Conductivity of 0.001 M aqueous solution of | vedclass

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(B) The molar conductivity $\Lambda_m$ is calculated as: $\Lambda_m = \frac{1000 	imes \kappa}{C} = \frac{1000 	imes 2.6 	imes 10^{-3}}{0.001} = 26

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) The molar conductivity $\Lambda_m$ is calculated as: $\Lambda_m = \frac{1000 	imes \kappa}{C} = \frac{1000 	imes 2.6 	imes 10^{-3}}{0.001} = 260 \ S \ cm^2 \ mol^{-1}$.According to Kohlrausch's law,$\Lambda_m^{\infty} = 2\lambda_m^{\infty}(Na^+) + \lambda_m^{\infty}(SO_4^{2-})$.Substituting the values: $260 = 2(50) + \lambda_m^{\infty}(SO_4^{2-})$.$260 = 100 + \lambda_m^{\infty}(SO_4^{2-})$.$\lambda_m^{\infty}(SO_4^{2-}) = 260 - 100 = 160 \ S \ cm^2 \ mol^{-1}$.

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