JEE Main | 2014ElectrochemistryHard
Question
Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m-1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is
Options
A.5 × 10-4
B.5 × 10-3
C.5 × 103
D.5 × 102
Solution
For 0.2 M solution
R = 50 Ω
σ = 1.4 Sm-1 = 1.4 × 10-2 S cm-1]
⇒
Ω cm
Now, R = ρ
⇒
= 50 × 1.4 × 10-2
For 0.5 M solution
R = 280 Ω
σ = ?
= 50 × 1.4 × 10-2
⇒ R = ρ
⇒
⇒ ρ
× 50 × 1.4 × 10-2
× 70 × 10-2
= 2.5 × 10-3 S cm-1
Now, λm
= 5 S cm2 mol-1
= 5 × 10-4 S m2 mol-1
R = 50 Ω
σ = 1.4 Sm-1 = 1.4 × 10-2 S cm-1]
⇒
Ω cm Now, R = ρ
⇒
= 50 × 1.4 × 10-2 For 0.5 M solution
R = 280 Ω
σ = ?
= 50 × 1.4 × 10-2 ⇒ R = ρ
⇒
⇒ ρ
× 50 × 1.4 × 10-2 × 70 × 10-2= 2.5 × 10-3 S cm-1
Now, λm
= 5 S cm2 mol-1
= 5 × 10-4 S m2 mol-1
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