Λ∞NH4Cl=Λ∞NH4++Λ∞Cl−=130 .......(i) Λ∞NaOH=Λ∞Na++Λ∞OH−=217 ........(ii) Λ∞NaCl=Λ∞Na++Λ∞Cl−=109 ......(iii) On adding Eq. (i) and (ii) and subtracting the Eq. (iii) Λ∞NH4++Λ∞OH−=130+217−109 Λ∞NH4OH=347−109 =238ohm−1cm2equiv−1