JEE Main 2017 — Solutions Question with Solution
From: JEE Main 2017 (Online) 9th April Morning Slot
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 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.325
Step-by-step explanation
Mole fractions can be calculated as
and
We know that
Mole fraction of CHCl3 in vapour form can be calculated as
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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.