dibasic acid completely neutralizes ‘X’ litres of 0.1 M triacidic base. The value of ‘X’ is
2 litres of 0.2 M dibasic acid completely neutralizes ‘X’ litres of 0.2 M triacidic base. The value of ‘X’ is
2 litres of 0.2 M dibasic acid completely neutralizes ‘X’ litres of 0.2 M triacidic base. The value of ‘X’ is
2 litres of 0.1 M dibasic acid completely neutralizes ‘X’ litres of 0.2 M triacidic base. The value of ‘X’ is
‘X’ litres of 0.2 M dibasic acid completely neutralizes 2 litres of 0.4 M triacidic base. The value of ‘X’ is
‘X’ litres of 0.2 M dibasic acid completely neutralizes 2 litres of 0.4 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
‘X’ ml of 0.25 M dibasic acid completely neutralizes 600 ml of 0.5 M triacidic base. The value of ‘X’ is
2 litres of 1.5 M dibasic acid completely neutralizes 12 litres of ‘X’ M triacidic base. The value of ‘X’ is
1 litre of 2.5 M dibasic acid completely neutralizes 5 litres of ‘X’ M triacidic base. The value of ‘X’ is
4 litres of 2.5 M dibasic acid completely neutralizes 20 litres of ‘X’ M triacidic base. The value of ‘X’ is
150 ml of 1 M dibasic acid completely neutralizes 450 ml of ‘X’ M triacidic base. The value of ‘X’ is
100 ml of ‘X’ M dibasic acid completely neutralizes 600 ml of 2 M triacidic base. The value of ‘X’ is
100 ml of ‘X’ M dibasic acid completely neutralizes 300 ml of 4 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
37.5 ml of ‘X’ M dibasic acid completely neutralizes 50 ml of 3 M triacidic base. The value of ‘X’ is
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 1 M Sulphuric acid completely neutralizes ‘X’ ml of 1 M Aluminium hydroxide. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
500 ml of 0.2 M Sulphuric acid completely neutralizes ‘X’ ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
1 litre of 0.2 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.2 M Sulphuric acid completely neutralizes ‘X’ litres of 0.2 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Sulphuric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.2 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Sulphuric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 2 M Sulphuric acid completely neutralizes 1 litre of 0.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 3 M Sulphuric acid completely neutralizes 300 ml of 1.5 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 1.5 M Sulphuric acid completely neutralizes 12 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
1 litre of 2.5 M Sulphuric acid completely neutralizes 5 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
1 litre of 1.5 M Sulphuric acid completely neutralizes 6 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
4 litres of 2.5 M Sulphuric acid completely neutralizes 20 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
4 litres of 2.5 M Sulphuric acid completely neutralizes 20 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
40 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
80 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
80 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
80 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
80 ml of 5 M Sulphuric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 600 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of ‘X’ M Sulphuric acid completely neutralizes 300 ml of 4 M Aluminium hydroxide. The value of ‘X’ is
37.5 ml of ‘X’ M Sulphuric acid completely neutralizes 50 ml of 3 M Aluminium hydroxide. The value of ‘X’ is
37.5 ml of ‘X’ M Sulphuric acid completely neutralizes 50 ml of 3 M Aluminium hydroxide. The value of ‘X’ is
37.5 ml of ‘X’ M Sulphuric acid completely neutralizes 50 ml of 3 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 4 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 4 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 4 M Aluminium hydroxide. The value of ‘X’ is
400 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 4 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
100 ml of 2 M Hydrochloric acid completely neutralizes ‘X’ ml of 0.5 M Aluminium hydroxide. The value of ‘X’ is
1 litre of 0.2 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.1 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.2 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.2 M Aluminium hydroxide. The value of ‘X’ is
2 litres of 0.1 M Hydrochloric acid completely neutralizes ‘X’ litres of 0.2 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.2 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 0.1 M Hydrochloric acid completely neutralizes 2 litres of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ litres of 2 M Hydrochloric acid completely neutralizes 1 litre of 0.5 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.1 M Hydrochloric acid completely neutralizes 200 ml of 0.4 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.25 M Hydrochloric acid completely neutralizes 300 ml of 0.75 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.25 M Hydrochloric acid completely neutralizes 300 ml of 0.75 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.25 M Hydrochloric acid completely neutralizes 300 ml of 0.75 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
‘X’ ml of 0.3 M Hydrochloric acid completely neutralizes 600 ml of 0.6 M Aluminium hydroxide. The value of ‘X’ is
4 litres of 2.5 M Hydrochloric acid completely neutralizes 20 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
1 litre of 1.5 M Hydrochloric acid completely neutralizes 6 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
1 litre of 2.5 M Hydrochloric acid completely neutralizes 5 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
4 litres of 2.5 M Hydrochloric acid completely neutralizes 20 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
0.5 litres of 2.5 M Hydrochloric acid completely neutralizes 10 litres of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
40 ml of 5 M Hydrochloric acid completely neutralizes 80 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
150 ml of 1 M Hydrochloric acid completely neutralizes 900 ml of ‘X’ M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
250 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 20 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
250 ml of ‘X’ M Hydrochloric acid completely neutralizes 25 ml of 20 M Aluminium hydroxide. The value of ‘X’ is
15 ml of ‘X’ M Hydrochloric acid completely neutralizes 7.5 ml of 2 M Aluminium hydroxide. The value of ‘X’ is
Equal volumes of 0.1 M HCl and 0.4 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M HCl and 0.4 M HNO 3 are mixed with each other. The solution is diluted to one litre. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M HCl and 0.4 M HNO 3 are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M HCl and 0.4 M HNO 3 are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M HCl and 0.4 M HNO 3 are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
100 mL of 0.2 M HCl and 400 mL of 0.4 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
200 mL of 0.2 M HCl and 800 mL of 0.4 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
200 mL of 0.2 M HCl and 800 mL of 0.4 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
200 mL of 0.2 M HCl and 800 mL of 0.4 M HNO 3 are mixed with each other. Molarity of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M HCl and 2 litres 0.4 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M HCl and 2 litres 0.4 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.6 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M HNO 3 and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M HNO 3 and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.2 M HNO 3 and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.2 M HNO 3 and 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M HNO 3 and 2 litres 0.4 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M HNO 3 and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.3 M HNO 3 and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.6 M HNO 3 and 3 litres 0.4 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.6 M HNO 3 and 3 litres 0.4 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.3 M HNO 3 and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. Normality of the resultant mixture is —– N
Equal volumes of 0.1 M KOH and 0.2 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M KOH and 0.4 M NaOH are mixed with each other. The solution is diluted to one litre. Molarity of the resultant mixture is —– M
Equal volumes of 0.1 M KOH and 0.4 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
100 mL of 0.2 M KOH and 400 mL of 0.4 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M KOH and 0.4 M NaOH are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M KOH and 0.4 M NaOH are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
100 mL each of 0.1 M KOH and 0.4 M NaOH are mixed with each other. The solution is diluted to half litre. Molarity of the resultant mixture is —– M
200 mL of 0.2 M KOH and 800 mL of 0.4 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
200 mL of 0.2 M KOH and 800 mL of 0.4 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
200 mL of 0.2 M KOH and 800 mL of 0.4 M NaOH are mixed with each other. Molarity of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
Equal volumes of 0.1 M HCl and 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
1 litre of 0.6 M HCl and 3 litres 0.4 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
1 litre of 0.6 M HCl and 3 litres 0.4 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
2 litres of 0.3 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
2 litres of 0.3 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
Equal volumes of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
Equal volumes of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
Equal volumes of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.2 M HCl and 2 litres 0.4 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
Equal volumes of 0.1 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.6 M HCl and 3 litres 0.2 M H 2 SO 4 are mixed with each other. Proton concentration of the resultant mixture is —– M
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. Hydroxyl ion concentration of the resultant mixture is —– M
0.2 litres each of 1 M HCl and 2 M H 2 SO 4 are mixed with each other. The solution is diluted to one litre. Proton concentration of the resultant mixture is —– M
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
0.3 litres each of 1 M HCl and 2 M H 2 SO 4 are mixed with each other. The solution is diluted to one litre. Proton concentration of the resultant mixture is —– M
0.5 litres of 1 M HCl and 0.3 litres of 1 M H 2 SO 4 are mixed with each other. The solution is diluted to two litres. Proton concentration of the resultant mixture is —– M
1 litre each of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre each of 0.2 M NaOH and 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
2 litres of 0.6 M NaOH and 3 litres 0.2 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
1 litre of 0.2 M NaOH and 2 litres 0.4 M Ba(OH) 2 are mixed with each other. The solution is diluted to ten litres. Hydroxyl ion concentration of the resultant mixture is —– M
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 2 M HCl and 1 litre of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 2 M HCl and 1 litre of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
6 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
6 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
4 litres of 5 M HCl and 2 litres of 5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
4 litres of 5 M HCl and 1 litre of 5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
4 litres of 5 M HCl and 1 litre of 5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
400 ml of 1 M HCl and 100 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
400 ml of 1 M HCl and 100 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 4 M HCl and 200 ml of 0.5 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
4 litres of 1 M HCl and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
2 litres of 2 M HCl and 1 litre of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
800 ml of 1 M HCl and 200 ml of 2 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
800 ml of 1 M HCl and 200 ml of 2 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
800 ml of 1 M HCl and 200 ml of 2 M NaOH are mixed with each other. Normality of the resultant mixture is —– N
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
300 ml of 3 M HCl and 200 ml of 1 M NaOH are mixed with each other. Proton concentration of the resultant mixture is —– M
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
2 litres of 1 M H 2 SO 4 and 2 litres of 0.5 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
5 litres of 3 M H 2 SO 4 and 2 litres of 1 M NaOH are mixed with each other.The solution is diluted to ten litres. Normality of the resultant mixture is —– N
Van’t Hoff factor of 10% dissociated CH 3 COOH solution is
Van’t Hoff factor of 5% dissociated CH 3 COOH solution is
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 35% dissociated CH 3 COOH solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 70% dissociated NaCl solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 70% dissociated MSO 4 solution is
Van’t Hoff factor of 5% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 60% dissociated bi-bivalent electrolyte solution is
Van’t Hoff factor of 70% dissociated MSO 4 solution is
Van’t Hoff factor of 70% dissociated MSO 4 solution is
Van’t Hoff factor of 70% dissociated MSO 4 solution is
Van’t Hoff factor of 65% dissociated MSO 4 solution is
Van’t Hoff factor of 70% dissociated MSO 4 solution is
Van’t Hoff factor of 95% dissociated MSO 4 solution is
Van’t Hoff factor of 95% dissociated MSO 4 solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 10% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated uni-bivalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 80% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 95% dissociated bi-univalent electrolyte solution is
Van’t Hoff factor of 5% dissociated Na 2 SO 4 solution is
Van’t Hoff factor of 10% dissociated K 2 SO 4 solution is
Van’t Hoff factor of 15% dissociated BaCl 2 solution is
Van’t Hoff factor of 20% dissociated Calcium nitrate solution is
Van’t Hoff factor of 25% dissociated Ba(NO 3 ) 2 solution is
Van’t Hoff factor of 35% dissociated Na 2 SO 4 solution is
Van’t Hoff factor of 15% dissociated BaCl 2 solution is
Van’t Hoff factor of 30% dissociated CaCl 2 solution is
Van’t Hoff factor of 10% dissociated K 2 SO 4 solution is
Van’t Hoff factor of 50% dissociated SrCl 2 solution is
Van’t Hoff factor of 30% dissociated CaCl 2 solution is
Van’t Hoff factor of 70% dissociated MgCl 2 solution is
Van’t Hoff factor of 80% dissociated BaCl 2 solution is
Van’t Hoff factor of 85% dissociated Calcium nitrate solution is
Van’t Hoff factor of 75% dissociated Barium nitrate solution is
Van’t Hoff factor of 90% dissociated K 2 SO 4 solution is
Van’t Hoff factor of 95% dissociated Sodium sulphate solution is
Van’t Hoff factor of 5% dissociated Aluminium fluoride solution is
Van’t Hoff factor of 20% dissociated Potassium phosphate solution is
Van’t Hoff factor of 25% dissociated AlF 3 solution is
Van’t Hoff factor of 30% dissociated Na 3 PO 4 solution is
Van’t Hoff factor of 75% dissociated K 3 PO 4 solution is
Van’t Hoff factor of 45% dissociated Sodium phosphate solution is
Van’t Hoff factor of 20% dissociated Potassium phosphate solution is
Van’t Hoff factor of 5% dissociated Aluminium fluoride solution is
Van’t Hoff factor of 30% dissociated Na 3 PO 4 solution is
Van’t Hoff factor of 60% dissociated Aluminium fluoride solution is
Van’t Hoff factor of 75% dissociated K 3 PO 4 solution is
Van’t Hoff factor of 70% dissociated Na 3 PO 4 solution is
Van’t Hoff factor of 85% dissociated Sodium phosphate solution is
Van’t Hoff factor of 90% dissociated AlF 3 solution is
Van’t Hoff factor of 90% dissociated AlF 3 solution is
Van’t Hoff factor of 90% dissociated AlF 3 solution is
Van’t Hoff factor of 80% dissociated Potassium phosphate solution is
Van’t Hoff factor of 10% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 5% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 10% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 10% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 10% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 50% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 35% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 35% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 55% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 45% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 55% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 55% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 55% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 55% dissociated tri-univalent electrolyte solution is
Van’t Hoff factor of 50% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 85% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 85% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 85% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 50% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 85% dissociated uni-trivalent electrolyte solution is
Van’t Hoff factor of 5% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 10% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 5% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 25% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 45% dissociated K 3 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 45% dissociated K 3 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 35% dissociated K 3 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 25% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 25% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 25% dissociated Potassium ferricyanide solution is
Van’t Hoff factor of 45% dissociated K 3 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 20% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 90% dissociated [Co(NH 3 ) 6 ]Cl 3 solution is
Van’t Hoff factor of 5% dissociated Potassium ferrocyanide solution is
Van’t Hoff factor of 10% dissociated Potassium ferrocyanide solution is
Van’t Hoff factor of 20% dissociated Aluminium sulphate solution is
Van’t Hoff factor of 15% dissociated Aluminium sulphate solution is
Van’t Hoff factor of 25% dissociated Calcium phosphate solution is
Van’t Hoff factor of 20% dissociated Aluminium sulphate solution is
Van’t Hoff factor of 30% dissociated Calcium phosphate solution is
Van’t Hoff factor of 35% dissociated [Co(NH 3 ) 6 ] 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 35% dissociated [Co(NH 3 ) 6 ] 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 50% dissociated [Co(NH 3 ) 6 ] 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 55% dissociated K 4 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 55% dissociated K 4 [Fe(CN) 6 ] solution is
Van’t Hoff factor of 65% dissociated Ca 3 (PO 4 ) 2 solution is
Van’t Hoff factor of 75% dissociated Al 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 80% dissociated Al 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 85% dissociated [Co(NH 3 ) 6 ] 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 85% dissociated [Co(NH 3 ) 6 ] 2 (SO 4 ) 3 solution is
Van’t Hoff factor of 95% dissociated Aluminium sulphate solution is
Van’t Hoff factor of 95% dissociated Aluminium sulphate solution is
Van’t Hoff factor of 70% dissociated Ca 3 (PO 4 ) 2 solution is
5% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
10% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
20% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
20% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
60% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
85% of a solute is dimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
40% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
90% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
If D T and D 0 are the theoretical and observed vapour densities at a definite temperature and α be the degree of dissociation of a substance. Then, α in the terms of D o , D T and n (number of moles of products formed from 1 mole reactant) is calculated by the formula :
The vapour pressure of a given liquid will decrease if:
The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 19 mm Hg. If enthalpy of vaporisation is 40.67 kJ/ mol, then temperature T would be
A sample of liquid H 2 O of mass 18.0 g is injected into an evacuated 7.6 L flask maintained at 27.0 o C. lf vapoure pressure of H 2 O at 27 o C is 24.63 mm Hg. What weight percentage of the water will be vapourised when the system comes to equilibrium? Assume water vapours behaves as an ideal gas. The volume occupied by the liquid water is negligible compared to the volume of the container:
For a binary ideal liquid solution, the total pressure of the solution is given as:
The boiling point of C 6 H 6 , CH 3 OH, C 6 H 5 NH 2 and C 6 H 5 NO 2 are 80 o C, 65 o C, 184 o C and 212 o C respectively Which will show highest vapour pressure at room temperature :
6.0 g of urea (molecular weight : 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P o , the vapour pressure of solution is:
Calculate the weight of non-volatile solute having molecular weight 40, which should be dissolved in 57 gm octane to reduce its vapour pressure to 800%:
Equal weight of a solute are dissolved in equal weight of two solvents A and B and formed very dilute solution. The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for the solution A. lf M A and M B are the molecular weights of solvents A and B respectively, then:
An ideal solution has two components A and B.A is more volatile than B i.e., P A ∘ > P B ∘ and also P A ∘ > P total . If X A and Y A are mole fractions of components A in liquid and vapour phases, then:
Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm H8. Determine the vapour pressure of A and B in their pure states (in mm Hg):
Calculate solubility (mol/m 3 ) of saturated solution of Ag 3 PO 4 , if its dilute aqueous solution at ? K has a vapour pressure of 730 mm Hg and vapour pressure of pure H 2 O is 740 mm Hg at TK.
The vapour pressure curves of the same solute in the same solvent are shown. The curves are parallel to each other and do not intersect. The concentrations of solutions are in order of:
For ideal binary liquid solutions, select incorrect statement.
The boiling point of an azeotropic mixture of water-ethanol is less than that of both water and ethanol. Then
Total vapour pressure of mixture of l mol X P X ∘ = 150 torr ) and 2 mol Y P y = 300 torr is 240 torr. In this case:
Which will form maximum boiling azeotrope ?
Azeotropic mixture of water and C 2 H 5 OH boils at 351 K. By distilling the mixture it is possible to obtain :
One mole of a solute A is dissolved in a given volume of a solvent. The association of the solute take place as follows: nA ⇌ A n If α is the degree of association of A, the van’t Hoff factor i is expressed as:
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase ?
Chloroform, CHCI 3 , boils at 61.7 o C. If the K b for chloroform is 3.63 o C/molal, what is the boiling point of a solution of 15.0 kg of CHCl 3 and 0.616 kg of acenaphthalene, C 12 H 10 ?
One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of 1.518 K. The degree of association for dimerization of the acid in benzene is (K b for benzene = 2.53 K kg mol -1 ) :
Which one of the following aqueous solutions will exhibit highest boiling point:
If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in solvent (K b = xK.kg.mo -1 ) is y K, then the depression in freezing point of solution of same concentration would be: K f of the solvent = zK ⋅ kg ⋅ mol − 1
When a solution containing non-volatile solute freezes, which equilibrium would exist?
Bromoform has a normal freezing point of 7.734 o C/m and it’s K f =14.4 o C/m. A solution of 2.60 g of an unknown in 100 g of bromoform freezes at 5.43 o C. What is the molecular weight of the unknown?
The freezing point of a solution of 2.40 g of biphenyl (C 12 H 10 ) in 75.0 g of benzene (C 6 H 6 ) is 4.40 o C. The normal freezing point of benzene is 5.50 o C. What is the molal freezing point constant ( o C/m) for benzene?
Camphor is often used in molecular mass determination because:
For 1molal solution of each compound minimum freezing point will be assuming complete ionisation in each case:
For 1 molal solution of each compound maximum freezing point will be assuming complete ionisation in each case:
Based on the given diagram, which of the following statements regarding the homogenous solutions of two volatile liquids are correct? (1) Plots 4D and BC show that Raoult’s law is obeyed for the solution in which B is a solvent and A is the solute and as well as for that in which A is solvent and B is solute. (2) Plot CD shows that Dalton’s law of partial pressures is obeyed by the binary solution of components A and B. (3) EF + EG = EH; and AC and BD correspond to the vapour pressures of the pure solvents A and B respectively. Select the correct answer using the options given below:
Formation of a solution from two components can be considered as : (i) Pure solvent separated solvent molecules, ∆ H 1 (ii) Pure solute separated solute molecules, ∆ H 2 (iii) separated solvent and solute molecules solution, ∆ H 3 Solution so formed will be ideal if :
In a mixture of A and B, components show Positive deviation when:
An azeotropic mixture of two liquids has a boiling point higher than either of them when it:
The azeotropic mixture of water (8.P.= 100 o C) and HCI (B.P : 86 o C) boils at about 120 o C. During fractional distillation of this mixture it is possible to obtain :
Which of the following is not a colligative property?
24.5 grams of Sulphuric acid is dissolved in a one litre solution. Normality of the solution is…………N
0.1 moles of Barium hydroxide is dissolved in a 400 ml solution. Normality of the solution is—–N
0.1 moles of Barium hydroxide is dissolved in a 400 ml solution. Normality of the solution is—–N
Normality of 0.2 M potassium permanganate solution when it acts like oxidizing agent in neutral medium is
0.1 moles of Barium hydroxide is dissolved in a 400 ml solution. Normality of the solution is—–N
Normality of 0.2 M potassium permanganate solution when it acts like oxidizing agent in neutral medium is
Henry’s constant values for four gases dissolved in water at 298 K and 1 bar are given below. The gas with highest solubility of the four will have a value of……..kPa
10ml of decamolar sulphuric acid is diluted to one litre. Proton concentration of the solution is
Colligative properties depend on the number of solute particles present in the solution. Osmotic pressure of 30% ionized 0.1M K 2 SO 4 solution at 27 0 c is
The solution with highest vapour pressure of the following is
Volume of centimolar sulphuric acid which can exactly neutralize 300ml of decimolar caustic potash solution is
The liquid mixture which shows positive deviations from Raoult’s law is
A mixture can be homogeneous or heterogeneous. Bronze is an example of
Value of cryoscopic constant depends on
0.73 grams of HCl is dissolved in 800 ml solution. Molarity of the solution is
Van’t Hoff factor of 33% dissociated AlF 3 solution is
0.8 grams of NaOH is dissolved in a 50 ml solution. Normality of the solution is —– N
Which one of the following statement is correct regarding minimum boiling azeotrope ?
0.1 gram equivalents of KOH is dissolved in a six litre solution. Molarity of the solution is
1.2 litres of 1 M Barium hydroxide is diluted to 6 litre solution. Normality of the resultant solution is —- N
4.5 litres of 6 M H 2 SO 4 is diluted to 1.5 N solution. Final volume of the solution is ——- litres
450 ml of 0.9 M Sulphuric acid is diluted to 0.45 N solution. Volume of water added is —— mL
0.1 litre of 1.25 M Phosphorous acid is diluted to 1 litre solution. Normality of the resultant solution is —- N
Molar mass of acetic acid measured by osmotic pressure experiments is 75 grams. Degree of dimerization of acetic acid dissolved in benzene is
20 ml of 1.8 M Hydrochloric acid completely neutralizes 50 ml of ‘X’ M Barium hydroxide solution. The value of ‘X’ is
Percentage of Nitrogen in the gaseous mixture of respiratory apparatus used by scuba divers is
30 ml of 0.15 M Sulphuric acid completely neutralizes ‘X’ ml of 0.1 M KOH solution. The value of ‘X’ is
Osmotic pressure is least for
Vanthoff factor of calomel in a solution where it is assumed to undergo 50% ionization is
Boiling point of 0.5 m NaCl is 373.52 K. Degree of ionization of NaCl is K b of water = 0 . 52 K . Kg . mol – 1
120 ml of 0.05 M Phosphorous acid completely neutralizes ‘X’ ml of 0.1 M KOH solution. The value of ‘X’ is
‘X’ ml of 0.8 M Sulphuric acid completely neutralizes 100 ml of 2.8 M Barium hydroxide solution. The value of ‘X’ is
‘X’ litres of 0.75 M tribasic acid completely neutralizes 1 litre of 2.25 M diacidic base. The value of ‘X’ is
36% of a solute is trimerised in a solution. Van’t Hoff factor of the solute in the solution is
3 litres of 1 M H 2 SO 4 and 1 litres of 2 M NaOH are mixed with each other.The solution is diluted to ten litres. Proton concentration of the resultant mixture is —– M
Which one of the following is not an ideal solution ?
Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 1 : 4. Vapour pressure of pure liquid ‘A’ is 800 mmHg and pure liquid ‘B’ is 100 mmHg. Mole fraction of ‘B’ in vapour phase will be
At 300 K, 0.15 M Fructose solution is isotonic with 0.1 M KCl solution. Percentage of dissociation of KCl in the solution is
Molarity of a solution of density 0.4 g/ml is 1 M. Molality of the same solution under similar conditions will be (Molar mass of solute – 100 g)
n-factor of potassium dichromate as oxidant in acidic medium is ‘X’. n-factor of stannous sulphate as reductant in acidic medium is ‘Y’. Product of X and Y is
25 ml of Ferrous sulphate solution reduces 75 ml of 0.04 M potassium permanganate solution in acidic medium. Molarity of Ferrous sulphate solution is
At 298 K , Henry’s law constant of a gas ‘X’ dissolved in water 1 . 8 × 10 – 5 Kbar . Number of moles of ‘X’ dissolved in 100 grams of water at 1 atm and 298 K will be
Degree of ionization of 0.1 m aqueous NaCl is α 1 . Degree of ionization of 0.1 m aqueous K 3 Fe CN 6 is α 2 . Both these solutions have the same boiling point when α 1 : α 2 is equal to
Molar mass of Na 2 SO 4 dissolved in water is determined from depression of freezing point experiment. Molar mass of Na 2 SO 4 observed can never be
Twelve grams of urea dissolved in 1 Kg of water showed a depression of freezing point of ‘X’. 36 grams of glucose dissolved in 500 grams of water showed a depression of freezing point of ‘Y’. Correct relation between ‘X’ and ‘Y’ is
At certain temperature, when 20 grams of a non-volatile solute is dissolved in 180 grams of water, the vapour pressure of water lowered from 81 mm Hg to 80.19 mm Hg. Molar mass of non volatile solute is nearly
54 ml of liquid ‘A’ is mixed with 54 ml of liquid ‘B’. The volume of the mixture formed is 108 ml. The two liquids are most likely to be
Molar mass of a diacidic base is 10g. Density of its aqueous solution is 1 g/ cc. More concentrated solution among the following is
Vapour pressure of a liquid depends on many factors. Correct statement of the following is
Relative lowering of vapour pressure of non aqueous solution containg non volatile solute is 0.1. Molality of the solution will be ——(MW of solvent=50)
80 grams of solute ( molar mass = 160 g ) is dissolved in 100 ml solution of density 1.8 g/ml. Molality of the solution is
Freezing point of aqueous solution of potassium ferrocyanide can be
Van’t Hoff factor of ‘X’ molal Sodium chloride solution is least when ‘X’ is
Molarity of 2 N Potassium ferricyanide solution is
Mole fraction of solute in 10%(w/w) aqueous NaOH solution is
The composition of bronze is
The composition of german silver is
The exact composition of brass is
The exact composition of german silver is
which of the fallowing one determine the chemical properties of the solution?
which of the fallowing component in lesser component in the solution?
Homogeneous mixture means
Homogeneous mixture have
Solution have
Sodium fluoride is mostly used in
Match Column I with Column II Column I Column II A 1ppm of F – in water i tooth to become mottled B 1.5 ppm of F – in water ii water become poisonous C Grater than than 1.5 ppm of F – in water iii used in rat poison iv prevent tooth decay Choose the correct answer from the options given. A B C 1 ii i iv 2 i ii ii 3 iv i ii 4 ii iv iii
Match Column I with Column II Column I Column II A 1ppm of F – in water i tooth to become mottled B 1.5 ppm of F – in water ii water become poisonous C Grater than than 1.5 ppm of F – in water iii used in rat poison iv prevent tooth decay Choose the correct answer from the options given. A B C 1 ii i iv 2 i ii ii 3 iv i ii 4 ii iv iii
Match Column I with Column II Column I Column II A 1ppm of F – in water i tooth to become mottled B 1.5 ppm of F – in water ii water become poisonous C Grater than than 1.5 ppm of F – in water iii used in rat poison iv prevent tooth decay Choose the correct answer from the options given. A B C 1 ii i iv 2 i ii ii 3 iv i ii 4 ii iv iii
Match Column I with Column II Column I Column II A 1ppm of F – in water i tooth to become mottled B 1.5 ppm of F – in water ii water become poisonous C Grater than than 1.5 ppm of F – in water iii used in rat poison iv prevent tooth decay Choose the correct answer from the options given. A B C 1 ii i iv 2 i ii ii 3 iv i ii 4 ii iv iii
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
How much percentage of fluoride ions in water causes poisonous?
1 part per million (ppm) of fluoride ions in water
How much percentage of fluoride ions in water causes the tooth to become mottled?
Statement I: Intravenous injections are always dissolved in water containing salts at particular ionic concentrations Statement II: At particular ionic concentrations that match with blood plasma concentrations
Match Column I with Column II Column I Column II A A solution of gas in gas i Mixture of oxygen and nitrogen B A solution of solid in liquid ii O 2 dissolved in water C A solution of liquid in liquid iii Amalgam of mercury with sodium D A solution of liquid in solid iv Glucose in water v Ethanol in water Choose the correct answer from the options given. A B C D 1 i iv v iii 2 iii ii iv v 3 i iii iv ii 4 i iv ii v
What (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine?
35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At this concentration the antifreeze lowers the freezing point of water to
Statement I: 35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. Statement II: At this concentration the antifreeze lowers the freezing point of water to 255.4K (–17.6°C).
Which of the fallowing compound an antifreeze, is used in cars for cooling the engine?
Statement I: Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity is a function of temperature. Statement II: This is because volume depends on temperature and the mass does not
Which of the fallowing concentration term is independent on temperature?
