ExamBro
ExamBro
MHT CET · Chemistry · Electrochemistry

The molar conductivity of 0.02 M KCl solution is \(410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Calculate its conductivity?

  1. A \(8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
  2. B \(2.8 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
  3. C \(4.1 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
  4. D \(5.4 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\Lambda_{\mathrm{m}} =\frac{1000 \mathrm{k}}{\mathrm{c}} \)
\( \mathrm{k} =\frac{\Lambda_{\mathrm{m}} \times \mathrm{c}}{1000}=\) \(\frac{410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.02 \mathrm{~mol} \mathrm{dm}^{-3}}{1000 \mathrm{~cm}^3 \mathrm{dm}^{-3}} \)
\( =8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)