Calculate the depression in the freezing point of water when $10 \ g$ of $CH_3CH_2CHClCOOH$ is added to $250 \ g$ of water. $K_a = 1.4 \times 10^{-3}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$

(D) Molar mass of $CH_3CH_2CHClCOOH = (3 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) + 1 = 122.5 \ g \ mol^{-1}$.
Number of moles of $CH_3CH_2CHClCOOH = \frac{10 \ g}{122.5 \ g \ mol^{-1}} = 0.0816 \ mol$.
Molality $(m)$ of the solution $= \frac{0.0816 \ mol}{0.250 \ kg} = 0.3264 \ mol \ kg^{-1}$.
For the dissociation $CH_3CH_2CHClCOOH \leftrightarrow CH_3CH_2CHClCOO^{-} + H^{+}$,the dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha} \approx C \alpha^2$.
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} = 0.0655$.
The van't Hoff factor $i = 1 + \alpha = 1 + 0.0655 = 1.0655$.
The depression in freezing point $\Delta T_f = i \cdot K_f \cdot m = 1.0655 \times 1.86 \ K \ kg \ mol^{-1} \times 0.3264 \ mol \ kg^{-1} = 0.647 \ K \approx 0.65 \ K$.