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Saturday, 31 October

JEE Main 2024 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 8 April 2024, Shift 1, Chemistry Q84

Question 84 of 90 in this shift, Chemistry question 24 of 30, Section B.

A solution containing 10 g of an electrolyte AB2\mathrm{AB_2} in 100 g of water boils at 100.52∘100.52^\circC. The degree of ionization of the electrolyte (α\alpha) is __________ × 10−1\times\ 10^{-1}. (nearest integer) [Given : Molar mass of AB2\mathrm{AB_2} = 200 g mol−1^{-1}, KbK_b (molal boiling point elevation const. of water) = 0.52 K kg mol−1^{-1}, boiling point of water = 100∘100^\circC ; AB2\mathrm{AB_2} ionises as AB2→A2++2B−\mathrm{AB_2 \rightarrow A^{2+} + 2B^-}]

Official answer

5

NTA final key.

Chapter
Solutions
Topic
Colligative Properties
Idea tested
Degree of Dissociation from Van't Hoff Factor

Same idea in other shifts

Asked 6× in all
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  5. 17 Mar 2021, Shift 2 · Q54A 1 molal K4Fe(CN)6\mathrm{K_4Fe(CN)_6} solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.