Question

The resistance of a conductivity cell containing $0.01$ M KCl solution at $298\ K$ is $1750\ Ω$. If the conductivity of $0.01$ M KCl solution at $298\ K$ is $0.152×10^{–3}$ S cm–1, then the cell constant of the conductivity cell is ______ $×10^{–3}$ cm–1.

Answer 1

Molarity of $KCl$ solution $= 0.1$ M
Resistance $=$ $1750\ Ω$
Conductivity $= 0.152×10^{–3}$ S cm–1
Conductivity $= \frac {\text {Cell\ constant}}{\text{Resistance}}$
∴ Cell constant $= 0.152×10^{–3}×1750$
$= 266×10^{–3}$ cm–1

So, the answer is $266$.