Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Ω 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

A
$12.4 \times 10 - 4 S m^{2} {mol}^{- 1}$
B
$124 \times 10^{- 4} S m^{2} {mol}^{- 1}$
C
$1240 \times 10^{- 4} S m^{2} {mol}^{- 1}$
D
$1.24 \times 10^{- 4} S m^{2} {mol}^{- 1}$

Solution:

Given $R = 100 Ω$ and $κ = 1.29 S m^{- 1}$

Cell constant $K_{cell} = κ R = (1.29 S m^{- 1}) (100 Ω) = 129 m^{- 1}$

For 0.02 M solution, we have $κ = \frac{K}{R} = \frac{129 m^{- 1}}{520 Ω}$ and $Λ = \frac{κ}{c} = \frac{(129 / 520) S m^{- 1}}{0.02 \times 10^{3} mol m^{- 3}} = 124 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

Solution:

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