Resistance of 0.2 M solution of an electrolyte is 50 Ω. The conductivity 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

A
$5 \times 10^{2}$
B
$5 \times 10^{- 4}$
C
$5 \times 10^{- 3}$
D
$5 \times 10^{3}$

Solution:

Cell constant $K_{cell} = R κ = (50 Ω) (1.4 S m^{- 1}) = 70 m^{- 1}$ Conductivity of 0.5 M solution and molar conductivity are

$κ = \frac{κ}{R} = \frac{70 m^{- 1}}{280 Ω} = \frac{1}{4} {Sm}^{- 1}; Λ_{m} = \frac{κ}{c} = \frac{(1 / 40) {Sm}^{- 1}}{(0.5 mol {dm}^{- 3})} = \frac{(1 / 4) {Sm}^{- 1}}{(0.5 \times 10^{3} {molm}^{- 3})} = 5 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

Resistance of 0.2 M solution of an electrolyte is $50 Ω$ . The conductivity 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 $S m^{2} {mol}^{- 1}$ is

Solution:

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