Question

The resistance of a conductivity cell filled with 0.1 mol L\(^{-1}\) KCl solution is 100 \( \Omega \). If the resistance of the same cell when filled with 0.02 mol L\(^{-1}\) KCl solution is 520 \( \Omega \), find the conductivity and molar conductivity of the 0.02 mol L\(^{-1}\) KCl solution. The conductivity of a 0.1 mol L\(^{-1}\) KCl solution is 1.29 S/m.

Hint:

Find the cell constant \( \kappa R \) from the 0.1 M data, reuse it for the 0.02 M cell, then \( \Lambda_m=\kappa/c \).

Answer 1

Step 1: The cell geometry is fixed, so the ratio (length/area), the cell constant, is the same for both fillings. Get it from the known solution: cell constant \(=\kappa R=1.29\times100=129\ m^{-1}=1.29\ cm^{-1}\).
Step 2: Apply \(\kappa=\dfrac{\text{cell constant}}{R}\) to the dilute solution in cm units: \(\kappa=\dfrac{1.29\ cm^{-1}}{520\ \Omega}=2.48\times10^{-3}\ S\,cm^{-1}\).
Step 3: Express concentration in mol per cm\(^3\): \(c=0.02\ mol\,L^{-1}=2\times10^{-5}\ mol\,cm^{-3}\).
Step 4: Molar conductivity \(\Lambda_m=\dfrac{\kappa}{c}=\dfrac{2.48\times10^{-3}}{2\times10^{-5}}=124\ S\,cm^{2}\,mol^{-1}\) (equivalently \(\kappa=0.248\ S\,m^{-1}\)).
\[ \boxed{\kappa=0.248\ S\,m^{-1},\ \Lambda_m=124\ S\,cm^{2}\,mol^{-1}} \]