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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

The molar conductivity of 0.02 M electrolytic solution is 220\ ^ -1...

The molar conductivity of 0.02 M electrolytic solution is 220 Ω−1 cm2 mol−1220\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} at 25\ ^{}^{\circ}\text{C}. Find its conductivity.

Options

  1. A0.0044 Ω−1 cm−10.0044\ \Omega^{-1}\text{ cm}^{-1}
  2. B0.0022 Ω−1 cm−10.0022\ \Omega^{-1}\text{ cm}^{-1}
  3. C0.0011 Ω−1 cm−10.0011\ \Omega^{-1}\text{ cm}^{-1}
  4. D0.0033 Ω−1 cm−10.0033\ \Omega^{-1}\text{ cm}^{-1}

Correct answer

A. 0.0044 Ω−1 cm−10.0044\ \Omega^{-1}\text{ cm}^{-1}

Step-by-step solution

Given:

Molar conductivity, Λm=220 Ω−1 cm2 mol−1\Lambda_m = 220\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}

Concentration, C=0.02 MC = 0.02\text{ M}

The relation between molar conductivity and conductivity is given by:

Λm=κ×1000C\Lambda_m = \dfrac{\kappa \times 1000}{C}

Rearranging the formula to solve for conductivity κ\kappa:

κ=Λm×C1000\kappa = \dfrac{\Lambda_m \times C}{1000}

Substituting the given values:

κ=220×0.021000\kappa = \dfrac{220 \times 0.02}{1000}

κ=4.41000\kappa = \dfrac{4.4}{1000}

κ=0.0044 Ω−1 cm−1\kappa = 0.0044\ \Omega^{-1}\text{ cm}^{-1}

Answer: 0.0044 Ω−1 cm−10.0044\ \Omega^{-1}\text{ cm}^{-1}
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