Specific conductance of 0.1 MCH3COOH at 25∘ is 3.9×10−4ohm−1cm−1. If λ∞(H3O+) and λ∞(VH3COO−) at 25∘C are 349.0 and 41.0 ohm−1cm2mol−1 respectively, degree of ionization of CH3COOH at the given concentration is :
1.0%
λm(CH3COOH)=1000k0.1=1000×3.9×10−40.1=3.9 ohm−1cm2mol−1αCH3COOH=λmλ∘m=3.9λ∘CH3COO−+λ∘H+=3.941+349=1.0×10−2=1%