To find the solubility product $(K_{sp})$ of $$BaSO$_4,$ we need to follow these steps:$\\$1. Calculate the molar solubility of $$BaSO$_4: \\$Given solubility of $$BaSO$_4$ in water is $2.42 \times 10^{-3}$ g/L.$\\$Molar mass of $$BaSO$_4 = 233$ g/mol.$\\$Molar solubility $(s)$ is given by:$\\ s = \frac{Solubility \ in \ g/L}{Molar \ mass \ in \ g/mol} \\ s = \frac{2.42 \times 10^{-3}}{233}$ mol/L$\\$Calculate $s: \\ s = 1.039 \times 10^{-5}$ mol/L$\\$2. Write the dissociation equation for $$BaSO$_4: \\$BaSO$_4 (s) \rightleftharpoons$Ba$^{2+} (aq) +$SO$_4^{2-} (aq) \\$At equilibrium, the concentration of $$Ba$^{2+}$ and $$SO$_4^{2-}$ is equal to $s. \\$3. Write the expression for the solubility product $(K_{sp}): \\ K_{sp} = [$Ba$^{2+}][$SO$_4^{2-}] \\$Since both ions have the same concentration $s,$ we have:$\\ K_{sp} = s \times s = s^2 \\$4. Substitute the value of $s$ into the $K_{sp}$ expression:$\\ K_{sp} = (1.039 \times 10^{-5})^2 \\ K_{sp} = 1.08 \times 10^{-10}$ mol$^2$ L$^{-2} \\$Therefore, the solubility product $K_{sp}$ of $$BaSO$_4$ is $1.08 \times 10^{-10}$ mol$^2$ L$^{-2}. \\$This corresponds to Option 3.