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MHT CET · Chemistry · Electrochemistry

The conductivity of 0.005 M NaI solution at \(25^{\circ} \mathrm{C}\) is \(6.07 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\). Calculate its molar conductivity

  1. A \(121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  2. B \(110.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  3. C \(201.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
  4. D \(241.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\Lambda_{\mathrm{m}} =\frac{1000 \mathrm{k}}{\mathrm{c}}=\) \(\frac{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1} \times 6.07 \times 10^{-4} \Omega^{-1} \cdot \mathrm{~cm}^{-1}}{0.005 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\( =121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)