3) (iii) Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance 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.

[Given : Conductivity of 0.1 M KCl solution is 1.29 S/m]

Dear Student,

Cell constant = Conductivity×Resistanceor, cell constant = 1.29 Sm-1×100Ω = 129 m-1Thus, conductivity of 0.02 M KCl =Cell constantResistance=129 m-1520 Ω= 0.248 Sm-1Now, concentration = 0.02 molL-1= 1000×0.02 molm-3=20molm-3Thus, molar conductivity λm=κC=0.248 Sm-120molm-3= 124×10-4 Sm2mol-1

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