Chemistry Help -11 U- Solutions- Finding g/L

At 303K, 9.00g of solid KCl will exactly saturate 25.0 mL of water. The solubility of KCl in grams per 1.0 L of water at 303 K is?

If y'all could help me out with this, I'd be much obliged. Thanks.

(9.00 g) / (0.0250 L) = 360 g/L