STANDARD Chemistry Molar Conductivity MCQ Question
Molar conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹, the degree of dissociation and dissociation constant will be (Given : λ°_H⁺ = 349.6 S cm² mol⁻¹ and λ°_HCOO⁻ = 54.6 S cm² mol⁻¹)
11.4%, 3.67 × 10⁻⁴ mol L⁻¹
22.8%, 1.83 × 10⁻⁴ mol L⁻¹
52.2%, 4.25 × 10⁻⁴ mol L⁻¹
1.14%, 3.67 × 10⁻⁶ mol L⁻¹
Correct Answer
Detailed Explanation
The degree of dissociation (α) is calculated using the formula α = Λ_m / Λ°_m. The dissociation constant (K_a) is calculated using K_a = Cα² / (1-α).
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