The resistance of a conductivity cell with cell constant 1.14 cm , containing 0.001 M KCl at 298 K i — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the given data** - Cell constant (G*) = 1.14 cm⁻¹ - Resistance (R) = 1500 Ω - Molarity (C) = 0.001 M - Temperature = 298 K **Step 2: Calculate conductivity (κ)** Use the formula: κ = G*/R κ = 1.14/1500 = 0.00076 S cm⁻¹ **Step 3: Understand molar conductivity formula** Molar conductivity (Λₘ) is defined as: Λₘ = κ/C where κ is conductivity and C is molarity. **Step 4: Convert units appropriately** κ = 0.00076 S cm⁻¹ C = 0.001 M = 0.001 mol cm⁻³ **Step 5: Calculate molar conductivity** Λₘ = κ/C = 0.00076/(0.001) Λₘ = 0.76 S cm² mol⁻¹ **Step 6: Convert to required units** Λₘ = 0.76 × 1000 = 760 S cm² mol⁻¹ Therefore, the answer is 760.00.