How many milliliters of an aqueous solution of 0.180 M iron(III) chloride is needed to obtain 14.7 grams of the salt?

Molar mass of FeCl₃ = (55.85 + 35.45×3) g/mol = 162.2 g/mol
No. of moles of FeCl₃ = (14.7 g) / (162.2 g/mol) = 0.0906 mol

Volume of 0.180 M FeCl₃ solution needed = (0.0906 mol) / (0.180 mol/L) = 0.503 L = 503 mL