Correct Answer - C
For a general reaction,
`A_(x)B_(y) hArr xA^(y+)+yB^(x-)`
Solubility product `(K_(sp))=[A^(y+)]^(x)[B^(x-)]^(y)`
For `BaSO_(4)` (binary solute giving two ions)
`BaSO_(4)(s) hArr Ba^(2+)underset(S)((aq))+underset(S)(SO_(4)^(2-))(aq)`
`therefore K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=(S)(S)=S^(2) " " ["where, S=Solubility"]`
Given, `S=2.42xx10^(-3) gL^(-1)`
Molar mass of `BaSO_(4)=233 g "mol"^(-1)`
`therefore` Solubility of `BaSO_(4)`
`(S)=(2.42xx10^(-3))/(233) "mol" L^(-1)`
`=1.04xx10^(-5) "mol" L^(-1)`
On substituting the value of S in Eq. (i), we get
`K_(sp)=(1.04xx10^(-5) "mol" L^(-1))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`