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$x\times {10}^{-3}$.

The value of x (Round off to the Nearest Integer) is

[Given : molal depression constant of water = $1.85Kkg{\text{mol}}^{-1}$]
Freezing point of pure water = $0^{o}C$

• A. 50.00
• B. 50.0
• C. 50

Answer 1

Answer: (A), (B), (C)

$HA\rightleftharpoons H^{+}+A^{-}$
$1-\alpha \alpha \alpha$
($\alpha$-degree of dissociation)
$\text{van't Hoff factor}\left(i\right)=1+\alpha$
$\Delta T_f=i{K}_{f}m$
$m$ is molality
$\Delta T_f$ is depression in freezing point
$K_f$ is cryoscopic constant
$i$ is van't Hoff factor
$\Rightarrow 3.885=(1+\alpha )\times 1.85\times 2$
$\Rightarrow (1+\alpha )=1.05$
$\alpha =0.05=50\times {10}^{-3}$
$x=50$