The value of Ksp for silver chromate, Ag2CrO4, is 9.0×10−12. Calculate the solubility of Ag2CrO4 in grams per liter.?

                 Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)    Ksp = 9.0 × 10⁻¹²
Initial (M):                            0                  0
Change (M):                     +2s               +s
Eqm (M):                            2s                 s

At equilibrium:
Ksp = [Ag⁺] [CrO₄²⁻]
9.0 × 10⁻¹² = (2s)²(s)
4s³ = 9.0 × 10⁻¹²
s = ³√{(9.0 × 10⁻¹²)/4}
s = 1.31 × 10⁻⁴

Molar mass of Ag₂CrO₄ = (107.9×2 + 52.0 + 16.0×4) g/mol = 331.8 g/mol
Solubility of Ag₂CrO₄ = (1.31 × 10⁻⁴ mol/L) × (331.8 g/mol) = 0.0435 g/L