A KCl solution of conductivity 0.14 S m⁻¹ shows a resistance of 4.19Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03Ω. The conductivity of the HCl solution is ____________ × 10⁻² S m⁻¹. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 3 steps

Same cell means the same cell constant, so conductivity times resistance stays fixed.

1. Cell constant. For any cell, κ = G^*/R, where G^* = l/A is the cell constant. The cell is not changed, so G^* is the same for both solutions.
2. Find it from KCl. G^* = κ R = 0.14 S m⁻¹ × 4.19 Ω = 0.5866 m⁻¹.

   κ_(HCl) = G^*/R_(HCl) = (0.5866 m⁻¹)/(1.03 Ω) = 0.5695 S m⁻¹

3. So. κ_(HCl) = 56.95 × 10⁻² S m⁻¹, which rounds to 57.

The common slip: 3. Writing κ_(HCl) = 0.14 × 1.03/4.19 (ratio upside down) gives 3.44 × 10⁻², so 3. That is wrong because a smaller resistance must give a larger conductivity.

The idea to keep. In one conductivity cell, κ and R are inversely proportional: κ₁ R₁ = κ₂ R₂. Lower resistance always means higher conductivity.

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