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MHT CET · Chemistry · Electrochemistry

The conductivity of \(0.20 \mathrm{M} \mathrm{KCl}\) solution at \(300 \mathrm{~K}\) is \(0.0248 \Omega^{-1} \mathrm{~cm}^{-1}\). What is its molar conductivity?

  1. A \(124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  2. B \(93 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  3. C \(62 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  4. D \(186 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} \Lambda_{\mathrm{m}} & =\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & =\frac{0.0248 \times 1000}{0.2} \\ & =124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)