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

Correct Answer: 266

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 \(266\).