JEE Main Chemistry question
The equivalent conductance of 0.1 N acetic acid is 5 cm²ohm⁻¹eq⁻¹ at 298 K, while λ∞ is 390 cm²ohm⁻¹eq⁻¹. The degree of dissociation of 0.1 N acetic acid will be:
Answer: (B)
Solution
α=Λ/Λ∞=5/390=0.013.
The equivalent conductance of 0.1 N acetic acid is 5 cm²ohm⁻¹eq⁻¹ at 298 K, while λ∞ is 390 cm²ohm⁻¹eq⁻¹. The degree of dissociation of 0.1 N acetic acid will be:
α=Λ/Λ∞=5/390=0.013.