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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 $\mathrm{Cl}=35.5)$
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Verified Answer
The correct answer is:
$73.5 \%$
$73.5 \%$
$$
\text { Mol.wt. }=\frac{k_f \times w \times 1000}{\Delta T_f \times W}=\frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55 \mathrm{gm}
$$
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, $i=\frac{M_{\text {normal }}}{M_{\text {observed }}}=\frac{136.3}{55}=2.47$
$$
\begin{aligned}
& \underset{1-\alpha}{\mathrm{ZnCl}_2} \leftrightharpoons \underset{\alpha}{\mathrm{Zn}^{++}}+\underset{2 \alpha}{2 \mathrm{Cl}^{-}} \\
& \text {Van't Hoff factor ( } i \text { ) } \\
& =\frac{1-\alpha+\alpha+2 \alpha}{1}=2.47 \\
& \therefore \quad \alpha=0.735=73.5 \% \\
&
\end{aligned}
$$
\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}
$$
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