JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 16th March Morning Shift
Question
AB2 is 10% dissociated in water to A2+ and B. The boiling point of a 10.0 molal aqueous solution of AB2 is __________C. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water Kb = 0.5 K kg mol1 boiling point of pure water = 100C]
[Given : Molal elevation constant of water Kb = 0.5 K kg mol1 boiling point of pure water = 100C]
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Show full solutionCorrect answer: 106
Correct answer
106
Step-by-step explanation
AB2 A+ + 2B
For AB2, n = 3
i = 1 + (n 1)
= 1 + (3 1) 0.1
= 1.2
Now, Tb = Kb (im)
Tb T = 1.2 0.5 10
Tb 100 = 6
Tb = 106
For AB2, n = 3
i = 1 + (n 1)
= 1 + (3 1) 0.1
= 1.2
Now, Tb = Kb (im)
Tb T = 1.2 0.5 10
Tb 100 = 6
Tb = 106
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This is a previous-year question from JEE Main 2021, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.