Molar ionic conductivities of a divalent cati | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the mol

Vedclass · Accepted Answer

$130 \ S \ cm^2 \ mol^{-1}$

Vedclass · Answer

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar ionic conductivities of its constituent ions.For an electrolyte $CA$ where $C$ is a divalent cation $(C^{2+})$ and $A$ is a divalent anion $(A^{2-})$,the molar conductivity is given by:$\Lambda_m^\circ = \lambda_+ + \lambda_-$Given:$\lambda_{C^{2+}} = 57 \ S \ cm^2 \ mol^{-1}$$\lambda_{A^{2-}} = 73 \ S \ cm^2 \ mol^{-1}$Therefore:$\Lambda_m^\circ = 57 + 73 = 130 \ S \ cm^2 \ mol^{-1}$

Molar ionic conductivities of a divalent cation and a divalent anion are $57 \ S \ cm^2 \ mol^{-1}$ and $73 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of a solution of an electrolyte containing these ions will be:

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