A solution containing $0.85 \mathrm{~g}$ of $\mathrm{ZnCl}_2$ in $125.0 \mathrm{~g}$ of water freezes at $-0…

A solution containing $0.85 \mathrm{~g}$ of $\mathrm{ZnCl}_2$ in $125.0 \mathrm{~g}$ of water freezes at $-0.23^{\circ} \mathrm{C}$. The apparent degree of dissociation of the salt is $\left(K_f\right.$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, atomic mass: $\mathrm{Zn}=65.3$ and $\mathrm{Cl}=35.5)$
  1. $1.36 \%$
  2. $73.5 \%$
  3. $7.35 \%$
  4. $2.47 \%$

Solution

$ \text { Mol.wt. }=\frac{k_f \times w \times 1000}{\Delta T_f \times W}=\frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55 \mathrm{gm} $ Where $ \begin{aligned} & w=0.85 \mathrm{~g} \\ & W=125 \mathrm{~g} \\ & \Delta T_f=0^{\circ} \mathrm{C}-\left(-23^{\circ} \mathrm{C}\right)=23^{\circ} \mathrm{C} \end{aligned} $ Now, $i=\frac{M_{\text {normal }}}{M_{\text {observed }}}=\frac{136.3}{55}=2.47$ $ \begin{aligned} & \underset{1-\alpha}{\mathrm{ZnCl}_2} \leftrightharpoons \underset{\alpha}{\mathrm{Zn}^{++}}+\underset{2 \alpha}{2 \mathrm{Cl}^{-}} \\ & \text {Van't Hoff factor ( } i \text { ) } \\ & =\frac{1-\alpha+\alpha+2 \alpha}{1}=2.47 \\ & \therefore \quad \alpha=0.735=73.5 \% \\ & \end{aligned} $

Asked in: JEE Main 2012 (12 May Online)

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