Question

About 6 × 109 g of gold is thought to be dissolved in the oceans of the world. If the total volume of the oceans is 1.5 × 1021 L, what is the average molar concentration of gold in seawater?

Answer 1

First, determine the number of moles of gold.

Number of moles =
`(given mass in g)/(molar mass)`

Given mass of gold =
`6 * 10^(9) g`

Molar mass of gold = 196.97 g/mol

Put the values,

Number of moles of gold =
`(6 * 10^(9) g)/(196.97 g/mol)`

=
`0.03046*  10^(9) mole` or
`3.046*  10^(7) moles`

Now, molarity =
`(moles of solute)/(volume of the solution in liters)`

Put the values, volume of ocean =
`1.5 * 10^(21) L`

Molarity =
`(3.046* 10^(7) moles)/(1.5 * 10^(21) L)`

=
`2.03* 10^(-14) M`

Thus, average molar concentration =
`2.03* 10^(-14) M`