Earths oceans have an average depth of 3800 m, a total area of 3.63 * 10^8 km^2, and an average concentration of dissolved gold of 5.8*10^-9 g/L. If a recent price of gold was $883.00 troy/oz, what is the value of gold in the ocean? (1 troy oz=31.1g)

Question

Luca Marquez

Mathematics

20 December, 05:07

Earths oceans have an average depth of 3800 m, a total area of 3.63 * 10^8 km^2, and an average concentration of dissolved gold of 5.8*10^-9 g/L. If a recent price of gold was $883.00 troy/oz, what is the value of gold in the ocean? (1 troy oz=31.1g)

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Answers (1)

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Giuliana Case · Answer 1

1 m² = 1000 L

Volume = 3800 m * 3.63 * 10 ^14 m² = 13,794 * 10 ^14 m³ =

= 13,794 * 10^17 L

Mass (gold) = 13,794 * 10 ^17 L * 5.8 * 10 ^ (-9) g / L =

= 80,005.2 * 10 ^12 g : 31.1 = 2,572.5 * 10 ^12 troy oz

The value of gold in the ocean:

$883.00 * 2,572.5 * 10^12 = $2271517.5 * 10 ^12 ≈ $ 2.27 * 10^18