Question

The electrical resistance of a column of 0.05 M NaOH solution of area 0.8 cm$^2$ and length 40 cm is 5 $\times$ 10$^3$ ohms. Calculate its resistivity, conductivity, and molar conductivity.

Hint:

Resistivity is a property of the material itself, while conductivity is the ability of the material to conduct electricity. Molar conductivity is the conductivity for one mole of solute.

Answer 1

First, we can calculate the resistivity \(\rho\) using the formula: \[ R = \rho \times \frac{L}{A} \] Where: - \(R\) is the resistance = 5 $\times$ 10$^3$ ohms - \(L\) is the length = 40 cm = 0.40 m - \(A\) is the area = 0.8 cm$^2$ = \(0.8 \times 10^{-4}\) m$^2$ Rearranging the formula for resistivity: \[ \rho = \frac{R \times A}{L} = \frac{5 \times 10^3 \times 0.8 \times 10^{-4}}{0.40} = 1.0 \times 10^{-1} \, \Omega \, \text{m} \] Next, we can calculate the conductivity \(\kappa\) using the reciprocal of resistivity: \[ \kappa = \frac{1}{\rho} = \frac{1}{1.0 \times 10^{-1}} = 10 \, \text{S/m} \] Finally, the molar conductivity \(\Lambda_m\) is given by: \[ \Lambda_m = \kappa \times \frac{1000}{C} \] Where \(C = 0.05 \, \text{mol/L}\). \[ \Lambda_m = 10 \times \frac{1000}{0.05} = 200 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \]