Question

A KCl solution of conductivity 0.14 S m^–1 shows a resistance of 4.19 W in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 W. The conductivity of the HCl solution is ...... x 10^–2 S m–1. (Round off to the Nearest Integer).

Answer 1

k = \(\frac{1}{R}.G*\)

For same conductivity cell, G* is constant and hence k.R. = constant.

∴ 0.14 × 4.19 = k × 1.03

or, k of HCl solution = \(\frac{0.14\times4.19}{1.03}\)

= 0.5695 Sm^–1 

= 56.95 × 10^–2 Sm^–1

= 57 × 10^–2 Sm^–1