Given data:Ore from gold mines contains about 0.02 wt.% gold.Seawater contains gold too, at about 5 parts per trillion.To find: The tons of seawater required to produce 1 oz of gold.Solution:1 ounce is equivalent to 0.0283495 kilograms. 1 ton is equivalent to 1000 kilograms.The concentration of gold in seawater is 5 parts per trillion.1 part per trillion is equivalent to 1 microgram per liter (μg/L) or 1 nanogram per liter (ng/L).So, 5 parts per trillion is equivalent to 5 ng/L.Therefore, the weight of gold per liter of seawater (d) is: d = (5 × 10⁻⁹ g) / (1 L)The amount of gold in 1 ton of seawater will be:Amount of gold = (d) × (1000 L) = (5 × 10⁻⁹ g/L) × (1000 L) = 5 × 10⁻⁶ g or 5 μgTo obtain 1 ounce of gold, the amount of seawater needed is:1 ounce = 28.3495 grams5 μg of gold is obtained from 1 ton of seawater. Therefore, the amount of seawater needed to obtain 28.3495 g of gold will be:Amount of seawater = (28.3495 g) / (5 × 10⁻⁶ g) = 5,669,900,000 or 5.67 × 10⁹ tonsHence, the tons of seawater required to produce 1 oz of gold are approximately 5.67 × 10⁹ tons.