JEE Main 2020ChemistrySolutionsMediumMCQ

JEE Main 2020 — Solutions Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile 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]

Choose an option

Show full solutionCorrect option: C
Correct answer
CA>B>C

Step-by-step explanation

Number of moles = wtgmMW=10MW

Relative lowering in vapour pressure depends on no. of mole of solute greater the no. of moleof solute greater in RLVP and smaller will be vapour pressure.

So order of vapour pressure is A>B>C.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Solutions chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

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.