The limiting molar conductivities wedge ^0 fo | vedclass

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(C) According to Kohlrausch's law of independent migration of ions:$\wedge _{NaCl}^0 = \lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0 = 126 \ S \ cm^2 \ mo

Vedclass · Accepted Answer

$128$

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(C) According to Kohlrausch's law of independent migration of ions:$\wedge _{NaCl}^0 = \lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0 = 126 \ S \ cm^2 \ mol^{-1}$ $(1)$$\wedge _{KBr}^0 = \lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0 = 152 \ S \ cm^2 \ mol^{-1}$ $(2)$$\wedge _{KCl}^0 = \lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0 = 150 \ S \ cm^2 \ mol^{-1}$ $(3)$To find $\wedge _{NaBr}^0 = \lambda _{Na^{+}}^0 + \lambda _{Br^{-}}^0$,we perform the operation: $(1)$ + $(2)$ - $(3)$$\wedge _{NaBr}^0 = (\lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0) + (\lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0) - (\lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0)$$\wedge _{NaBr}^0 = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

The limiting molar conductivities $\wedge ^0$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge ^0$ for $NaBr$ is ............ $S \ cm^2 \ mol^{-1}$.

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