MHT CET · Chemistry · Electrochemistry
The conductivity of \(0.3 \mathrm{M}\) solution of \(\mathrm{KCl}\) at \(298 \mathrm{~K}\) is \(0.0627 \mathrm{~S} \mathrm{~cm}^1\). What is its molar conductivity?
- A \(104 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(188 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(209 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(109 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(209 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & \mathrm{k}=0.0627 \mathrm{~S} \mathrm{~cm}^1 \\ & \mathrm{M}=0.3 \\ & =\frac{0.0627 \times 1000}{0.3} \\ & =209 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}\)
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