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Saturday, 31 October

JEE Main 2022 · ChemistryNumerical answerNumerical valueHardCalculation

JEE Main 29 July 2022, Shift 1, Chemistry Q81

Question 81 of 90 in this shift, Chemistry question 21 of 30, Section B.

Resistance of a conductivity cell (cell constant 129 m−1\mathrm{m^{-1}}) filled with 74.5 ppm solution of KCl is 100 Ω\Omega (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω\Omega (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 i.e. Λ1Λ2=x×10−3\frac{\Lambda_1}{\Lambda_2} = x\times10^{-3}. The value of xx is __________. (Nearest integer) Given, molar mass of KCl is 74.5 g mol−1\mathrm{mol^{-1}}.

Official answer

1000

NTA final key.

Chapter
Electrochemistry
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

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.