Question

. Mg2+ ions are present in seawater, and the metal is often prepared by "harvesting" these ions and converting them to neutral Mg metal. The average magnesium content of the oceans is about 1270 g Mg2+ per ton of seawater, and the density of seawater is about 1.02 g/mL. (Some additional unit conversion factors are given below.) 1 ton = 2000 pounds 1 gallon = 3.7854 L 454 g = 1 pound

(6 pts) (a). How many gallons of seawater would be required to supply 123 kg of magnesium?

(6 pts) (b). Use information from above to find the molarity of Mg2+ ions in seawater?

Answer 1

a) given 1270 g Mg2+ per ton of sea water

1270 g Mg2+ per 2000 x 454 g of sea water          ( 1 ton = 2000 pond , 1 pond = 454 g)

1270 g Mg2+ per 908000 g

1.27 kg Mg2+   per 908 kg sea water          ( 1kg = 1000 g)

we need 123 kg of Mg2+

sea water needed = ( 123 ) x ( 908 /1.27) = 87940 kg

b) we have 1270 g Mg2+ per 908 kg sea water

sea water volume = mass / density = 908 kg / ( 1.02 kg/L)   = 890.2 L           ( since 1.02 g/ml = 1.02 kg/L)

moles of Mg2+ = mass of Mg2+ / Molar mass of Mg2+

            = 1270 g /( 24.3 g/mol ) = 52.26

Molarity = moles of Mg2+ / ( solution vol in L)

             = 52.26 / 890.2 = 0.0587 mol/L = 0.0587 M