Question

The resistance of a conductivity cell, filled with 0.1 mol L\(^{-1}\) KCl solution is 100 \(\Omega\). If the resistance of this cell is 500 \(\Omega\) on filling 0.02 mol L\(^{-1}\) KCl solution, then calculate the conductivity and molar conductivity of 0.02 mol L\(^{-1}\) KCl solution. The conductivity of 0.1 mol L\(^{-1}\) KCl solution is 1.29 S m\(^{-1}\).

Hint:

Always calculate cell constant from a standard solution first, then apply it to other resistances.

Answer 1

Step 1: Calculate cell constant.
Conductivity (\(\kappa\)) = Cell constant / Resistance.
For 0.1 M KCl: \[ 1.29 = \frac{\text{Cell constant}}{100} \quad \Rightarrow \quad \text{Cell constant} = 129 \, m^{-1} \] Step 2: Find conductivity for 0.02 M KCl.
\[ \kappa = \frac{\text{Cell constant}}{R} = \frac{129}{500} = 0.258 \, S \, m^{-1} \] Step 3: Find molar conductivity.
\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.258 \times 1000}{0.02 \times 1000} = 12.9 \, S \, cm^2 \, mol^{-1} \] Step 4: Conclusion.
Conductivity = \(\mathbf{0.258 \, S \, m^{-1}}\), Molar conductivity = \(\mathbf{12.9 \, S \, cm^2 \, mol^{-1}}\).