Question

Gold occurs in sea water at an average concentration of 1.1*10^-2 PPb. How many liters of sea water must be processed to recover one troy ounce of gold, assuming 81.5% efficiency (d of seawater = 1.025g/mol ; 1 troy ounce = 31.1g)?

Short answer

3.38*10⁹L of sea water must be processed to recover one troy ounce of gold.

Step-by-step solution

moles of Gold

To calculate the mole of gold, first, calculate the efficiency in ppb by using the given concentration of gold.Multiply the concentration of gold with given % efficiency divided by 100.

$efficiency=1.1\times {10}^{-2}ppb\times \frac{81.5}{100}$

=8.965$\times {10}^{-3PPb}$

Now we can calculate the moles of solute by using the following formula;

$ppb=\frac{\left(massofsolute\right(g\left)\right)}{\left(massofsolution\right(g\left)\right)}\times {10}^9$

Here, solute is gold and solution is sea water.

Hence,

$$\begin{gathered} massofseawater=\frac{massofAu}{ppb}\times {10}^9 \\ =\frac{31.1g}{8.965\times {10}^{-3}}\times {10}^9 \\ =3.47\times {10}^{12}g \end{gathered}$$

volume of sea water

Given the density of sea water is 1.025g/mL.

To calculate the volume of sea water we can use the following formula which shows relation between volume, density and mass.

Hence,

$$\begin{gathered} volumeofseawater=\frac{massofseawater}{densityofseawater} \\ =\frac{3.47\times {10}^{12}g}{1.025g/mL}=3.38\times {10}^{12}mL \\ =3.38\times {10}^{9}L \end{gathered}$$