Question

(a) How many grams of magnesium phosphate, Mg3(PO4)2, would you need if you were required to make 215.0 mL of a magnesium phosphate solution with a concentration of 0.110 mol L-1? (5 marks) Molar masses: Mg3(PO4)2 = 262.9 g mol-1 1 L = 1000 mL g Mg3(PO4)2

and What is the concentration of PO43- ions in the 0.110 mol L-1 solution of magnesium phosphate? (2 marks) mol L-1

Answer 1

Moles of Mg3(PO4)2 = Volume in Liters * Molarity = 0.215 L * (0.110 mol L-1) = 0.02365 mole

Mass of Mg3(PO4)2 = Moles * Molar mass = 0.02365 mole * 262.9 g mol -1 = 6.22 g

Concentration of PO43- = [ PO43- ] = 0.110 mol L-1 * 2 = 0.220 mol L-1

(Each molecule of Mg3(PO4)2 produces 2 molecules of PO43- in aqueous solution)