The limiting molar conductivities of the give | vedclass

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Question

(A) The limiting molar conductivities of the given electrolytes are calculated using Kohlrausch's law: $\lambda^{\circ}_{KCl} = \lambda^{\circ}_{K^{+}

Vedclass · Accepted Answer

$KCl < CaCl_{2} < K_{2}SO_{4}$

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(A) The limiting molar conductivities of the given electrolytes are calculated using Kohlrausch's law: $\lambda^{\circ}_{KCl} = \lambda^{\circ}_{K^{+}} + \lambda^{\circ}_{Cl^{-}} = 73.5 + 76.3 = 149.8 \ S \ cm^{2} \ mol^{-1}$ $\lambda^{\circ}_{CaCl_{2}} = \lambda^{\circ}_{Ca^{2+}} + 2\lambda^{\circ}_{Cl^{-}} = 119.0 + 2 	imes 76.3 = 271.6 \ S \ cm^{2} \ mol^{-1}$ $\lambda^{\circ}_{K_{2}SO_{4}} = 2\lambda^{\circ}_{K^{+}} + \lambda^{\circ}_{SO_{4}^{2-}} = 2 	imes 73.5 + 160.0 = 307.0 \ S \ cm^{2} \ mol^{-1}$ Comparing the values,the order is: $\lambda^{\circ}_{KCl} < \lambda^{\circ}_{CaCl_{2}} < \lambda^{\circ}_{K_{2}SO_{4}}$.

The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

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