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^{−1}KCl solution. The conductivity of 0.1 mol L^{−1} KCl solution is 1.29 × 10^{−}^{2} Ω^{−1} cm^{−1}.

#### Solution

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^{−1}cm^{−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 = \[\frac{Cell \ constant}{Resistance}\] \[\frac{G *}{R} = \frac{129 m^{- 1}}{520 ohm} = 0 . 248 {Sm}^{- 1}\]

^{−1}

^{−3}= 20 mol m

^{−3}

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

^{−1}KCl solution is \[124 \times {10}^{- 4} S m^2 {mol}^{- 1}\]