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?
- A \(8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- B \(2.8 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- C \(4.1 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
- D \(5.4 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Lambda_{\mathrm{m}} =\frac{1000 \mathrm{k}}{\mathrm{c}} \)
\( \mathrm{k} =\frac{\Lambda_{\mathrm{m}} \times \mathrm{c}}{1000}=\) \(\frac{410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.02 \mathrm{~mol} \mathrm{dm}^{-3}}{1000 \mathrm{~cm}^3 \mathrm{dm}^{-3}} \)
\( =8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
\( \mathrm{k} =\frac{\Lambda_{\mathrm{m}} \times \mathrm{c}}{1000}=\) \(\frac{410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.02 \mathrm{~mol} \mathrm{dm}^{-3}}{1000 \mathrm{~cm}^3 \mathrm{dm}^{-3}} \)
\( =8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\)
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