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Friday, 9 October

JEE Main 2021 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 3 August 2021, Shift 2, Chemistry Q51

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

Chlorine is prepared according to the following equation : 4HCl+MnO2(s)⟶2H2O(l)+MnCl2(aq)+Cl2(g)\mathrm{4HCl + MnO_{2(s)} \longrightarrow 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}}. 10g sample of MnO2\mathrm{MnO_2} produces 2.24L of chlorine under SATP, the percentage purity of the MnO2\mathrm{MnO_2} sample is ‾\underline{\hspace{2em}}. (Nearest integer) [Atomic weight = H : 1.0, O : 16.0, C : 12.0, Cl : 35.5, Mn : 55.0] [SATP : T = 298K, P = 10510^5 Pa]

Official answer

87

NTA final key.

Chapter
Some Basic Concepts of Chemistry
Topic
Mole Concept
Idea tested
Percentage Purity of Samples

Same topic in other shifts

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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 42 of the 174 JEE Main shifts held from 2017 to 2026.