JEE Main 2018 — Solutions Question with Solution
From: JEE Main 2018 (Online) 16th April Morning Slot
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 to 75%, is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C150 g
Step-by-step explanation
Molar mass of octane
(C8 H18) = 8 12 + 18 = 114 g/mol
Let, is the mass of solute.
Relative lowering vapour pressure,
=
=
= 150 g
(C8 H18) = 8 12 + 18 = 114 g/mol
Let, is the mass of solute.
Relative lowering vapour pressure,
=
=
= 150 g
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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.