1

### AIPMT 2007

Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL$-$1. Volume of acid required to make one litre of 0.1 M H2SO4 solution is
A
16.65 mL
B
22.20 mL
C
5.55 mL
D
11.10 mL

## Explanation

Normality = ${{98 \times 1.8 \times 10} \over {49}}$ = 36 N

N2 = 0.1 $\times$ 2 = 0.2 N

N2V2 = N1V1

$\Rightarrow$ 36 $\times$ V = 0.2 $\times$ 1000

$\Rightarrow$ V = ${{0.2 \times 1000} \over {36}}$ = 5.55 mL
2

### 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
3

### 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
4

### 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