The molar conductivities (lambda_{m}^0) at in | vedclass

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(B) According to Kohlrausch's law of independent migration of ions:$\lambda_{m}^0(KBr) = \lambda_{K^+} + \lambda_{Br^-} = 120.5 \ S \ cm^2 \ mol^{-1}$

Vedclass · Accepted Answer

$390.5 \ S \ cm^2 \ mol^{-1}$

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(B) According to Kohlrausch's law of independent migration of ions:$\lambda_{m}^0(KBr) = \lambda_{K^+} + \lambda_{Br^-} = 120.5 \ S \ cm^2 \ mol^{-1}$$\lambda_{m}^0(HBr) = \lambda_{H^+} + \lambda_{Br^-} = 420.6 \ S \ cm^2 \ mol^{-1}$$\lambda_{m}^0(KNH_2) = \lambda_{K^+} + \lambda_{NH_2^-} = 90.48 \ S \ cm^2 \ mol^{-1}$We need to find $\lambda_{m}^0(NH_3) = \lambda_{H^+} + \lambda_{NH_2^-}$.Using the given values:$\lambda_{m}^0(NH_3) = \lambda_{m}^0(HBr) + \lambda_{m}^0(KNH_2) - \lambda_{m}^0(KBr)$$\lambda_{m}^0(NH_3) = 420.6 + 90.48 - 120.5 = 390.58 \ S \ cm^2 \ mol^{-1}$.Rounding to the nearest provided option,the value is $390.5 \ S \ cm^2 \ mol^{-1}$.

The molar conductivities $(\lambda_{m}^0)$ at infinite dilution of $KBr$,$HBr$,and $KNH_2$ are $120.5$,$420.6$,and $90.48 \ S \ cm^2 \ mol^{-1}$ respectively. Find the value of $\lambda_{m}^0$ for $NH_3$.

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