JEE Main 2017ChemistrySolutionsMediumMCQ

JEE Main 2017 — Solutions Question with Solution

JEE Main 2017 (09 Apr Online)

Question

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is: Molar mass of Cl=35.5 g mol-1

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Show full solutionCorrect option: C
Correct answer
C0.325

Step-by-step explanation

mole of CH2Cl2 in liquid phase=8.585=0.1

mole of CHCl3 in liquid phase=11.95119.5=0.1

mole fraction of CH2Cl2 in liquid phase=0.10.2=12

mole fraction of CHCl3 in liquid phase=0.10.2=12

PT=XCH2Cl2×  vapour pressureCH2Cl2+ XCHCl3× vapour pressureCHCl3

=415× 1 2 +200× 1 2 =307.5

P T ( X A ) VP = ( X A ) LP ×   Vapour pressure of CHCl 3

307.5×  ( X CHC l 3 ) VP =200× 1 2

XCHCl3=100307.5=0.325

 

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About this question

This is a previous-year question from JEE Main 2017, 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.