$\mathrm{A} 0.0020 \mathrm{m}$ aqueous solution of an ionic compound $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\left({\mathrm{NO}}_2\right)\right]\mathrm{Cl}$  freezes at $-{0.00732}^{\circ }\mathrm{C}$. The number of moles of ions which one mole of ionic compound produces on being dissolved in water will be :

Depression in Freezing point is given by

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ \mathrm{where}, \\ \mathrm{i}=\mathrm{Vant} \mathrm{Hoff}\text{'}\mathrm{s} \mathrm{factor} \\ \mathrm{i}=1+\mathrm{\alpha }\left(\mathrm{n}-1\right) \mathrm{for} \mathrm{dissociation}, \\ {\mathrm{K}}_{\mathrm{f}}=\mathrm{Cryoscopic} \mathrm{constant} \\ \left(\mathrm{For} \mathrm{water}, \mathrm{value} \mathrm{of} {\mathrm{K}}_{\mathrm{f}}=1.86\right) \\ \mathrm{m}=\mathrm{molality} \mathrm{of} \mathrm{solution} \\ \mathrm{Here} ∆{\mathrm{T}}_{\mathrm{f}}=0-\left(-0.00732\right)=0.{00732}^{\circ }\mathrm{C} \\ \mathrm{i}=\frac{∆{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}}=\frac{0.00732}{1.86\times 0.0020}=1.86\simeq 2 \end{gathered}$$ 

So, this compound, $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\left({\mathrm{NO}}_2\right)\right]\mathrm{Cl}$ in solution will dissociate into ${\mathrm{Cl}}^{-}$ and ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\left({\mathrm{NO}}_2\right)\right]}^{+}$

$\mathrm{n}=2$

Therefore, two moles of ions are formed.