Answer:
The volume of seawater needed to extract [tex]8.0\times 10^4[/tex] tons of magnesium is [tex]5.420\times 10^{10} L[/tex].
Explanation:
Concentration magnesium in sea water = 1.3 g /kg of seawater
Extracted mass of magnesium ,= [tex]8.0\times 10^4 tons[/tex]
(1 ton = 907185 g)
[tex]8.0\times 10^4 tons=8.0\times 10^4\times 907185 g[/tex]
[tex]=7.257\times 10^{10} g[/tex]
If 1 kg of sea water contains 1.3 grams of magnesium. Then [tex]7.257\times 10^{10} g[/tex] of magnesium will be contained by:
[tex]\frac{1}{1.3} kg\times 7.257\times 10^{10}=5.582\times 10^{10} kg[/tex] sea water.
Mass of sea water,m =[tex]5.582\times 10^{10} kg=5.582\times 10^{13} g[/tex]
Volume of sea water = v
Density of sea water ,d= 1.03 g/mL
[tex]v=\frac{m}{d}=\frac{5.582\times 10^{13} g}{1.03 g/mL}=5.420\times 10^{13} mL[/tex]
1 mL = 0.001 L
[tex]v = 5.420\times 10^{10} L[/tex]
The volume of seawater needed to extract [tex]8.0\times 10^4[/tex] tons of magnesium is [tex]5.420\times 10^{10} L[/tex].
