A 5.0 m mot dm aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell — Electrochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Molar Conductivity of KCl Solution **Step 1: Identify the formula for molar conductivity** Molar conductivity (Λₘ) is given by: $$Λ_m = \frac{\kappa}{c}$$ Where κ is conductivity and c is molar concentration. **Step 2: Convert conductance to conductivity** Conductivity is related to conductance by: $$\kappa = G × k$$ Where G is conductance (0.55 mS) and k is cell constant (1.3 cm⁻¹) $$\kappa = 0.55 × 1.3 = 0.715 \text{ mS·cm}^{-1}$$ **Step 3: Convert conductivity to standard units** Convert mS·cm⁻¹ to S·m⁻¹: $$\kappa = 0.715 \text{ mS·cm}^{-1} = 0.0715 \text{ S·m}^{-1}$$ **Step 4: Convert concentration to standard units** Given concentration: 5.0 m mol dm⁻³ = 5.0 mol m⁻³ **Step 5: Calculate molar conductivity** $$Λ_m = \frac{0.0715}{5.0} = 0.0143 \text{ S·m}^2·\text{mol}^{-1}$$ **Step 6: Convert to mS·m²·mol⁻¹** $$Λ_m = 0.0143 × 1000 = 14.3 \text{ mS·m}^2·\text{mol}^{-1}$$ Rounding to the nearest integer: **14** Therefore, the answer is 14.00.