1

### AIPMT 2010 Prelims

A solution of sucrose (molar mass = 342 g mol$-$) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol$-$1)
A
$-$ 0.372oC
B
$-$ 0.520oC
C
+ 0.372oC
D
$-$ 0.570oC

## Explanation

Depression in freezing point,

$\Delta$Tf = Kf $\times$ m

m = ${{{w_B}} \over {{M_B}}} \times {{1000} \over {{W_A}}}$

= ${{68.5 \times 1000} \over {342 \times 1000}}$

$\Delta$Tf = 1.86 $\times$ ${{68.5} \over {342}}$

= 0.372 oC

$\therefore$ Tf = 0 - 0.372 oC = - 0.372 oC
2

### AIPMT 2009

A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at $-$ 0.00732oC. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = $-$1.86oC/m)
A
3
B
4
C
1
D
2

## Explanation

The number of moles of ions produced by 1 mol of ionic compound = i

Given, m = 0.0020 m

$\Delta$Tf = 0oC – 0.00732oC = – 0.00732oC

Kf = – 1.86 oC

$\Delta$Tf = ikfm

$\Rightarrow$ i = ${{0.00732} \over {1.86 \times 0.0020}}$ = 2
3

### 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
4

### 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