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If resistivity of $0.8 \mathrm{M} \mathrm{KCl}$ solution is $2 \cdot 5 \times 10^{-3} \Omega \mathrm{cm} .$ Calculate molar conductivity of solution?
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The correct answer is:
$5 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
(A)
$\begin{array}{l}
\text { Conductivity }(\mathrm{k})=\frac{1}{\text { resistivity }}=\frac{1}{2.5 \times 10^{-3} \Omega \mathrm{cm}}=0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1} \\
\text { Molar conductivity } \quad\left(\Lambda_{\mathrm{m}}\right)=\frac{1000 \mathrm{k}}{\mathrm{C}}=\frac{1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1} \times 0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1}}{0.8 \mathrm{~mol} \mathrm{~L}^{-1}}
\end{array}$
$\begin{array}{l}
\text { Conductivity }(\mathrm{k})=\frac{1}{\text { resistivity }}=\frac{1}{2.5 \times 10^{-3} \Omega \mathrm{cm}}=0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1} \\
\text { Molar conductivity } \quad\left(\Lambda_{\mathrm{m}}\right)=\frac{1000 \mathrm{k}}{\mathrm{C}}=\frac{1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1} \times 0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1}}{0.8 \mathrm{~mol} \mathrm{~L}^{-1}}
\end{array}$
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