Solutions · Question 39

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol-1, the lowering in freezing point of the solution is

1. 0.56 K
2. 1.12 K
3. -0.56 K
4. -1.12 K

During osmosis, flow of water through a semipermeable membrane is

1. from solution having lower concentration only
2. from solution having higher concentration only
3. from both sides of semipermeable membrane with equal flow rates
4. from both sides of semipermeable membrane with unequal flow rates.

A solution of acetone in ethanol

1. obeys Raoult's law
2. shows a negative deviation from Raoult's law
3. shows a positive deviation from Raoult's law
4. behaves like a near ideal solution.

1.00 g of a non-electrolyte solute (molar mass 250 g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by

1. 0.2 K
2. 0.4 K
3. 0.3 K
4. 0.5 K

A solution containing 10 g per dm3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5% solution of a nonvolatile solute. The molecular mass of this nonvolatile solute is

1. 200 g mol-1
2. 250 g mol-1
3. 300 g mol-1
4. 350 g mol-1

The mole fraction of the solute in one molal aqueous solution is

1. 0.009
2. 0.018
3. 0.027
4. 0.036

The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be

1. 72 torr
2. 140 torr
3. 68 torr
4. 20 torr

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be

1. 0.200
2. 0.549
3. 0.786
4. 0.478

A solution of urea (mol. mass 60 g mol-1) boils at 100.18 at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at

1. 0.654
2. -0.654
3. 6.54
4. -6.54

Formation of solution from two components can be considered as (i) Pure solvent separated solvent molecules, H1 (ii) Pure solute separated solute molecules, H2 (iii) Separated solvent and solute molecules solution, H3 Solution so formed will be ideal if

1. Hsoln = H1 + H2 + H3
2. Hsoln = H1 + H2 - H3
3. Hsoln = H1 - H2 - H3
4. Hsoln = H3 - H1 - H2