Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in the manufacture of chemicals. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Referring to Problem 1.87, calculate the volume of seawater (in liters) needed to extract 8.0 x 10^4 tons of Mg, which is roughly the annual production in the United States.

6.0 * 10¹⁰ L

Explanation:

We know that 1 ton = 10⁶ g. 8.0 * 10⁴ tons of Mg are:

8.0 * 10⁴ ton * (10⁶ g / 1 ton) = 8.0 * 10¹⁰ g

There is about 1.3 g of Mg for every kilogram of seawater. The mass of seawater that contains 8.0 * 10¹⁰ grams of Mg is:

8.0 * 10¹⁰ g Mg * (1 kg seawater / 1.3 g Mg) = 6.2 * 10¹⁰ kg seawater

The average density of seawater is 1.025 kg/L. The volume corresponding to 6.2 * 10¹⁰ kg is:

6.2 * 10¹⁰ kg * (1 L / 1.025 kg) = 6.0 * 10¹⁰ L

the volume of sea water is 85866.407 L, necessary to extract 8.0 E4 tons of Mg

Explanation:

%p/p = 1.3 g Mg / 1000g seawater) * 100 = 0.13 %

⇒ 8.0 E4 ton Mg * (0.00110231 Kg / ton) = 88.1848 Kg Mg

⇒ 0.13 % = (88.1848 Kg Mg / Kg seawater) * 100

⇒ (0.13 / 100) = 88.1848 / Kg seawater

⇒ Kg seawater = 88.1848 / 1 E-3

⇒ Kg seawater = 88184.8 Kg

⇒ Vseawater = mass seawater / density seawater

∴ density seawater = 1027 Kg / m³ ... from literature

⇒ V seawater = 88184.8 Kg / 1027 Kg/m³

⇒ V seawater = 85.866 m³ * (1000L / m³)

⇒ V seawater = 85866.407 L