JEE Main 2021 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 31 August 2021, Shift 1, Chemistry Q51

Question 51 of 90 in this shift, Chemistry question 21 of 30, Section B.

The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (

\mathrm{H_2C_2O_4\cdot 2H_2O}

) in 250 mL of water in mol

\mathrm{L^{-1}}

x\times10^{-2}

. The value of

x

is __________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0 ]

Official answer

NTA final key.

Chapter: Some Basic Concepts of Chemistry
Topic: Mole Concept
Idea tested: Molarity Calculation

Same idea in other shifts

Asked 16× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.