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Q. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100 $\Omega $. The conductivity of this solution is 1.29 S m^{-l}. Resistance of the same cell when filled with 0.02M of the same solution is 52 $\Omega $. The molar conductivity of 0.02M solution of the electrolyte will be

1) 12.4 x S 10^{-4 }m^{2}. mol^{-l} ^{ }

2) 124 x 10^{-4 }S m2 . mol^{-l}

3) 1240 x 10^{-4 }S m^{2}. mol^{-l}

4) 1.24 x 10^{-4} S m^{2}. mol-l

Cell constant can be calculated using 0.1 M KCl

= κ × R

= 0.0129 * 100

= 1.29 cm^{-1}

Calculating conductivity of 0.02 M KCl solution

κ = Cell constant / Resistance

= 1.29 / 520

= 2.48 * 10^{-3} S cm^{-1}

Therefore, molar conductivity:

Λm = 10^{3} κ / M

= 10^{3} * 2.48 * 10^{-3}/0.02

= 124 S cm^{2} mol-1

= 124 x 10^{-4 }S m^{2} . mol^{-1}

So, option (2) is correct

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