Question

Magnesium is lighter than other structural metals, so it is increasingly important in the design of more efficient vehicles. Mg2+ ions are present in seawater, and the metal is often prepared by "harvesting" these ions and converting them to 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 1.03 g/ml. What is the molarity of mg2+ ions in seawater? The design for a concept car calls for 125 kilograms of magnesium per vehicle. How many gallons of seawater would be required to supply enough magnesium to build one of these cars?

Answer 1

The design for a concept car calls for 125 kilograms of magnesium per vehicle. How many gallons of seawater would be required to supply enough magnesium to build one of these cars?

In each ton of sea water the mass of Mg+2 = 1270 grams

one ton = 907185 grams

so volume of one ton of sea water = mass / density = 907185 / 1.03

= 880762 mL

Volume of sea water in L = 880.760 L

atomic mass of Mg+2 = 24g/ mole

So moles of Mg+2 in sea water = mass / molar mass = 1270 g / 24 = 52.92 moles

molarity = moles of solute / volume of solution in litres = 52.92 / 880.760 = 0.06 M

PArt 2:

As in each litre of sea water the mass of Mg+2 = 1270g / 880.760 L = 1.442 g/L

So for 1 grams we need = 0.693 L of sea water

For 125 kilograms = 125 X 0.693 X 1000 L

We know that 1L = 0.2642 gallons

so 125 X 0.693 X 1000 L = 0.2642 X 125 X 0.693 X 1000 L gallons

= 22886.33 gallons of sea water will be required