The following limiting molar conductivities a | vedclass

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Question

(D) According to Kohlrausch's law,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent

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$\frac{x-y}{2} + z$

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(D) According to Kohlrausch's law,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.$1. \lambda_{m(H_{2}SO_{4})}^{0} = 2\lambda_{H^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = x$$2. \lambda_{m(K_{2}SO_{4})}^{0} = 2\lambda_{K^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = y$$3. \lambda_{m(CH_{3}COOK)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{K^{+}}^{0} = z$We need to find $\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{H^{+}}^{0}$.From the given equations:$\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{m(CH_{3}COOK)}^{0} + \frac{1}{2}\lambda_{m(H_{2}SO_{4})}^{0} - \frac{1}{2}\lambda_{m(K_{2}SO_{4})}^{0}$Substituting the values:$\lambda_{m(CH_{3}COOH)}^{0} = z + \frac{x}{2} - \frac{y}{2} = \frac{x-y}{2} + z \ S \ cm^{2} mol^{-1}$.

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