JEE Main 2015ChemistrySolutionsMediumMCQ

JEE Main 2015 — Solutions Question with Solution

JEE Main 2015 (04 Apr)

Question

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC , its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:

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Show full solutionCorrect option: C
Correct answer
C64

Step-by-step explanation

Δ P = 1 8 5 - 1 8 3 = 2 torr

MCH3-CO||-CH3=15×2+16+12=58 g/mol

ΔPP0=185-183185=2185=XB=1.2M1.2M+10058

1.2 M << 1 0 0 5 8

⇒   2 1 8 5 = 1.2 M × 5 8 1 0 0

M = 5 8 × 1.2 1 0 0 × 1 8 5 2

=64.38≈64 g/mol

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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.