Question

Resistance of a conductivity cell filled with 0.2 mol L⁻¹ $$\mathrm{KCl}$$ solution is 200 Ω. If the resistance of the same cell when filled with 0.05 mol L⁻¹ $$\mathrm{KCl}$$ solution is 620 Ω, calculate the conductivity and molar conductivity of 0.05 mol L⁻¹ $$\mathrm{KCl}$$ solution. The conductivity of 0.2 mol L⁻¹ $$\mathrm{KCl}$$ solution is 0.0248 S cm⁻¹.

Answer 1

Cell constant = conductivity × resistance

= 0.0248 × 200

= 0.0496 cm⁻¹

For 0.05 mol L⁻¹ $$\mathrm{KCl}$$ solution:

Cell constant = conductivity × resistance

${\displaystyle \text{Conductivity}=\frac{\text{cell constant}}{\text{resistance}}}$

Conductivity = ${\displaystyle \frac{0.0496}{620}}$

= 8 × 10⁻⁵ ohm⁻¹ cm⁻¹

Molar conductivity = ${\displaystyle \frac{\text{conductivity}}{\text{concentration}}}$

= ${\displaystyle \frac{8\times {10}^{-5}}{0.05\times {10}^{-3}}}$

= 160 × 10⁻²