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
- A \(121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(110.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(201.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(241.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Lambda_{\mathrm{m}} =\frac{1000 \mathrm{k}}{\mathrm{c}}=\) \(\frac{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1} \times 6.07 \times 10^{-4} \Omega^{-1} \cdot \mathrm{~cm}^{-1}}{0.005 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\( =121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
\( =121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
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