MHT CET · Chemistry · Electrochemistry
The molar conductivity of \(0.4 \mathrm{M} \mathrm{KCl}\) solution is \(2.5 \times 10^5 \Omega^{-1}\) \(\mathrm{cm}^2 \mathrm{~mol}^{-1}\). What is the resistivity of solution?
- A \(2.1 \times 10^2\)
- B \(2.5 \times 10^2\)
- C \(1 \times 10^{-2}\)
- D \(2.8 \times 10^{-2}\)
Answer & Solution
Correct Answer
(C) \(1 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\
& 2.5 \times 10^5=\frac{\mathrm{k} \times 1000}{0.4} \\
& \text {Conductivity }(\mathrm{k})=\frac{1}{\rho(\text { resistivity })} \\
& =\frac{2.5 \times 10^5 \times 0.4}{1000} \\
& =100 \Omega^{-1} \mathrm{~cm}^{-1} \\
& \rho=\frac{1}{\mathrm{k}}=\frac{1}{100}=10^{-2} \Omega \mathrm{cm}
\end{aligned}
\)
\begin{aligned}
& \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\
& 2.5 \times 10^5=\frac{\mathrm{k} \times 1000}{0.4} \\
& \text {Conductivity }(\mathrm{k})=\frac{1}{\rho(\text { resistivity })} \\
& =\frac{2.5 \times 10^5 \times 0.4}{1000} \\
& =100 \Omega^{-1} \mathrm{~cm}^{-1} \\
& \rho=\frac{1}{\mathrm{k}}=\frac{1}{100}=10^{-2} \Omega \mathrm{cm}
\end{aligned}
\)
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