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A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which $1 \mathrm{~mol}$ of ionic compound produces on being dissolved in water will be $\left(\mathrm{k}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$
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The correct answer is:
2
Given,
$$
\begin{aligned}
\text { molality, } \mathrm{m} & =0.0020 \mathrm{~m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =0^{\circ} \mathrm{C}-0.00732^{\circ} \mathrm{C}=0.00732^{\circ} \mathrm{C} \\
\mathrm{k}_{\mathrm{f}} & =1.86^{\circ} \mathrm{C} / \mathrm{m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.0020} \\
& =1.96 \approx 2
\end{aligned}
$$
Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, $i$. Hence, 2 moles of ions are produced.
$$
\begin{aligned}
\text { molality, } \mathrm{m} & =0.0020 \mathrm{~m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =0^{\circ} \mathrm{C}-0.00732^{\circ} \mathrm{C}=0.00732^{\circ} \mathrm{C} \\
\mathrm{k}_{\mathrm{f}} & =1.86^{\circ} \mathrm{C} / \mathrm{m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.0020} \\
& =1.96 \approx 2
\end{aligned}
$$
Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, $i$. Hence, 2 moles of ions are produced.
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