Question

Consider two solutions filled with the same conductivity cell where the solution $1$ has $0.1\mathrm{mol}{\mathrm{L}}^{-1}$$\mathrm{KCl}$ with resistance $100\mathrm{\Omega }$ and conductivity $1.29\mathrm{S}{\mathrm{m}}^{-1}$and the second solution has $0.02\mathrm{M}$$\mathrm{KCl}$, resistance is $520$ ohm. Calculate the conductivity and molar conductivity of the second solution.

Answer 1

Step 1: Analyzing given data

Molarity of $\mathrm{KCl}=0.1\mathrm{mol}{\mathrm{L}}^{-1}$

Resistance of $0.1\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}$ solution $=100\mathrm{ohm}$

Conductivity of $0.1\mathrm{M}\mathrm{KCl}=1.29\mathrm{S}{\mathrm{m}}^{-1}$

Molarity of $\mathrm{KCl}$ solution whose conductivity and molar conductivity have to be calculated$=0.02\mathrm{M}$

Resistance of $0.02\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}=520\mathrm{ohm}$

Step 2: Calculating conductivity and molar conductivity

Cell constant$=$ Conductivity$\times$Resistance

For $0.1\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}$ solution, Cell constant$=1.29\times {10}^{-2}{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}\times 100\mathrm{\Omega }\Rightarrow 1.29{\mathrm{cm}}^{-1}$

For $0.02\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}$ solution, Conductivity(k)$=\frac{\mathrm{Cell}\mathrm{constant}}{\mathrm{Resistance}}$$=\frac{1.29{\mathrm{cm}}^{-1}}{520\mathrm{\Omega }}\Rightarrow 0.0248{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

Molar conductivity$=\frac{\mathrm{Conductivity}\left(\mathrm{k}\right)\times 1000{\mathrm{cm}}^3{\mathrm{L}}^{-1}}{\mathrm{Molarity}}$

$=\frac{0.0248{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}\times 1000{\mathrm{cm}}^3{\mathrm{L}}^{-1}}{0.02\mathrm{Mol}{\mathrm{L}}^{-1}}\Rightarrow 124{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$

Therefore, the conductivity and molar conductivity of $0.02\mathrm{M}$ $\mathrm{KCl}$ is $0.0248{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$and $124{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$.