The conductivity of $0.20 \, M$ solution of $KCl$ at $298 \, K$ is $0.0248 \, S \, cm^{-1}$. Calculate its molar conductivity.
$(124 S CM^2 MOL^-1)$ Given,conductivity $\kappa = 0.0248 \, S \, cm^{-1}$.
Concentration $c = 0.20 \, M$.
Formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{c}$.
Substituting the values: $\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$.
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \, S \, cm^{2} \, mol^{-1}$.
Concentration $c = 0.20 \, M$.
Formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{c}$.
Substituting the values: $\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$.
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \, S \, cm^{2} \, mol^{-1}$.
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