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Solutions · Question 42

AIPMT 2006MCQEasy📈 Topic in 38% papers+4−1
1

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.

Answer

(3) shows a positive deviation from Raoult's law

Solution

As per NCERT, in a solution of ethanol and acetone, the molecules of acetone get in between the hydrogen-bonded ethanol molecules and break some of the hydrogen bonds.

⇒ The A–B (acetone–ethanol) interactions are weaker than the ethanol–ethanol interactions.

⇒ Molecules escape more easily, so the vapour pressure is higher than predicted by Raoult's law ( mixH > 0, mixV > 0).

∴ The solution shows a positive deviation from Raoult's law.

Answer: (3)

AIPMT 2006

Keep practising — Solutions

2

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
AIPMT 2006
3

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
AIPMT 2006
4

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

  1. 0.009
  2. 0.018
  3. 0.027
  4. 0.036
AIPMT 2005
5

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
AIPMT 2005
6

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
AIPMT 2005
7

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
AIPMT 2005
8

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
AIPMT 2003
9

A solution containing components A and B follows Raoult's law when

  1. A-B attraction force is greater than A-A and B-B
  2. A-B attraction force is less than A-A and B-B
  3. A-B attraction force remains same as A-A and B-B
  4. volume of solution is different from sum of volumes of solute and solvent
AIPMT 2002
10

A solution contains non-volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?

  1. M2 = ( m2 )VRT
  2. M2 = ( m2V ) RT
  3. M2 = ( m2V ) RT
  4. M2 = ( m2V ) RT
AIPMT 2002
11

Pure water can be obtained from sea water by

  1. centrifugation
  2. plasmolysis
  3. reverse osmosis
  4. sedimentation
AIPMT 2001

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