1

### AIPMT 2007

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
A
0.56 K
B
1.12 K
C
$-$ 0.56 K
D
$-$ 1.12 K

## Explanation

HX H+ + X-
Initially 1 0 0
At equilibrium 1 - $\alpha$ $\alpha$ $\alpha$

Total moles = 1 – $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$

$\therefore$ i = ${{1 + \alpha } \over 1}$

Given, $\alpha$ = 20% = 0.2

$\therefore$ i = 1 + $\alpha$ = 1 + 0.2 = 1.2

$\Delta$Tf = ikf m = 1.2 × 1.86 × 0.5 = 1.12 K
2

### AIPMT 2006

A solution containing 10 g per dm3 of urea (molecular mass = 60 g mol$-$1) is isotonic with a 5% solution of a nonvolatile solute is
A
200 g mol$-$1
B
250 g mol$-$1
C
300 g mol$-$1
D
350 g mol$-$1

## Explanation

For isotonic solution,

osmotic pressure of urea = osmotic pressure of nonvolatile solute

${{10} \over {60 \times 1000}}$ = ${5 \over {m \times 100}}$

$\Rightarrow$ m = 300 g mol–1
3

### AIPMT 2006

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
A
0.2 K
B
0.4 K
C
0.3 K
D
0.5 K

## Explanation

Molality of non-electrolyte solute

= ${1 \over {250 \times 0.0512}}$ = 0.0781 m

$\Delta$Tf = kf m

= 5.12 × 0.0781 = 0.4 K
4

### AIPMT 2006

A solution of acetone in ethanol
A
obeys Raoult's law
B
shows a negative deviation from Raoult's law
C
shows a positive deviation from Raoult's law
D
behaves like a near ideal solution.

## Explanation

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.