Earth’s oceans have an average depth of 3800 m, a total surface area of 3.63×108km2, and an average concentration of dissolved gold of 5.8×10−9g/l.
assuming the price of gold is $1595/troy oz, what is the value of gold in the oceans
(1 troy oz 5 31.1 g; d of gold 5 19.3 g/cm3)?

The value of gold in the oceans is 4.07x10¹⁴ $

Explanation:

A problem with relation between the units.

This are the data, 3800 m (depth of the ocean)

The total surface 3.63×10⁸ km² (total surface)

5.8×10⁻⁹g/L (gold concentration)

$1595 (value of each troy oz)

31.1 g (mass of each troy oz in grams)

19.3 g/cm³ (density of gold)

As we have a depth (a kind of height) and the total surface we can know the volume that the ocean occupies. This height is in m, the surface in km².

We should convert eveything in dm to work with concentration.

3800 m to cm = 3800 . 10 = 3800 → 3.8 x10⁴ dm

3.63×10⁸ km² to dm² = 3.63×10⁸ . 1x10⁸ = 3.63×10¹⁶ dm²

(1 km² = 1x10⁸dm)

3.8 x10⁴ dm  . 3.63×10¹⁶ dm² = 1.37 x10²¹ dm³

This is the volume that the ocean occupies. By using concentration, we can know the mass of gold in all the ocean.

1L = 1 dm³

1L 5.8×10⁻⁹g

1.37 x10²¹ L 7.9 x10¹² g

So 1 troy oz pays $1595 and 1 troy oz is 31.1 grams, so 31.1 grams pay $1595.

The final rule of three will be

31.1 g __ pay ___ $1595

7.9 x10¹² g ___ pay (8 x10¹¹ g . $1595) / 31.1 g = 4.07x10¹⁴ $

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