Compute the cell constant and molar conductance:

Problem:

The resistance of a conductivity cell was 702 ohms when filled with 0.1 M KCl when filled with 0.1 M KCl solution (K = 0.14807 ohm^-1 m^-1) and 6920 ohm when filled with 0.01M acetic acid solution. Compute the cell constant and molar conductance for the acid solution.

Answer:

Cell constant = KR = (0.14807 ×702) (ohm-1 m-1) (ohm) =103.94 m^-1   = 1.039 cm^-1

Conductivity of acetic acid K = (1/R) (l/A)

= (1 / 6920 ?) (1.039) cm^-1

= 1.501 × 10-4 ?^-1 cm^-1

= 1.501 × 10-2 ?^-1 m^-1

Concentration = 0.01 M = 0.01 mol dm³ = 0.01 × 103 mol m^-3

Λ_m = K/c = 1.501×10^-2 ?^-1 m^-1/ 0.01×10³ mol m^-3

= 1.501 × 10^-3 mol^-1 ?^-1 m²