Resistance of a conductivity cell (cell constant 129 m ) filled with 74.5 ppm solution of KCl is 100 — Electrochemistry Chemistry Question
Question
Resistance of a conductivity cell (cell constant 129 m ) filled with 74.5 ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. = x × 10 . The value of x is _____. (Nearest integer) Given, molar mass of KCl is 74.5 g mol –1 –3 –1
💡 Solution & Explanation
**Step 1: Calculate conductivity (κ) for each solution** Using κ = (cell constant × 1000) / Resistance For solution 1: κ₁ = (129 × 1000) / 100 = 1290 S/m For solution 2: κ₂ = (129 × 1000) / 50 = 2580 S/m **Step 2: Convert ppm to molarity** For solution 1 (74.5 ppm KCl): - Molarity M₁ = (74.5 ppm) / (74.5 g/mol) = 1 M For solution 2 (149 ppm KCl): - Molarity M₂ = (149 ppm) / (74.5 g/mol) = 2 M **Step 3: Calculate molar conductivity using Λₘ = κ / M** For solution 1: Λₘ₁ = 1290 / 1 = 1290 S·m²/mol For solution 2: Λₘ₂ = 2580 / 2 = 1290 S·m²/mol **Step 4: Find the ratio** Λₘ₁ / Λₘ₂ = 1290 / 1290 = 1 **Step 5: Express in form x × 10⁻³** 1 = x × 10⁻³ x = 1000 Therefore, the answer is 1000.