Question

The specific conductance of a saturated solution of AgCl in water is `1.826xx10^(-6)"ohm"^(-1)cm^(-1)` at `25^(@)C`. Calculate its solubility in water at `25^(@)C`. [Given `Lambda_(m)^(oo)(Ag^(+))=61.92" ohm"^(-1)cm^(2)mol^(-1) "and" Lambda_(m)^(oo)(Cl^(-))=76.34" ohm"^(-1)cm^(2)mol^(-1)]`.

Answer 1

`Lambda_(m)^(oo)(AgCl)=lambda_(m)^(oo)(Ag^(+))+lambda_(m)^(oo)(Cl^(-))=61.92" ohm"^(-1)cm^(2)mol^(-1)+76.34" ohm"^(-1)cm^(2) mol^(-1)`
`=138.26" ohm"^(-1)cm^(2)mol^(-1)`
`k=1.826xx10^(-6)"ohm"^(-1)cm^(-1)`
`"Solubility" ("in" mol^(-1))=(kxx1000 cm^(3) L^(-1))/(Lambda_(m)^(oo))=((1.826xx10^(-6)"ohm"^(-1) cm^(-1))xx(1000 cm^(3)L^(-1)))/((138.26" ohm"^(-1)cm^(2)mol^(-1)))`
`=1.32xx10^(-5)" mol "L^(-1)`.