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 rightarrow A 2+ +2 B - ]

Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 8 April, Shift 1 — Question 80

A solution containing 10 g of an electrolyte $\mathrm{AB}_{2}$ in 100 g of water boils at $100.52^{\circ} \mathrm{C}$ . The degree of ionization of the

electrolyte $(\alpha)$ is $\qquad$ $\times 10^{-1}$ . (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

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

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{AB}_{2} \rightarrow \mathrm{~A}^{+2}+2 \mathrm{~B}$

$\mathrm{i}=1+(3-1) \alpha$ $i=1+2 \alpha$ $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \mathrm{im}$

$0.52=0.52(1+2 \alpha) \frac{\frac{10}{200}}{\frac{100}{1000}}$ $1=(1+2 \alpha) \frac{10}{20}$ ]

$2=1+2 \alpha$ $\alpha=0.5$ Ans. $\alpha=5 \times 10^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Abnormal Colligative Properties - van't Hoff Factor