Resistance of a conductivity cell (cell constant 129 m⁻¹) 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. ∧₁/∧₂=x × 10⁻³. The value of x is __________. (Nearest integer)
Given, molar mass of KCl is 74.5 g mol⁻¹.

Solution

Chemistry ma’am · 3 steps

Doubling the KCl also doubles the conductivity, so molar conductivity does not change at all.

1. Conductivity of each. Conductivity is cell constant divided by resistance, κ = G^*/R. So κ₁ = 129/100 = 1.29 S m⁻¹ and κ₂ = 129/50 = 2.58 S m⁻¹.
2. Concentration of each. 74.5 ppm means 74.5 g of KCl in 10⁶ g of solution, which is about 1000 L of a dilute solution. That is 1 mol in 1000 L, so c₁ = 10⁻³ mol L⁻¹. In the same way, 149 ppm gives c₂ = 2 × 10⁻³ mol L⁻¹.

   Λ₁/Λ₂ = (κ₁ / c₁)/(κ₂ / c₂) = κ₁/κ₂ × c₂/c₁ = 1/2 × 2 = 1

3. So. Λ₁/Λ₂ = 1 = 1000 × 10⁻³, so x = 1000.

The common slip: 500. Comparing only the conductivities gives κ₁/κ₂ = 0.5 and x = 500. You still have to divide each by its concentration.

The idea to keep. Molar conductivity is κ/c. When you compare two solutions, take the ratio of κ and the ratio of c separately. Units and the exact ppm conversion cancel out.

Answer from the NTA final answer key. Solution by PrepWiser.