NEET Chemistry Conductance MCQ Question
Λₘ⁰ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm² mol⁻¹, respectively. If the conductivity of 0.001 M HA is 5 × 10⁻⁵ S cm⁻¹, degree of dissociation of HA is
0.25
0.50
0.75
0.125
Correct Answer
Detailed Explanation
The degree of dissociation (α) can be calculated using the formula α = Λ/Λₘ⁰. Given Λ = 5 × 10⁻⁵ S cm⁻¹ and Λₘ⁰ = 126.4 S cm² mol⁻¹, α = (5 × 10⁻⁵) / (126.4 × 10⁻²) = 0.25.
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