# 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 m2. mol-l 2) 124 x 10-4 S m2 . mol-l 3) 1240 x 10-4 S m2. mol-l 4) 1.24 x 10-4 S m2. 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 = 103 κ / M

= 103 * 2.48 * 10-3/0.02

= 124 S cm2 mol-1

124 x 10-4 S m2 . mol-1

So, option (2) is correct

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