Question

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract 8.0 x 104 tons of Mg, which is roughly the annual production in the United States. (rho seawater = 1.03 g/ml).

Answer 1

The volume of seawater needed to extract
`8.0* 10^4` tons of magnesium is
`5.420* 10^(10) L`.

Step-by-step explanation:

Concentration magnesium in sea water = 1.3 g /kg of seawater

Extracted mass of magnesium ,=
`8.0* 10^4 tons`

(1 ton = 907185 g)

`8.0* 10^4 tons=8.0* 10^4* 907185 g`

`=7.257* 10^(10) g`

If 1 kg of sea water contains 1.3 grams of magnesium. Then
`7.257* 10^(10) g` of magnesium will be contained by:

`(1)/(1.3) kg* 7.257* 10^(10)=5.582* 10^(10) kg` sea water.

Mass of sea water,m =
`5.582* 10^(10) kg=5.582* 10^(13) g`

Volume of sea water = v

Density of sea water ,d= 1.03 g/mL

`v=(m)/(d)=(5.582* 10^(13) g)/(1.03 g/mL)=5.420* 10^(13) mL`

1 mL = 0.001 L

`v = 5.420* 10^(10) L`

The volume of seawater needed to extract
`8.0* 10^4` tons of magnesium is
`5.420* 10^(10) L`.