Answer to Question #156067 in Organic Chemistry for Akshat Kotnala

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

Solution.

$\lambda = \frac{\kappa \times 1000}{N}$

$\kappa = \frac{l}{R \times a}$

$\kappa = \frac{0.5}{20.2} = 0.025 \ ohm^{-1} \ cm^{-1}$

$\lambda = \frac{0.025 \times 1000}{0.02} = 1250 \ ohm^{-1} \ cm^2 \ (g-eq)^{-1}$

Answer:

$\lambda = 1250 \ ohm^{-1} \ cm^2 \ (g-eq)^{-1}$