Solutions · Question 4

What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution? [Given: Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol-1 respectively.]

1. 0.825
2. 0.032
3. 0.867
4. 0.967

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is kg mol-1. (Nearest integer) (Given : g = 10 m s-2, R = 8.3 kPa dm3 K-1 mol-1, density of solution = 1000 kg m-3)

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is . (Nearest integer) [Given: boiling point of the pure solvent = 319.5 K, Kb of the solvent = 0.3 K kg mol-1]

At 27 °C, 0.1 M, 1 L K4[Fe(CN)6] aqueous solution and 0.1 M, 1 L FeCl3 aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct? [figure]

1. Blue color is formed on both sides.
2. Ionic solutes in aqueous solution can pass through semi-permeable membrane.
3. Solution on side 'y' is hypotonic.
4. To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side 'x'.

19.5 g of fluoro acetic acid (molar mass = 78 g mol-1) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1°C. What is Ka of fluoro acetic acid ? (For water, Kf = 1.86 K kg mol-1). Assume molarity and molality to have same values.

1. 10-6
2. 4 10-4
3. 3 10-5
4. 3 10-3

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