The resistance of a conductivity cell with cell constant 1.14 cm⁻¹, containing 0.001 M KCl at 298 K is 1500 Ω. The molar conductivity of 0.001 M KCl solution at 298 K in S cm² mol⁻¹ is ____________. (Integer answer)

Solution

Chemistry ma’am · 3 steps

Resistance and cell constant give you κ first; then divide by concentration in mol per cm³.

1. Conductivity first. Conductivity is the cell constant divided by the resistance: κ = G^*/R = (1.14 cm⁻¹)/(1500 Ω) = 7.6 × 10⁻⁴ S cm⁻¹.
2. Match the units. κ is per cm, but molarity is per litre. One litre is 1000 cm³, so the factor 1000 converts the concentration to mol per cm³.

   Λₘ = (κ × 1000)/C = (7.6 × 10⁻⁴ × 1000)/0.001

3. So. Λₘ = 760 S cm² mol⁻¹, so the answer is 760.

The common slip: 0.76. Dividing κ by 0.001 without the factor 1000 gives 0.76, not 760.

The idea to keep. Always go R → κ → Λₘ. With κ in S cm⁻¹ and C in mol L⁻¹, use Λₘ = κ × 1000/C to get S cm² mol⁻¹.

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