Question

Earth's oceans have an average depth of 3800 m, a total surface area of 3.63 x 108 km2, and an average concentration of dissolved gold of 5.8 x 10?9 VI_ (a) How many grams of gold are in the oceans? {b} How many cubic meters of gold are in the oceans? (c). Assuming the price of gold is $1595/troy oz, what is the value of gold in the oceans (1 tray oz = 311 g: dof gold =19.3 g/cm3)?

Answer 1

Answer:

a) Mass of gold in ocean = 8.1*10¹² g

b) Volume of gold = 4.2*10¹¹ cm³

c) Value of gold = $ 4.1*10¹³

Explanation:

a) Depth of ocean = 3800 m

Surface area of ocean = 3.63*10⁸ km²

Now, 10⁶ m² = 1 km²

Therefore, the area in units of m² = 3.63*10¹⁴ m²

$Volume\ of\ ocean = Depth*Area = 3800m*3.63*10^{14} m2=1.4*10^{18} m3$

Average concentration of gold in ocean = 5.8*10⁻⁹ g/L

Now, 1 g/L = 1000 g/m³

Therefore, concentration of gold in g/m³ = 5.8*10⁻⁶ g/m³

$Mass\ of\ gold =1.4*10^{18} m^{3} *5.8*10^{-6} g/m3 = 8.1*10^{12} g$

b) Mass of gold present = 8.1*10¹² g

Density of gold = 19.3 g/cm³

$Volume\ of\ gold = \frac{Mass}{density} =\frac{8.1*10^{12} }{19.3} =4.2*10^{11} cm^{3}$

c) Price of gold = $1595/troy oz

Unit conversion

1 troy oz = 311 g

Therefore, 8.1*10¹² g of gold is equivalent to:

$\frac{8.1*10^{12} g* 1\ troy\ oz}{311g} =2.6*10^{10} \ troy\ oz$

The value of gold in the ocean is:

$\frac{2.6*10^{10} troy\ oz*1595\ dollars}{1\ troy\ oz} =4.1*10^{13} \ dollars$