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(A) According to Kohlrausch's law of independent migration of ions:$\wedge_m^0(NH_4OH) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$We are given:$1. \wedge

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$249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

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(A) According to Kohlrausch's law of independent migration of ions:$\wedge_m^0(NH_4OH) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$We are given:$1. \wedge_m^0(Ba(OH)_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{OH^-}^0 = 520 \ \Omega^{-1} \ cm^2 \ mol^{-1}$$2. \wedge_m^0(BaCl_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{Cl^-}^0 = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$$3. \wedge_m^0(NH_4Cl) = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0 = 129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$To get $\wedge_m^0(NH_4OH)$,we perform the operation: $\frac{1}{2} \wedge_m^0(Ba(OH)_2) + \wedge_m^0(NH_4Cl) - \frac{1}{2} \wedge_m^0(BaCl_2)$$= \frac{1}{2}(520) + 129 - \frac{1}{2}(280)$$= 260 + 129 - 140$$= 249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

Calculate the molar conductivity of $NH_4OH$ at infinite dilution if the molar conductivities of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ at infinite dilution are $520$,$280$,and $129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

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