MHT CET · Chemistry · Electrochemistry
The conductivity of \(0.012 \mathrm{M} \mathrm{NaBr}\) solution is \(2.67 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}\). What is it's molar conductivity?
- A \(26.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(32.04 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(12.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \quad \mathrm{K}=2.67 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \\ & \mathrm{M}=0.012 \mathrm{M} \\ & \Lambda_{\mathrm{m}}=\frac{2.67 \times 10^{-4} \times 1000}{0.012} \\ & =22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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