Question

2 molal solution of a weak acid HA has a freezing point of $3.885{}^{o}C$. The degree of dissociation of this acid is _________ × ${10}^{-3}$. (Round off to the Nearest Integer). [Given: Molal depression constant of water = $1.85Kkgmo{l}^{-1}$
Freezing point of pure water = $0^{o}C$]

Answer 1

$T_{f}sol.=–3.885°C$

$\Delta T_f=+3.885=i\times k_f\times molality$

$3.885=i\times 1.85\times 2$

For, $HX\rightleftharpoons H^{+}+X^{-}$

$i=\frac{\text{No. of moles after dissociation}}{\text{No. of moles before dissociation}}$

$i=1+\alpha$

$i=\frac{3.885}{1.85\times 2}=[1+\alpha ]$

$\alpha =\frac{3.885}{3.7}-1=0.05=50\times {10}^{-3}$

Answer: 50