Molar ionic conductivities of divalent cation and anion are 57 S cm 2 mol -1 and 73 S cm 2 mol -1 respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :

Chemistry · Electrochemistry

JEE Main 2024 — 5 April, Shift 1 — Question 75

Molar ionic conductivities of divalent cation and anion are $57 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $73 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :

Option A:
$65 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Option B:
$130 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Correct
Option C:
$187 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Option D:
$260 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Answer: B

Step-by-step solution

$\quad \Lambda_{\mathrm{C}}^{+2}=57 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$ $\Lambda_{\mathrm{A}}^{+2}=73 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\Lambda_{\text {Solution }}=\lambda_{\mathrm{C}}^{+2}+\Lambda_{\mathrm{A}}^{-2}$

$=57+73=130$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law

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Answer: B

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