Conductivity of 2.5 × 10⁻⁴ M methanoic acid is 5.25 × 10⁻⁵ Scm⁻¹. Calculate its molar conductivity and degree of dissociation. (Given: λ⁰(H⁺) = 349.5 Scm²mol⁻¹ and λ⁰(HCOO⁻) = 50.5 Scm²mol⁻¹)

We know molar conductivity, (λm) = 1000 × conductivity(k) / concentration(c)
∴ λm = 1000 × 5.25 × 10⁻⁵ / 2.5 × 10⁻⁴ = 210 Scm²mol⁻¹
λ⁰CHCOOH = λ⁰H⁺ + λ⁰(HCOO⁻) = (349.5 + 50.5) = 400 Scm²mol⁻¹
∴ α = λm / λ⁰ = 210 / 400 = 0.52 or, α = 52.5%