JEE Main 2020 — Solutions Question with Solution
From: JEE Main 2020 (Online) 6th September Evening Slot
Question
A set of solutions is prepared using 180 g of
water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative
lowering of vapour pressure in the presence of
these solutes are in the order :
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
Choose an option
Show full solutionCorrect option: C
Correct answer
CA > B > C
Step-by-step explanation
Relative lowering in vapour pressure (RLVP)
=
n moles of solute
N moles of solvent
(RLVP)A =
(RLVP)B =
and (RLVP)C =
(RLVP)A > (RLVP)B > (RLVP)C
So, A > B > C
=
n moles of solute
N moles of solvent
(RLVP)A =
(RLVP)B =
and (RLVP)C =
(RLVP)A > (RLVP)B > (RLVP)C
So, A > B > C
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This is a previous-year question from JEE Main 2020, 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.