According to Kohlrausch's law, ∧∘ for CH3COOH=λCH3COO−∘+λH+∘ ∧∘ for NaCl=λNa+∘+λCl−∘ =125mhocm2mol−1....(i) ∧∘ for HCl=λH+∘+λCl−∘ =425mhocm2mol−1...(2) ∧∘ for CH3COONa=λCH3COO−∘+λNa+∘ =90mhocm2mol−1 Adding Eqs. (ii) and (iii) and subtracting (i), we get
λH+∘+λCl−∘+λCH3COO−∘+λNa+∘−λNa+∘−λCl−∘
=425+90−125 =390mhocm2mol−1 or∧∘CH3COOH=λCH3COO−+λH+∘ =390mhocm2mol−1 Thus, the molar conductivity of CH3COOH at infinite dilution ∧∘=390mhocm2mol−1 The molar conductivity of 0.1MCH3COOH solution (∧mc) =7.8mhocm2mol−1 Degree of dissociation (α)=