How many liters of water are required to dissolve 1.00 g of silver chromate?

At 25.0∘C , the molar solubility of silver chromate in water is 1.10×10−12M. We calculate as follows:

1.10 x 10^-10 mol/L times 291.747 g/mol = 3.21 x 10^-8 g/L 

1.00 g divided by 3.21 x 10^-8 g/L = 3.12 x 10^7 L