A 453 mL solution has a chloride concentration of 0.0253 M. This solution was prepared by adding solid FeCl3 to water. How many grams of FeCl3 are required?
mols Cl^- = M x L = approx 0.011 mols.
0.011 mol Cl^- x (1 mol FeCl3/3 mols Cl^-) = approx 0.004
Then grams = mols FeCl3 x molar mass FeCl3 = ?