Calculate molar conductivity at infinite dilu | vedclass

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Question

(A) According to Kohlrausch's law of independent migration of ions: $\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$ $\wedge^0_{NaCl} = \lambda

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

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(A) According to Kohlrausch's law of independent migration of ions: $\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$ $\wedge^0_{NaCl} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ $\wedge^0_{KBr} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ $\wedge^0_{KCl} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ To obtain $\wedge^0_{NaBr}$,we perform the operation: $\wedge^0_{NaBr} = \wedge^0_{NaCl} + \wedge^0_{KBr} - \wedge^0_{KCl}$ $\wedge^0_{NaBr} = 126 + 152 - 150 = 128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

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