Question

Magnesium occurs in seawater to the extent of 1.4g magnesium per kilogram of seawater. The annual production of magnesium in the US is about 105 tons. If all of the Magnesium were extracted from seawater, what volume of seawater, in m3, would have to be processed? Assume density of seawater to be 1.025g/cm3.

Answer 1

10^5 tons in g = 10^5 x 9071985 g = 9.07 x 10^11 g Mg in seawater

Now :

1.4 g Mg ---------------------->1 kg of water

9.07 x 10^11 ------------------> 9.07 x 10^11 x 1 / 1.4 = 6.48 x 10^11 kg of water = 6.48 x 10^14 g

      denisty = 1.025 g / cm^3

volume = mass / density

V =    6.48 x 10^14 / 1.025

V = 6.3 x 10^7 m^3   ------------------------------> answer