Question:
2 molal solution of a weak acid HA has a freezing point of $3.885^{\circ} \mathrm{C}$. The degree of dissociation of this acid is_______ $\times 10^{-3}$. (Round off to the Nearest Integer).
[Given : Molal depression constant of water $=$ $1.85 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ Freezing point of pure water $\left.=0^{\circ} \mathrm{C}\right]$
Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \mathrm{K}_{\mathrm{f}} \mathrm{m}$
$\alpha=0.05=50 \times 10^{-3}$