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The conductivity of a solution of AgCl at 298 K is found to be `1.382xx10^(-6)Omega^(-1)cm^(-1)` the ionic conductance of `Ag^(+)` and `Cl^(-)` at infinite dilution are `61.9Omega^(-1)cm^(2)col^(-1)` ad `76.3Omega^(-1)cm^(2)mol^(-1)` respectively the solubility of AgCl isA. `1.4xx10^(-5)molL^(-1)`B. `1xx10^(-2)molL^(-1)`C. `1xx10^(-5)molL^(-1)`D. `1.9xx10^(-5)molL^(-1)` |
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Answer» Correct Answer - C `K=1.382xx10^(-6)scm^(-1)` `^^_(AgCl)=61.9+76.3=138.2=(1000xx1.382xx10^(-6))/(S)` `S=10^(-5)M`. |
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