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The conductivity of a saturated solution of AgCl at 288 K is found to be 1.382xx10^(-6)" S "cm^(-1). Find its solubility. Given ionic conductances of Ag^(+) and Cl^(-) at infinite dilution are 61.9 S cm^(2)mol^(-1) and 76.3"S "cm^(2)mol^(-1). respectively |
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Answer» SOLUTION :`wedge_(m)^(@)(AgCl)=lamda_(ag^(+))^(@)+lamda_(Cl^(-))^(@)=61.9+76.3=138.2" S "cm^(2)mol^(-1)` Solubility `=(kappaxx1000)/(wedge_(m)^(@))=(1.382xx10^(-6)xx1000)/(138.2)=10^(-5)" mol "L^(-1)=10^(-5)xx143.5" G "L^(-1)=1.435xx10^(-3)g" "L^(-1)`. |
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