A $0.0020 \mathrm{M}$ aqueous solution of an ionic compound…
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 mole 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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Solution
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \cdot \mathrm{m}$
$\begin{aligned}
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.002}=\frac{0.00732}{0.00372} \\
\mathrm{i} & =2
\end{aligned}$
Compound will be $\left.\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right] \mathrm{NO}_5 \mathrm{NO}_2\right] \mathrm{Cl}$
Total possible ions $=2$