Question

 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

K = Conductivity = G × l / A .

K = 1/ R × l / A

= 1/ 20.2 × 0.5 = 0.0247

Total Conductivity = 0.0247 + 7.1 × 10^-6

= 0.0247 ( approx ).

Molar conductivity = 1000 × K / C

= 1000 × 0.0247 / 0.02 = 1238 .