The solubility product of BaSO$$_4$$ is $$1 \times 10^{-10}$$ at 298 K. The solubility of BaSO$$_4$$ in 0.1 M K$$_2$$SO$$_4$$(aq) solution is _______ $$\times 10^{-9}$$ g L$$^{-1}$$ (nearest integer).
Given: Molar mass of BaSO$$_4$$ is 233 g mol$$^{-1}$$

We need to find the solubility of BaSO$$_4$$ in 0.1 M K$$_2$$SO$$_4$$ solution, given $$K_{sp}(\text{BaSO}_4) = 1 \times 10^{-10}$$ at 298 K.

The dissolution equilibrium is:

$$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$$

Now, K$$_2$$SO$$_4$$ is a strong electrolyte that completely dissociates, giving an initial SO$$_4^{2-}$$ concentration of 0.1 M (common ion effect). Let the molar solubility of BaSO$$_4$$ be $$s$$ mol/L.

At equilibrium, $$[\text{Ba}^{2+}] = s$$ and $$[\text{SO}_4^{2-}] = 0.1 + s \approx 0.1$$ (since $$s \ll 0.1$$).

Applying the solubility product expression:

$$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$$

$$1 \times 10^{-10} = s \times 0.1$$

$$s = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9}$$ mol/L

Converting to g/L using the molar mass of BaSO$$_4$$ = 233 g/mol:

$$\text{Solubility in g/L} = 1 \times 10^{-9} \times 233 = 233 \times 10^{-9}$$ g/L

So, the answer is $$233$$.