Question

Resistance of a conductivity cell filled with 0.1 mol L⁻¹ KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L⁻¹ KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L⁻¹KCl solution. The conductivity of 0.1 mol L⁻¹ KCl solution is 1.29 × 10⁻² Ω⁻¹ cm⁻¹.

Answer 1

Given that:
Concentration of the KCl solution = 0.1 mol L⁻¹
Resistance of cell filled with 0.1 mol L⁻¹ KCl solution = 100 ohm
Cell constant = G^* = conductivity ×× resistance
1.29×10⁻² ohm⁻¹cm⁻¹ × 100 ohm = 1.29 cm⁻¹ = 129 m⁻¹
Cell constant for a particular conductivity cell is a consant.

Conductivity of 0.02 mol L⁻¹ KCl solution = \[\frac{Cell \ constant}{Resistance}\] \[\frac{G *}{R} = \frac{129 m^{- 1}}{520 ohm} = 0 . 248 {Sm}^{- 1}\]

Concentration = 0.02 mol L⁻¹

= 1000××0.02 mol m⁻³ = 20 mol m⁻³

Now,

Molar conductivity = \[\Lambda_m = \frac{\kappa}{c} = \frac{248 \times {10}^{- 3} S m^{- 1}}{20 mol m^{- 3}} = 124 \times {10}^{- 4} S m^2 {mol}^{- 1}\]

Therefore, the molar conductivity of 0.02 mol L⁻¹ KCl solution is \[124 \times {10}^{- 4} S m^2 {mol}^{- 1}\]