Calculate the depression in the freezing point of water when

Molar mass of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}=15+14+13+35.5+12+16+16+1$

$=122.5 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ No. of moles present in $10 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}=\frac{10 \mathrm{~g}}{122.5 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=0.0816 \mathrm{~mol}$

It is given that $10 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}$ is added to $250 \mathrm{~g}$ of water.

$\therefore$ Molality of the solution, $=\frac{0.0186}{250} \times 1000$

$=0.3264 \mathrm{~mol} \mathrm{~kg}^{-1}$

Let $a$ be the degree of dissociation of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}$.

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}$ undergoes dissociation according to the following equation:

$\therefore K_{a}=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}$

$=\frac{C \alpha^{2}}{1-\alpha}$

Since α is very small with respect to 1, 1 − α ≈ 1

Now, $K_{a}=\frac{C \alpha^{2}}{1}$

$\Rightarrow \alpha=\sqrt{\frac{K_{a}}{C}}$

$=\sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} \quad\left(\because K_{a}=1.4 \times 10^{-3}\right)$

$=0.0655$

Again,

Total moles of equilibrium = 1 − α + α + α

= 1 + α

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

$=1+\alpha$

$=1+0.0655$

$=1.0655$

Hence, the depression in the freezing point of water is given as:

$\Delta T_{f}=i . K_{f} m$

$=1.0655 \times 1.86 \mathrm{Kkg} \mathrm{mol}^{-1} \times 0.3264 \mathrm{~mol} \mathrm{~kg}^{-1}$

$=0.65 \mathrm{~K}$