The freezing point of a $1.00 \mathrm{~m}$ aqueous solution of $\mathrm{HF}$ is found to be $-1.91^{\circ} \mathrm{C}$. The freezing point constant of water, $K_f$ is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The percentage dissociation of $\mathrm{HF}$ at this concentration is

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Aicharya · Accepted Answer

$\Delta T_{\mathrm{f}}=K_{\mathrm{f}} 	imes m 	imes i$ $ \begin{aligned} & i=\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} 	imes m}=\frac{1.91}{1.86 	imes 1}=1.02 \ & 	ext { For } \mathrm{HI} \leftrightharpoons \mathrm{H}^{+}+\mathrm{I}^{-} \ & \quad(1-\alpha) \quad \alpha \quad \alpha \ & 1-\alpha+\alpha+\alpha=i=1.027 \ & 1+\alpha=1.027 \ & \alpha=0.027 	ext { or } 2.7 \% \end{aligned} $

The freezing point of a $1.00 \mathrm{~m}$ aqueous solution of $\mathrm{HF}$ is found to be $-1.91^{\circ}…

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