Answer to Question #174728 in Organic Chemistry for Tuhin Subhra das

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

Cell constant,k= l/A= 0.5cm^-1

Molarity= 0.02M= 0.02mol/dm³= 0.02/1000 mol/cm³= 2.0x10^-5mol/cm³

R= 20.2

Specific conductance of water= 7.1x10^-6S/cm

Molar conductivity, ^= ?

Let's Firstly find the specific conductance of the cell solution.

K= 1/R x l/A

But l/A= k

K= 1/R x k

K= 1/20.2 x 0.5= 0.025

Total conductance= 0.025+7.1x10^-6= 0.025

^=K/M = 0.025/2x10^-5= 1250Scm²