Question

Resistance of a conductivity cell filled with 0.2 mol L⁻¹ \[\ce{KCl}\] solution is 200 Ω. If the resistance of the same cell when filled with 0.05 mol L⁻¹ \[\ce{KCl}\] solution is 620 Ω, calculate the conductivity and molar conductivity of 0.05 mol L⁻¹ \[\ce{KCl}\] solution. The conductivity of 0.2 mol L⁻¹ \[\ce{KCl}\] 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 \[\ce{KCl}\] solution:

Cell constant = conductivity × resistance

`"Conductivity" = "cell constant"/"resistance"`

Conductivity = `0.0496/620`

= 8 × 10⁻⁵ ohm⁻¹ cm⁻¹

Molar conductivity = `"conductivity"/"concentration"`

= `(8 xx 10^-5)/(0.05 xx 10^-3)`

= 160 × 10⁻²