The limiting molar conductivity Lambda^o for | vedclass

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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

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(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.$\Lambda^{\circ}_{NaCl} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$$\Lambda^{\circ}_{KBr} = \lambda^{\circ}_{K^+} + \lambda^{\circ}_{Br^-} = 152 \ S \ cm^2 \ mol^{-1}$$\Lambda^{\circ}_{KCl} = \lambda^{\circ}_{K^+} + \lambda^{\circ}_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$To find $\Lambda^{\circ}_{NaBr} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Br^-}$,we use the expression:$\Lambda^{\circ}_{NaBr} = \Lambda^{\circ}_{NaCl} + \Lambda^{\circ}_{KBr} - \Lambda^{\circ}_{KCl}$$\Lambda^{\circ}_{NaBr} = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

The limiting molar conductivity $\Lambda^o$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\Lambda^o$ for $NaBr$ is (in $S \ cm^2 \ mol^{-1}$).

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