A solution containing $10 \mathrm{~g}$ of an electrolyte $\mathrm{AB}_2$ in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is _______ $\times 10^{-1}$. (nearest integer) [Given : Molar mass of \(\mathrm{AB}_2=200…

Question

A solution containing $10 \mathrm{~g}$ of an electrolyte $\mathrm{AB}_2$ in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is _______ $\times 10^{-1}$. (nearest integer) [Given : Molar mass of $\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}$ (molal boiling point elevation const. of water) $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$ ionises as $\left.\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}\right]$

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{AB}_2 ightarrow \mathrm{A}^{+2}+2 \mathrm{~B}^{\ominus} \ & \mathrm{i}=1+(3-1) \alpha \ & \mathrm{i}=1+2 \alpha \ & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \mathrm{im} \ & 0.52=0.52(1+2 \alpha) \frac{\frac{10}{200}}{\frac{100}{1000}} \ & 1=(1+2 \alpha) \frac{10}{20} \ & 2=1+2 \alpha \ & \alpha=0.5\end{aligned}$ Ans. $\alpha=5 	imes 10^{-1}$

A solution containing \(10 \mathrm{~g}\) of an electrolyte \(\mathrm{AB}_2\) in \(100 \mathrm{~g}\) of water…

Solution