Question

The resistance of a conductivity cell when filled with 0.02 M KCl solution is 164 ohms at 298 K. When

the cell is filled with 0.05 M AgNO3 solution its resistance is found to be 78.5 ohms at 298 K Calculate

the molar conductivity of 0.05 M AgNO3 solution if the conductivity of 0.02 M KCl solution is 2.768 x

10-3 S cm-1.​

Answer 1

Answer:

Resistance= Specific conductivity1×Al

Cell constant= Al=1.29cm−1

R=520ohms

Conductivity or specific conductivity= ResistanceCell constant

                                                            =5201.29=0.00248mho/cm

Molar conductivity= Molarity1000×conductivity=0.021000×0.00248=124Scm2mol−1