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.
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Answer:
Resistance=
Specific conductivity
1
×
A
l
Cell constant=
A
l
=1.29cm
−1
R=520ohms
Conductivity or specific conductivity=
Resistance
Cell constant
=
520
1.29
=0.00248mho/cm
Molar conductivity=
Molarity
1000×conductivity
=
0.02
1000×0.00248
=124Scm
2
mol
−1
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