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Resistance of cell having 0.02 M KCl solution is 164 Omega. If 0.05 M AgNO_(3) is filled then resistance of cell become 75.8 Omega, then calculate following : [Conductivity of 0.02 M KCl=2.768xx10^(-3)Omega^(-1)cm^(-1)] (i) Conductivity of 0.05 M AgNO_(3). (ii) Molar conductivity of AgNO_(3) solution. |
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Answer» (II) `Lamda_(m)(AgNO_(3))=115.6Omega" "cm^(-1)mol^(-1)` |
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