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A $0.1$ molal aqueous solution of a weak acid is $30 \%$ ionised. If $K_{f}$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
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The correct answer is:
$-0.24^{\circ} \mathrm{C}$
Freezing point depression $(\Delta T)=i K_{f} m$ $\mathrm{H} \mathrm{A} \rightarrow \quad \mathrm{H}^{+}+\mathrm{A}-$
$$
\begin{array}{ll}
\mathrm{HA} \rightarrow & \mathrm{H}^{+} \\
1-\alpha & \alpha \\
1-0.3 & 0.3 \\
i=1-0.3+0.3+0.3 & \mathrm{~A}^{-} \\
i=1.3 & 0.3 \\
\therefore \Delta T_{f}=1.3 \times 1.86 \times 0.1=0.2418^{\circ} \mathrm{C} \\
T_{f}=0-0.2418^{\circ} \mathrm{C} \\
=-0.2418^{\circ} \mathrm{C}
\end{array}
$$
$$
\begin{array}{ll}
\mathrm{HA} \rightarrow & \mathrm{H}^{+} \\
1-\alpha & \alpha \\
1-0.3 & 0.3 \\
i=1-0.3+0.3+0.3 & \mathrm{~A}^{-} \\
i=1.3 & 0.3 \\
\therefore \Delta T_{f}=1.3 \times 1.86 \times 0.1=0.2418^{\circ} \mathrm{C} \\
T_{f}=0-0.2418^{\circ} \mathrm{C} \\
=-0.2418^{\circ} \mathrm{C}
\end{array}
$$
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