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  = $\frac{given mass in g}{molar mass}$

Given mass of gold  =$6 \times 10^{9} g$

Molar mass of gold  = 196.97 g/mol

Put the values,

Number of moles of gold  = $\frac{6 \times 10^{9} g}{196.97 g/mol}$

= $0.03046\times 10^{9} mole$ or $3.046\times 10^{7} moles$

Now, molarity  = $\frac{moles of solute}{volume of the solution in liters}$

Put the values, volume of ocean  =$1.5 \times 10^{21} L$

Molarity = $\frac{3.046\times 10^{7} moles}{1.5 \times 10^{21} L}$

= $2.03\times 10^{-14} M$

Thus, average molar concentration = $2.03\times 10^{-14} M$