Electrochemistry

The cell constant tells us about the cell's design and size. It can be found by using the known conductivity and resistance of 0.1 M KCl solution.$G^{\star}=\kappa \cdot R$So, $G^{\star}=1.29 \times 100 = 129\ \mathrm{m}^{-1}$ or $1.29\ \mathrm{cm}^{-1}$Step 2: Calculate the conductivity ( $\kappa$ ) for $0.02\ \mathrm{M}\ \mathrm{KCl}$ solutionWe use the cell constant and the new measured resistance for the 0.02 M solution.$\kappa = \; \frac{G^{\star}}{R} = \; \frac{1.29}{520} = 0.00248\ \mathrm{Sm}^{-1}$Step 3: Find the molar conductivity ( $\Lambda_{m}$ )Molar conductivity shows how well ions move in a solution. Use this formula:$\Lambda_{m} = \; \frac{\kappa \times 1000}{C}$$\Lambda_{m} = \; \frac{0.00248 \times 1000}{0.02}$$\; = 124\ \mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{mol}^{-1}$