Question

3) (iii) Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution.

[Given : Conductivity of 0.1 M KCl solution is 1.29 S/m]

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Cell}\mathrm{constant}=\mathrm{Conductivity}\times \mathrm{Resistance} \\ \mathrm{or},\mathrm{cell}\mathrm{constant}=1.29{\mathrm{Sm}}^{-1}\times 100\mathrm{\Omega }=129{\mathrm{m}}^{-1} \\ \mathrm{Thus},\mathrm{conductivity}\mathrm{of}0.02\mathrm{M}\mathrm{KCl}=\frac{\mathrm{Cell}\mathrm{constant}}{\mathrm{Resistance}}=\frac{129{\mathrm{m}}^{-1}}{520\mathrm{\Omega }}=0.248{\mathrm{Sm}}^{-1} \\ \mathrm{Now},\mathrm{concentration}=0.02{\mathrm{molL}}^{-1}=1000\times 0.02{\mathrm{molm}}^{-3}=20{\mathrm{molm}}^{-3} \\ \mathrm{Thus},\mathrm{molar}\mathrm{conductivity}{\mathrm{\lambda }}_{\mathrm{m}}=\frac{\mathrm{\kappa }}{\mathrm{C}}=\frac{0.248{\mathrm{Sm}}^{-1}}{20{\mathrm{molm}}^{-3}}=124\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1} \end{gathered}$$