What mass of magnesium bromide would be required to prepare 720?

Complete Question:
                               What mass of magnesium bromide would be required to prepare 720mL of a 0.0939M aqueous solution?

            Molarity is calculated using following formula,

                               Molarity  =  Moles / Volume of Solution  (1)

Data Give;
                 Molarity  =  0.0939 mol.L⁻¹

                 Volume  =  720 mL  =  0.72 L

Solving eq.1 for Moles,

                            Moles  =  Molarity × Volume

                            Moles  =  0.0939 mol.L⁻¹ × 0.72 L

                            Moles  =  0.0676 moles

Now convert Moles into Mass using following formula,

                            Moles  =  Mass / M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

                            Mass  =  0.0676 mol × 184.113 g.mol⁻¹

                            Mass  =  12.44 grams

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml