ElectrochemistryHard
Question
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 S m-1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω. The molar conductivity of 0.02 M solution of the electrolyte will be
Options
A.124 × 10-4S m2 mol-1
B.1240 × 10-4 S m2 mol-1
C.1.24 × 10-4 S m2 mol-1
D.12.4 × 10-4 S m2 mol-1
Solution
There is one mistake in Question paper.
Assuming concentration of solution is 0.2 M instead of 0.02 M. Since resistance of 0.2 M is 520 Ω.
R = 100 Ω
K =
1.29 =
= 129 m-1
R = 520Ω , C = 0.2 M
K =
(129) Ω-1 m-1
μ = K × Vin cm3
=
× 10-6
= 1.24 × 10-3
= 12.4 × 10-4
Assuming concentration of solution is 0.2 M instead of 0.02 M. Since resistance of 0.2 M is 520 Ω.
R = 100 Ω
K =
1.29 =
= 129 m-1R = 520Ω , C = 0.2 M
K =
(129) Ω-1 m-1μ = K × Vin cm3
=
× 10-6= 1.24 × 10-3
= 12.4 × 10-4
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