1

### AIPMT 2011 Prelims

The freezing point depression constant for water is $-$1.86o C m$-$1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by $-$3.82oC. Calculate the van't Hoff factor for Na2SO4.
A
2.05
B
2.63
C
3.11
D
0.381

## Explanation

According to depression in freezing point,

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

= i $\times$ Kf $\times$ ${{{w_B} \times 1000} \over {{m_B} \times {w_A}}}$

Given : $\Delta$Tf = 3.82, Kf = 1.86,

wB = 5, mB = 142, wA = 45

i = ${{\Delta T \times {m_B} \times {w_A}} \over {{K_f} \times {w_B} \times 1000}}$

= ${{3.82 \times 142 \times 45} \over {1.86 \times 5 \times 1000}}$

= 2.63
2

### AIPMT 2011 Mains

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be
A
$-$0.18oC
B
$-$0.54oC
C
$-$0.36oC
D
$-$0.24oC

## Explanation

We know that $\Delta$Tf = i × Kf × m

Here i is van’t Hoff’s factor.

i for weak acid is 1 + $\alpha$.

Here $\alpha$ is degree of dissociation i.e., 30/100 = 0.3

$\therefore$ i = 1 + $\alpha$ = 1 + 0.3 = 1.3

$\therefore$ Tf = i × Kf × m = 1.3 × 1.86 × 0.1 = 0.24

$\therefore$ Freezing point = – 0.24
3

### AIPMT 2010 Prelims

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
A
B
C
D

## Explanation

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles resulting in decreased surface area. However, addition of solvent, i.e., dilution increases the surface area of the liquid surface, thus results in increased vapour pressure. Hence, addition of water to the aqueous solution of (1 molal) KI results in increased vapour pressure.
4

### 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