The observed and normal molar masses of compound $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent…
Solution
Observed molar mass of $\mathrm{MX}_2=65.6 \mathrm{~g} \mathrm{~mol}^{-1}$
$\begin{aligned}
& \text { Van't Hoff factor }(\mathrm{i})=\frac{\text { Normal molar mass }}{\text { Observed molar mass }} \\
& =\frac{164}{65.6}=2.5
\end{aligned}$
If $\alpha$ is the degree of ionisation, then
$\begin{aligned}
& \underset{1-\alpha}{\mathrm{MX}_2} \rightleftharpoons \mathrm{M}_\alpha^{2+}+\underset{2 \alpha}{2 \mathrm{X}^{-}} \\
& \mathrm{i}=1-\alpha+\alpha+2 \alpha=1+2 \alpha \\
& 1+2 \alpha=2.5 \\
& \alpha=0.75
\end{aligned}$
$\therefore$ Percent degree of ionisation $=75 \%$
Asked in: JEE Main 2025 (24 Jan Shift 2)