The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 × 10⁻³ S cm⁻¹, then the cell constant of the conductivity cell is ____________ × 10⁻³ cm⁻¹.

Solution

Chemistry ma’am · 3 steps

Cell constant is just conductivity times resistance, so multiply the two numbers you are given.

1. Link the quantities. Resistance depends on the cell's geometry: R = ρ l/A. The ratio l/A is the cell constant G^*, and ρ = 1/κ, so R = G^*/κ.

   G^* = κ × R

2. Put in numbers. G^* = 0.152 × 10⁻³ S cm⁻¹ × 1750 Ω = 0.266 cm⁻¹. Since S × Ω = 1, only cm⁻¹ is left, the right unit for a cell constant.
3. So. G^* = 0.266 cm⁻¹ = 266 × 10⁻³ cm⁻¹, so the blank is 266.

The common slip: 0.266. Writing the cell constant as 0.266 ignores the × 10⁻³ already printed in the blank. Rewrite 0.266 cm⁻¹ as 266 × 10⁻³ cm⁻¹ to get 266.

The idea to keep. Use a standard KCl solution of known κ to calibrate a cell: G^* = κ R. Check the units, because S × Ω cancels to leave cm⁻¹.

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