JEE Main 2019 — Solutions Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -
Choose an option
Show full solutionCorrect option: C
Correct answer
C3A
Step-by-step explanation
For same freezing point,
(Tf)X = (Tf)Y
kf mx = kf my
M = 3.27A 3A
(Tf)X = (Tf)Y
kf mx = kf my
M = 3.27A 3A
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This is a previous-year question from JEE Main 2019, 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.