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The observed molar mass determined for $\mathrm{Na}_2 \mathrm{SO}_4$ by freezing point depression method is $50.0 \mathrm{~g} / \mathrm{mol}$. The molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4$ is 142 . What will be the degree of dissociation $\alpha$ for $\mathrm{Na}_2 \mathrm{SO}_4$ in water?
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The correct answer is:
0.92
van't Hoff factor, $\begin{aligned} i & =\frac{(\text { Molar mass })_{\text {calculated }}}{(\text { Molar mass })_{\text {observed }}} \\ & =\frac{142}{50}=2.84\end{aligned}$
$\mathrm{Na}_2 \mathrm{SO}_4$ is an electrolyte in water
$\Rightarrow \quad \alpha=\frac{i-1}{n-1}=\frac{2.84}{3-1}=0.92$
$\mathrm{Na}_2 \mathrm{SO}_4$ is an electrolyte in water
$\Rightarrow \quad \alpha=\frac{i-1}{n-1}=\frac{2.84}{3-1}=0.92$
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