Calculate the cell constant:

Problem:

A certain conductance cell was filled with 0.0100 M solution of KCl, whose conductivity that is 0.001409 ? ^-1 cm^-1 (S cm^-1) at 25 ^oC, it had a resistance of 161.8 ?, and when filled with 0.0050 M NaOH, it had a resistance of 190 ?. Calculate the cell constant, conductivity and molar conductivity of solution NaOH.

Answer:

Cell constant = KR = 0.001409 ?^-1 cm^-1× 161.8 ? = 0.228 cm^-1

κ _NaOH = Cell constant/R = 0.228 cm^-1/190 ? = 1.2 × 10^-3 ?^-1 cm^-1

Λ_m = K/c = 1.2 × 10^-3 ?^-1 cm^-1/ 0.0050 × 10^-3   mol cm^-3 = 240 ?^-1 cm²  mol^-1