NEETChemistryElectrochemistry
Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity.
Options
- A24.0 S/m ;10-4 S / m
- B124.0 S/m; 124 × 10-4 S m2
- C1.24 S/m; 100× 10-4 S m2 mol-1
- D0.248 S m-1 ; 124 × 10-4 S m2 mol-1
Correct answer
D. 0.248 S m-1 ; 124 × 10-4 S m2 mol-1
Step-by-step solution
The cell constant is given by the equation: Cell constant = G* = conductivity × resistance = 1.29 S/m × 100 Ω = 129 m-1 = 1.29 cm-1. Conductivity of 0.02 mol L-1 KCl solution = cell constant / resistance= G* R = 129 m-1 / 520 Ω= 0.248 S m-1 ; Molar conductivity = Λ m = 𝜅/C = 248 ×10-3 S m-1 / 20 mol m-3 = 124 × 10-4 S m2 mol-1.