Question

The electrical resistance of a column of 0.05 M NaOH solution of cell constant \( 50\ \text{cm}^{-1} \) is \( 4.5 \times 10^3\ \Omega \). Calculate its resistivity, conductivity and molar conductivity.

Hint:

Use \( \kappa = \frac{1}{R} \times \text{cell constant} \), and \( \Lambda_m = \frac{1000 \kappa}{C} \) to find molar conductivity.

Answer 1

Given: \[ R = 4.5 \times 10^3\ \Omega,\quad l/A = \text{cell constant} = 50\ \text{cm}^{-1} \] (i) Conductivity: \[ \kappa = \text{cell constant} \times \frac{1}{R} = 50 \times \frac{1}{4.5 \times 10^3} = 1.11 \times 10^{-2}\ \text{S cm}^{-1} \] (ii) Resistivity: \[ \rho = \frac{1}{\kappa} = \frac{1}{1.11 \times 10^{-2}} = 90.09\ \Omega\ \text{cm} \] (iii) Molar Conductivity: \[ \Lambda_m = \frac{1000 \times \kappa}{C} = \frac{1000 \times 1.11 \times 10^{-2}}{0.05} = 222\ \text{S cm}^2\ \text{mol}^{-1} \] \[ \boxed{\Lambda_m = 222,\quad \kappa = 1.11 \times 10^{-2},\quad \rho = 90.09} \]