MHT CET · Chemistry · Electrochemistry
What is the molar conductivity of \(0.005 \mathrm{M}\) \(\mathrm{NaI}\) solution if its conductivity is \(6.065 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\) ?
- A \(121.3 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(115.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(126.5 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(131.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(121.3 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\wedge_{\mathrm{m}}=\frac{1000 \mathrm{k}}{\mathrm{c}}\)
\(=\frac{1000 \times 6.065 \times 10^{-4}}{0.005}=121.3 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
\(=\frac{1000 \times 6.065 \times 10^{-4}}{0.005}=121.3 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
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