JEE Main 2015ChemistrySolutionsMediumMCQ

JEE Main 2015Solutions Question with Solution

JEE Main 2015 (10 Apr Online)

Question

A solution at 20oC is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively:

Choose an option

Show full solutionCorrect option: C
Correct answer
C38.0 torr and 0.589

Step-by-step explanation

XBenzene =1.55=0.3

XToluene=3.55=0.7
According to Raoult's Law

P=xAP°A +xBP°B

Ptotal=0.3×74.7+0.7×22.3

=22.41+15.61=38.02

38 Torr

By Dalton's law to vapour phase

Mole fraction of a gas in a mixture of gas = Partial pressure of gas P 1 Total no.moles in the mixture

XBenzeneVapour phase=0.3×74.738=22.4138

=0.589

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About this question

This is a previous-year question from JEE Main 2015, 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.