Question

You need to make an aqueous solution of 0.145 m iron(lI) chloride for an experiment in lab, using a 200 grams of water. How much solid iron(II) chloride should you add?y

Answer 1

Since m(H₂O) = 200 g, V_sol ≈ 200 mL = 0.2 L.

n(FeCl₂) = C(FeCl₂) × V(FeCl₂) = 0.145 M × 0.2 L = 0.029 mol.

m(FeCl₂) = M(FeCl₂) × n(FeCl₂) = 127 g/mol × 0.029 mol = 3.683 g.