If the molar conductivities (in S cm^2 mol^ | vedclass

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(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $KOH$ can be expressed as: $\lambda

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$274$

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(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $KOH$ can be expressed as: $\lambda^{\infty}_{KOH} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{OH^-}$ Given values: $\lambda^{\infty}_{NaCl} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$ $\lambda^{\infty}_{KCl} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$ $\lambda^{\infty}_{NaOH} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{OH^-} = 250 \ S \ cm^2 \ mol^{-1}$ To obtain $\lambda^{\infty}_{KOH}$,we perform the operation: $\lambda^{\infty}_{KOH} = \lambda^{\infty}_{KCl} + \lambda^{\infty}_{NaOH} - \lambda^{\infty}_{NaCl}$ $\lambda^{\infty}_{KOH} = 150 + 250 - 126$ $\lambda^{\infty}_{KOH} = 400 - 126 = 274 \ S \ cm^2 \ mol^{-1}$ Therefore,the correct option is $D$.

If the molar conductivities (in $S \ cm^2 \ mol^{-1}$) of $NaCl$,$KCl$,and $NaOH$ at infinite dilution are $126$,$150$,and $250$,respectively,the molar conductivity of $KOH$ (in $S \ cm^2 \ mol^{-1}$) is

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