Calculate the molar conductivity of $NH_4OH$ at infinite dilution by using the following data:
$[\Lambda _m^o(NH_4Cl) = 129.8, \Lambda _m^o(KOH) = 248.0$ and $\Lambda _m^o(KCl) = 126 \ S \ cm^2 \ mol^{-1}]$

According to Kohlrausch's law of independent migration of ions:
$NH_4OH \rightarrow NH_{4(aq)}^{+} + OH_{(aq)}^{-}$
$\Lambda_{m}^{\circ}(NH_4OH) = \lambda_{m}^{\circ}(NH_{4}^{+}) + \lambda_{m}^{\circ}(OH^{-})$
Given data:
$(I) \Lambda_{m}^{\circ}(NH_4Cl) = \lambda_{m}^{\circ}(NH_{4}^{+}) + \lambda_{m}^{\circ}(Cl^{-}) = 129.8 \ S \ cm^2 \ mol^{-1}$
$(II) \Lambda_{m}^{\circ}(KOH) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(OH^{-}) = 248.0 \ S \ cm^2 \ mol^{-1}$
$(III) \Lambda_{m}^{\circ}(KCl) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(Cl^{-}) = 126.0 \ S \ cm^2 \ mol^{-1}$
To obtain $\Lambda_{m}^{\circ}(NH_4OH)$,we perform the operation $(I) + (II) - (III)$:
$\Lambda_{m}^{\circ}(NH_4OH) = \Lambda_{m}^{\circ}(NH_4Cl) + \Lambda_{m}^{\circ}(KOH) - \Lambda_{m}^{\circ}(KCl)$
$\Lambda_{m}^{\circ}(NH_4OH) = 129.8 + 248.0 - 126.0$
$\Lambda_{m}^{\circ}(NH_4OH) = 251.8 \ S \ cm^2 \ mol^{-1}$