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MHT CET 202624 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
24 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 24 April · Morning Shift · Chemistry · Electrochemistry

A 0.02 M KCl solution has conductivity k=2.76×10^ -3 \ S cm^ -1 ....

A 0.02 M KCl solution has conductivity k=2.76×10−3 S cm−1k=2.76\times10^{-3}\ \text{S cm}^{-1}. Find molar conductivity.

Options

  1. A130 S cm2mol−1130\ \text{S cm}^2\text{mol}^{-1}
  2. B138 S cm2mol−1138\ \text{S cm}^2\text{mol}^{-1}
  3. C148 S cm2mol−1148\ \text{S cm}^2\text{mol}^{-1}
  4. D150 S cm2mol−1150\ \text{S cm}^2\text{mol}^{-1}

Correct answer

B. 138 S cm2mol−1138\ \text{S cm}^2\text{mol}^{-1}

Step-by-step solution

Molar conductivity Λm\Lambda_m is given by the formula:

Λm=k×1000M\Lambda_m = \dfrac{k \times 1000}{M}

Given:
k=2.76×10−3 S cm−1k = 2.76 \times 10^{-3}\ \text{S cm}^{-1}
M=0.02 MM = 0.02\ \text{M}

Substituting the values:

Λm=2.76×10−3×10000.02\Lambda_m = \dfrac{2.76 \times 10^{-3} \times 1000}{0.02}

Λm=2.760.02\Lambda_m = \dfrac{2.76}{0.02}

Λm=138 S cm2mol−1\Lambda_m = 138\ \text{S cm}^2\text{mol}^{-1}

Answer: 138 S cm2mol−1138\ \text{S cm}^2\text{mol}^{-1}
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