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NEET 2021 Chemistry Molar Conductivity MCQ Question

Type: MCQ-numerical-Medium-Class 12

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option. [Λ°_H⁺ = 350 S cm² mol⁻¹ Λ°_CH₃COO⁻ = 50 S cm² mol⁻¹]

A

1.75 × 10⁻⁴ mol L⁻¹

B

2.50 × 10⁻⁴ mol L⁻¹

C

1.75 × 10⁻⁵ mol L⁻¹

D

2.50 × 10⁻⁵ mol L⁻¹

Correct Answer

Option C

Detailed Explanation

The dissociation constant (K) can be calculated using the formula K = (Λ_m / Λ°)² × C, where Λ_m is the molar conductivity, Λ° is the limiting molar conductivity, and C is the concentration. Substituting the given values, we find K = 1.75 × 10⁻⁴ mol L⁻¹.

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