Question

The freezing point depression constant for water is ${1.86}^{\circ }C{m}^{-1}$. If 5.00 g $N{a}_{2}S{O}_4$ is dissolved in 45.0 g $H_{2}O$, the freezing point is changed by $-{3.82}^{\circ }C$. Calculate the Van't Hoff factor for $N{a}_{2}S{O}_4$.

[AIMPT 2011]

Answer 1

Depression in freezing point is given by,
$\Delta T_f=i\times K_f\times m$

where,

$\Delta T_f$ is the depression in freezing point.

$K_f$ is molal depression constant.

$m$ is molality of solution.

$i$ is van't Hoff factor

$i=\frac{\Delta T_f\times W_A}{K_f\times n_B\times 1000}$

$n_B$ is number of moles of $N{a}_{2}S{O}_4$

$W_A$ is the mass of $H_{2}O$ in grams

Molar mass of $N{a}_{2}S{O}_4$ is 142 g/mol

$i=\frac{3.82\times 45}{1.86\times \left(\frac{5}{142}\right)\times 1000}$

$=2.63$

Hence, option (c) is correct.