Question

The resistance of a conductivity cell filled with 0.1 mol L\(^{-1}\) KCl solution is 100 \( \Omega \). If the resistance of the same cell when filled with 0.02 mol L\(^{-1}\) KCl solution is 520 \( \Omega \), find the conductivity and molar conductivity of the 0.02 mol L\(^{-1}\) KCl solution. The conductivity of a 0.1 mol L\(^{-1}\) KCl solution is 1.29 S/m.

Hint:

Find the cell constant \( \kappa R \) from the 0.1 M data, reuse it for the 0.02 M cell, then \( \Lambda_m=\kappa/c \).

Answer 1

Step 1 (Cell constant): Cell constant \( G^{*} = \kappa \times R \). Using the 0.1 mol L\(^{-1}\) data: \[ G^{*} = 1.29\ \text{S m}^{-1} \times 100\ \Omega = 129\ \text{m}^{-1} \]
Step 2 (Conductivity of 0.02 mol L\(^{-1}\)): \( \kappa = \dfrac{G^{*}}{R} = \dfrac{129}{520} \).
Step 3 (Arithmetic): \[ \kappa = 0.248\ \text{S m}^{-1} \]
Step 4 (Molar conductivity formula): \[ \Lambda_m = \frac{\kappa \times 1000}{c} \] with \( \kappa \) in S cm\(^{-1}\) and \( c \) in mol L\(^{-1}\). Convert \( \kappa = 0.248\ \text{S m}^{-1} = 0.00248\ \text{S cm}^{-1} \).
Step 5 (Substitution): \[ \Lambda_m = \frac{0.00248 \times 1000}{0.02} = \frac{2.48}{0.02} \]
Step 6: \[ \boxed{\kappa = 0.248\ \text{S m}^{-1}, \quad \Lambda_m = 124\ \text{S cm}^{2}\,\text{mol}^{-1}} \]