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When $20 \mathrm{~g}$ of naphthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ is dissolved in $50 \mathrm{~g}$ of benzene, a freezing point depression of $2 \mathrm{~K}$ is observed. The vant Hoff factor (i) is $\left[\mathrm{K}_{\mathrm{f}}=1.72^{\circ} \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right]$
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The correct answer is:
$0.5$
$\begin{array}{l}
\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\
2=\mathrm{i} \times 1.72 \times \frac{20 \times 1000}{172 \times 50} \\
\therefore \mathrm{i}=\frac{2}{0.01 \times 2 \times 200}=\frac{1}{2}=0.5
\end{array}$
\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\
2=\mathrm{i} \times 1.72 \times \frac{20 \times 1000}{172 \times 50} \\
\therefore \mathrm{i}=\frac{2}{0.01 \times 2 \times 200}=\frac{1}{2}=0.5
\end{array}$
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