$0.6 \, mL$ of acetic acid $(CH_{3}COOH)$,having density $1.06 \, g \, mL^{-1}$,is dissolved in $1 \, L$ of water. The depression in freezing point observed for this strength of acid was $0.0205^{\circ} \, C$. Calculate the van't Hoff factor and the dissociation constant of acid.

Number of moles of acetic acid = $\frac{0.6 \, mL \times 1.06 \, g \, mL^{-1}}{60 \, g \, mol^{-1}} = 0.0106 \, mol$.
Molality $(m) = \frac{0.0106 \, mol}{1 \, kg} = 0.0106 \, mol \, kg^{-1}$.
Theoretical depression in freezing point $(\Delta T_{f, \text{calc}}) = K_{f} \times m = 1.86 \, K \, kg \, mol^{-1} \times 0.0106 \, mol \, kg^{-1} = 0.0197 \, K$.
van't Hoff Factor $(i) = \frac{\Delta T_{f, \text{obs}}}{\Delta T_{f, \text{calc}}} = \frac{0.0205 \, K}{0.0197 \, K} \approx 1.041$.
For dissociation $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$,$i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
$\alpha = i - 1 = 1.041 - 1 = 0.041$.
Dissociation constant $(K_{a}) = \frac{c \alpha^{2}}{1 - \alpha} = \frac{0.0106 \times (0.041)^{2}}{1 - 0.041} = \frac{0.0106 \times 0.001681}{0.959} \approx 1.86 \times 10^{-5}$.