JEE Main 2019ChemistrySolutionsHardMCQ

JEE Main 2019 — Solutions Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

Liquid M and liquid N form an ideal solution. The vapour pressures of pure liquids M and N are 450 and 700 mmHg, respectively, at the same temperature. Then correct statements is:
( xM=Mole fraction of 'M' in solution;
xN=Mole fraction of 'N'in solution;
yM=Mole fraction of 'M' in vapour phase;
yN=Mole fraction of 'N' in vapour phase; )

Choose an option

Show full solutionCorrect option: A
Correct answer
AxMxN>yMyN

Step-by-step explanation

PMo  is given as=450 mm of Hg.

PNo  is given as=700 mm of Hg.

We know that PM=xM×P MO→ Raoult’s law.

PN=xN×P NO→ Raoult’s law.

From Dalton’s law PM=yM×PT

PN=yM×PT

So,

xM×P Mo=yM×PT  …(i)

xM×PNo=yN×PT    …(ii)

(i) (ii) xMxN×P  MoP  No = yMyN

xMxN=yMyN×750450

xMxN>yMyN

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 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.