`K_(sp)` of `AgCI` is`1.8xx10^(-10).` Calculate molar solubility in 1M `AgNO_3`.

Explanation

`AgCI_((s))hArr Ag_((aq))^(+)+CI_((aq))^(-)`
`x="solubility of in " 1M AgNO_3`
`AgNO_(3(aq)) hArr underset(1M)(Ag_((aq))^(+))+underset(1M)(NO_(3(aq))^-)`
`[Ag^+]=x+1 cong 1M (because x ltlt 1)`
`[CI^-]=x`
`K_(sp)=[Ag^+][CI^-]`
`1.8xx10^(-10)=(1)(x)`
`x=1.8xx10^(-10)M`.