Solutions · Question 9

Solution A is prepared by dissolving 1 g of a protein (molar mass = 50,000 g mol-1) in 0.5 L of water at 300 K. Its osmotic pressure is x bar. Solution B is made by dissolving 2 g of the same protein in 1 L of water at 300 K. Osmotic pressure of solution B is y bar. Entire solution of A is mixed with entire solution of B at the same temperature. The osmotic pressure of resultant solution is z bar. x, y and z respectively are: (R = 0.083 L bar mol-1 K-1)

1. 9.96 10-4; 9.96 10-4; 9.96 10-4
2. 9.96 10-4; 9.96 10-4; 19.92 10-4
3. 4.98 10-4; 4.98 10-4; 9.96 10-4
4. 4.98 10-4; 4.98 10-4; 4.98 10-4

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

1. 400, 300
2. 600, 400
3. 500, 200
4. 300, 200

Consider the following aqueous solutions. I. 2.2 g glucose in 125 mL of solution. II. 1.9 g calcium chloride in 250 mL of solution. III. 9.0 g urea in 500 mL of solution. IV. 20.5 g aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be : [Given : Molar mass in g mol-1 : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]

1. III < I < II < IV
2. II < III < IV < I
3. II < III < I < IV
4. I < II < III < IV

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 g mol-1 and 0.9 g of a non-volatile non-electrolyte solute 'B' of molar mass 180 g mol-1 in 100 mL H2O at 27 . Osmotic pressure of the solution will be [Given: R = 0.082 L atm K-1 mol-1]

1. 2.46 atm
2. 0.82 atm
3. 1.47 atm
4. 1.23 atm

'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g mol-1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and Kb for water is 0.52 K kg mol-1, then the mole fraction of the electrolyte solute (x2) in the solution can be expressed as (Given : density of water = 1 g/mL and boiling point of water = 373 K)

1. 162.6 WM
2. 1.38 MW
3. 2.616 MW
4. 1.38 WM

At 298 K, the mole percentage of N2(g) in air is 80\%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2(g) in water at 298 K? (KH for N2 is 6.5 107 mmHg)

1. 9.35 10-5
2. 9.35 105
3. 1.23 10-7
4. 1.17 10-4

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m-2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

1. 0.5217
2. 0.480
3. 0.340
4. 0.663

Which one of the following graphs accurately represents the plot of partial pressure of CS2 vs its mole fraction in a mixture of acetone and CS2 at constant temperature?

1. [figure]
2. [figure]
3. [figure]
4. [figure]

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is . (Nearest integer)

Consider a solution of CO2(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2 in vapour phase above water) [y-axis] with log (mole fraction of CO2 in water) [x-axis] at 25 ?

1. [figure]
2. [figure]
3. [figure]
4. [figure]