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