Cell constant of the conductivity cell is — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate the conductivity of the KCl solution.** From the passage, a 0.02 M KCl solution has molar conductance $\lambda^\circ_{KCl} = 125\ \text{S cm}^2\text{mol}^{-1}$ at infinite dilution, and resistance in the cell $= 160\ \Omega$ (when 0.02 M KCl is used to calibrate). $$\kappa_{KCl} = \frac{\lambda^\circ \times C}{1000} = \frac{125 \times 0.02}{1000} = 0.0025\ \text{S cm}^{-1}$$ **Step 2: Calculate the cell constant.** The cell constant $G^*$ relates conductance and resistance: $$G^* = \kappa \times R = 0.0025 \times 160 = 0.4\ \text{cm}^{-1}$$ $$\boxed{\text{Answer: B — Cell constant} = 0.4\ \text{cm}^{-1}}$$ *Note: The resistance of the cell when filled with 0.02 M KCl is 160 Ω (from the passage context), giving a cell constant of 0.4 cm⁻¹.*