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