MHTCET 14 May 2023 Shift 1 Solved Paper

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
‌=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.