A solution containing 10 ~g of an electrolyte AB_2 in 100 ~g of water boils at 100.52^° C . The degree of ionization of the electrolyte (α) is _______ × 10^-1 . (nearest integer) [Given : Molar mass of AB_2=200 ~g ~mol^-1, ~K_b (molal boiling point elevation const. of water) =0.52 ~K ~kg ~mol^-1 , boiling point of water =100^° C ; AB_2 ionises as .AB_2 arrow A^2++2 ~B^- ]

Correct answer: 5. AB_2 arrow A^+2+2 ~B^ominus i=1+(3-1) α i=1+2 α Δ T_b=k_b im 0.52=0.52(1+2 α) (10/200)/(100/1000) 1=(1+2 α) 10/20 2=1+2 α α=0.5 Ans. α=5 × 10^-1

Q21 / 90 · JEE Main 2024 (08 Apr Shift 1) · Chemistry

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 $\left.\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}\right]$

Correct answer: 5

Solution

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

Hence, the correct answer is 5.

Solution

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