JEE Main 2018ChemistrySolutionsHardMCQ

JEE Main 2018 — Solutions Question with Solution

JEE Main 2018 (16 Apr Online)

Question

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75 %, is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C150 g

Step-by-step explanation

Molar mass of octane = 114 g/mol.

From the lowering of vapour pressure, we have,

∆PP=W2M2W2M2+W1M1

Where W2 and M2 are mass and molar mass of solute
and W1 and M1 are mass and molar mass of octane.

75100=W250 g/molW250 g/mol+114 g114 g/mol 

0.75=W250W250+1

W250+1=W250×0.75

W2=150g

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About this question

This is a previous-year question from JEE Main 2018, 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.