The limiting molar conductivities (Lambda_0) | vedclass

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(A) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar

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$128 \ S \ cm^2 \ mol^{-1}$

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(A) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.$\Lambda_{0(NaBr)} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$$\Lambda_{0(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$$\Lambda_{0(KBr)} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ S \ cm^2 \ mol^{-1}$$\Lambda_{0(KCl)} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$To find $\Lambda_{0(NaBr)}$,we perform the operation:$\Lambda_{0(NaBr)} = \Lambda_{0(NaCl)} + \Lambda_{0(KBr)} - \Lambda_{0(KCl)}$$\Lambda_{0(NaBr)} = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

The limiting molar conductivities $(\Lambda_0)$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. What is the $\Lambda_0$ of $NaBr$?

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