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Question

A conductivity cell filled with 0.01 M AgNO3 solution gives at 25oC a resistance of 1442 Ω. If the molar conductivity of 0.01 M AgNO3 solution is 124.8 Ω1 cm2 mol1, what is the cell constant?

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Solution

Resistance =484Ω conductivity =0.00141.Ω1am1
cell constant a=R.K
=484×0.00141=0.68244
Now
Resistance of save conductivity cell R=5490
conductivity k=GR=0.682445490
=1.293×104
Molar conductivity =kC
=1.243×1040.001=0.1243

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