Answer to Question #169498 in Physical Chemistry for Vk

Question

A conductivity cell having a cell constant of 0.5 cm –1 filled with 0.02 M solution of KCl at  298 K gave a resistance of 20.2 Ω. The water used for preparing the solution had a conductivity  of 7.1 × 10–6 S cm–1. Calculate the molar conductivity of 0.02 M KCl solution.

Answer 1

lmbda = k*1000/N

K = l/R*a

k = 0.5/20.2 = 0.025ohm^-1^cm^-1

lmbda = 0.025*1000/0.02 = 1250