# 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. How many grams of gold are in the oceans?

Sep 5, 2016

Over 8 million kilograms; i.e. $8 \times {10}^{9} \cdot g$

#### Explanation:

We need to find the volume of the ocean in ${m}^{3}$, and then multiply this volume by the average concentration in $g \cdot {L}^{-} 1$ knowing that there are $1000 \cdot L$ in a ${m}^{3}$.

$\text{Volume of the oceans}$ $=$ $3.63 \times {10}^{11} \cdot {m}^{2} \times 3800 \cdot m$ $=$ $1.38 \times {10}^{15} \cdot {m}^{3}$.

$\text{Mass of gold}$ $=$ $\text{Volume "xx" concentration}$

$=$ $1.38 \times {10}^{15} \cdot \cancel{{m}^{3}} \times 5.8 \times {10}^{-} 9 \cdot g \cdot \cancel{{L}^{-} 1} \times 1000 \cdot \cancel{L} \cdot \cancel{{m}^{-} 3}$

$=$ $8004000000$ $g$

$=$ $8004000 \cdot k g$

Anyway, go over my figures carefully. There is a lot of arithmetic here.