question_answer 9)

The molar conductivity of 0.025 mol L?1 methanoic acid is 46.1 S cm2 mol?1. Calculate its degree of dissociation and dissociation constant. Given  and .  

Answer:

C = 0.025 mol L?1                              = 349.6 + 54.6 = 404.2 S cm2 mol?1 Degree of dissociation, Dissociation constant, = 0.0003667 = 3.67 × 10?4.