JEE Main 2019ChemistryElectrochemistryHardMCQ

JEE Main 2019 — Electrochemistry Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

∧mo for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2mol-1 respectively. If the conductivity of 0.001 M HA is 5×10-5S cm-1, degree of dissociation of HA is

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Show full solutionCorrect option: A
Correct answer
A0.125

Step-by-step explanation

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.

λmoHA=λmoHCl+λmoNaA-λoNaCl

=425.9+100.5-126.4

=400

λmo=K×1000M=5×10-5×10310-3=50

α=50400=0.125

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About this question

This is a previous-year question from JEE Main 2019, covering the Electrochemistry 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.