Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100 $Ω$ . The conductivity of this solution is $1.29 S m^{- 1}$ . Resistance of the same cell when filled with 0.02M of the same solution is 520 $Ω$ . The molar conductivity of 0.02M solution in ${Sm}^{2} {mol}^{- 1}$

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$124 \times 10^{- 4}$

$1.24 \times 10^{- 4}$

$12.4 \times 10^{- 4}$

$1240 \times 10^{- 4}$

answer is B.

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Detailed Solution

Give $R = 100 Ω K = 1.29 {Sm}^{- 1}$
Cell constant $= K \times R = 1.29 \times 100 = 129 m^{- 1}$
$R = 520 Ω C = 0.02 M$
$K = \frac{cell consatant}{R} = \frac{129}{520} {Sm}^{- 1}$
molar conductivity $= \frac{K}{C \times 10^{3}} = \frac{129}{520 \times 0.02 \times 10^{3}}$
$= 124 \times 10^{- 4} S . m^{2} . {mol}^{- 1}$