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$0.1 \mathrm{M}$ solution is present in a conductivity cell with electrode of $100 \mathrm{~cm}^2$ area placed $1 \mathrm{~cm}$ apart and resistance observed is $5 \times 10^3 \mathrm{ohm}$. What is the molar conductivity of this solution?
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$0.02 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Molar conductivity, $\Lambda_m=\frac{\kappa \times 1000}{M}$
$M=$ molarity and $\kappa=$ specific conductivity
$\begin{aligned} \kappa & =\frac{1}{R} \times \text { cell constant } \\ & =\frac{1}{R} \times \frac{\ell}{A}=\frac{1}{5 \times 10^3 \Omega} \times \frac{1 \mathrm{~cm}}{100 \mathrm{~cm}^2} \\ & =2 \times 10^{-6} \Omega^{-1} \mathrm{~cm}^{-1} \\ \therefore \quad \Lambda_m & =\frac{2 \times 10^{-6} \times 1000}{0.1}=0.02 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
$M=$ molarity and $\kappa=$ specific conductivity
$\begin{aligned} \kappa & =\frac{1}{R} \times \text { cell constant } \\ & =\frac{1}{R} \times \frac{\ell}{A}=\frac{1}{5 \times 10^3 \Omega} \times \frac{1 \mathrm{~cm}}{100 \mathrm{~cm}^2} \\ & =2 \times 10^{-6} \Omega^{-1} \mathrm{~cm}^{-1} \\ \therefore \quad \Lambda_m & =\frac{2 \times 10^{-6} \times 1000}{0.1}=0.02 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
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