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Resistance of a conductivity cell fille...

Resistance of a conductivity cell filled with 0.1 M KCl is 100 ohm. If the resistence of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. Conductivity of 0.1 KCl solution is `1.29xx10^(-2)" ohm "^(-1)cm^(-1)`.

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Resistance of 0.1 M KCl solution `R =100 Omega`
Conductivity `K = 1.29 sm^(-1)`
Cell constant `G = kxxR = 1.29xx100 = 129 m^(-1)`
Resistance of 0.02 M KCl solution , `R= 520 Omega`
Conductivity , `K = ("cell constant")/(R ))= (129m^(-1)/(520Omega) = 0.248 Sm^(-1)`
Concentration , `C = 0.02molL^(-1) = 1000xx0.02molm ^(-3)= 20molm^(-3)`
Molar conductivity ,`wedge_(m) = (K)/(C) = (0.248Sm^(_1))/(20molm^(-3) = 0.0124 s M^(2)mol^(-1)`
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