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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?
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Verified Answer
The correct answer is:
$1 \times 10^{-2}$
$$
\begin{aligned}
& \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\
& 2.5 \times 10^5=\frac{\mathrm{k} \times 1000}{0.4} \\
& \text { Conductivity }(\mathrm{k})=\frac{1}{\rho(\text { resistivity })} \\
& =\frac{2.5 \times 10^5 \times 0.4}{1000} \\
& =100 \Omega^{-1} \mathrm{~cm}^{-1} \\
& \rho=\frac{1}{\mathrm{k}}=\frac{1}{100}=10^{-2} \Omega \mathrm{cm}
\end{aligned}
$$
\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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