The specific conductivity of the saturated aqueous solution of AgCl is `1.82xx10^(-6) ohm^(-1) cm^(-1)` at 298K. Calculate me solubility of AgCl.`[wedge_(m)^(@)(Ag^(+))=61.83 ohm^(-1) cm^(2) mol^(-1)]` and `wedge_(m)^(@) (CI)^(-)=76.41 ohm^(-1) cm^(2) mol^(-1)`

Explanation

`Lamda_(m)^(@) (AgCl)= lamda_(m)^(@) (Ag^(+)) + lamda_(m)^(@) (Cl^(-))`
`Lamda_(m)^(@) (AgCl) = 61.83 ohm^(-1) cm^(2) mol^(-1) + 76.41 ohm^(-1) cm^(2) mol^(-1) = 138.246 ohm^(-1) cm^(2) mol^(-1)`. Specific conductivity `=1.82 xx 10^(-6) ohm^(-1) cm^(1)`
Solubility `= (k xx 1000 cm^(3) L^(-1))/(Lamda_(m)) = (1.82 xx 10^(-6) ohm^(-1) cm^(-1) xx 1000 cm^(3) L^(-1))/(138.24 ohm^(-1) cm^(2) mol^(-1)) = 1.32 xx 10^(-5) mol L^(-1)`