NEETChemistry-Electrochemistry

NEET Chemistry Conductance MCQ Question

Type: MCQ-numerical-Medium-Class 12

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

A

0.25

B

0.50

C

0.75

D

0.125

Correct Answer

Option A

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.

Found an issue with this question?