JEE Main 2021ChemistrySolutionsMediumNumerical

JEE Main 2021 — Solutions Question with Solution

JEE Main 2021 (31 Aug Shift 1)

Question

The molarity of the solution prepared by dissolving 6.3 g of oxalic acid H2C2O4·2H2O in 250 mL of water in mol L-1 is x×10-2. The value of x is _________ . (Nearest integer)
[Atomic mass : H : 1.0, C : 12.0, 0 : 16.0 J]

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Show full solutionCorrect answer: 20
Correct answer
20

Step-by-step explanation

M=Wsolute×1000GMMsolute×Vml

=6.3×1000126×250=0.2 M
=20×10-2
So X=20

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About this question

This is a previous-year question from JEE Main 2021, 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.