The molar conductivity of 0.4 M KCl solutio | vedclass

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Question

(A) The formula for molar conductivity is $\Lambda_m = \frac{k 	imes 1000}{M}$.Given $\Lambda_m = 2.5 	imes 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ an

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$10 \ \Omega \ cm$

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(A) The formula for molar conductivity is $\Lambda_m = \frac{k 	imes 1000}{M}$.Given $\Lambda_m = 2.5 	imes 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $M = 0.4 \ M$.Substituting the values: $2.5 	imes 10^2 = \frac{k 	imes 1000}{0.4}$.$k = \frac{2.5 	imes 10^2 	imes 0.4}{1000} = \frac{100}{1000} = 0.1 \ \Omega^{-1} \ cm^{-1}$.Resistivity $ho$ is the reciprocal of conductivity $k$: $ho = \frac{1}{k} = \frac{1}{0.1} = 10 \ \Omega \ cm$.

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

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