According to Kohlrausch law, i. ∧0(Ba(OH)2)=λBa2+0+2λOH−0 ii. ∧0(BaCl2)=λBa2+0+2λCl−0 iii. ∧0(NH4‌Cl)=λNH4+0+λCl−0 Eq. (i) +‌
1
2
‌Eq (ii) −‌
1
2
‌Eq (iii) gives
∧0(NH4‌OH)‌=∧0(NH4‌Cl)+‌
1
2
∧0(Ba(OH)2)−‌
1
2
∧0(BaCl2) ∧0(NH4‌OH)‌=129+‌
1
2
520−‌
1
2
280
‌=249.0Ω−1cm2mol−1 When using Kohlrausch law, remember to multiply molar ionic conductivities of cation and anion with the number of cations and anions, respectively, as in the chemical formula of the compound.