A 5.0 m mol dm⁻³ aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm⁻¹. The molar conductivity of this solution is ___________ mSm² mol⁻¹. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 3 steps

Get conductivity from conductance and cell constant, then divide by concentration, all in metres.

1. Conductivity first. Conductivity is conductance times the cell constant: κ = G × G^* = 0.55 mS × 1.3 cm⁻¹ = 0.715 mS cm⁻¹.
2. Switch to SI units. The answer is wanted in mS m² mol⁻¹, so work in metres. Since 1 cm⁻¹ = 100 m⁻¹, κ = 71.5 mS m⁻¹. Since 1 m³ = 1000 dm³, c = 5.0 mmol dm⁻³ = 5.0 mol m⁻³.

   Λₘ = κ/c = (71.5 mS m⁻¹)/(5.0 mol m⁻³) = 14.3 mS m² mol⁻¹

3. So. The molar conductivity is 14.3 mS m² mol⁻¹, which rounds to 14.

The common slip: 14300. Using c = 5 × 10⁻³ (the value in mol dm⁻³) as if it were in mol m⁻³ gives 14 300, not 14. In SI units 5.0 mmol dm⁻³ is 5.0 mol m⁻³.

The idea to keep. Molar conductivity is κ / c, and both must be in the same length unit. Convert κ and c into the units of the answer before you divide.

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