MHT CET · Chemistry · Electrochemistry
The conductivity of \(0.20 \mathrm{M} \mathrm{KCl}\) solution at \(300 \mathrm{~K}\) is \(0.0248 \Omega^{-1} \mathrm{~cm}^{-1}\). What is its molar conductivity?
- A \(124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(93 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(62 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(186 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \Lambda_{\mathrm{m}} & =\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & =\frac{0.0248 \times 1000}{0.2} \\ & =124 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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