Questions: Earth's oceans have an average depth of 3800 m, a total area of 3.63 x 10^8 km^2, and an average concentration of dissolved gold of 5.8 x 10^-9 g/L. If a recent price of gold was 423.00 / troy oz, what is the value of gold in the oceans? (1 troy oz = 31.1 g) Be sure your answer has the correct number of significant figures.

Solution Steps

• Convert the total area of the oceans from \(\text{km}^2\) to \(\text{m}^2\): \[ 3.63 \times 10^{8} \, \text{km}^2 = 3.63 \times 10^{14} \, \text{m}^2 \]
• Calculate the volume of the oceans using the average depth: \[ \text{Volume} = \text{Area} \times \text{Depth} = 3.63 \times 10^{14} \, \text{m}^2 \times 3800 \, \text{m} = 1.3794 \times 10^{18} \, \text{m}^3 \]
• Convert the volume from \(\text{m}^3\) to \(\text{L}\) (since \(1 \, \text{m}^3 = 1000 \, \text{L}\)): \[ \text{Volume} = 1.3794 \times 10^{21} \, \text{L} \]

• Use the concentration of gold to find the total mass of gold: \[ \text{Mass of gold} = \text{Volume} \times \text{Concentration} = 1.3794 \times 10^{21} \, \text{L} \times 5.8 \times 10^{-9} \, \frac{\text{g}}{\text{L}} \] \[ \text{Mass of gold} = 7.99952 \times 10^{12} \, \text{g} \]

• Convert the mass of gold from grams to troy ounces (since \(1 \, \text{troy oz} = 31.1 \, \text{g}\)): \[ \text{Mass of gold in troy oz} = \frac{7.99952 \times 10^{12} \, \text{g}}{31.1 \, \text{g/troy oz}} = 2.571 \times 10^{11} \, \text{troy oz} \]
• Calculate the value of the gold using the price per troy ounce: \[ \text{Value of gold} = 2.571 \times 10^{11} \, \text{troy oz} \times \$423.00/\text{troy oz} \] \[ \text{Value of gold} = 1.087 \times 10^{14} \, \text{USD} \]

Final Answer