Gold occurs in seawater at an average concentration of \(1.1 \times 10^{-2}\) ppb. How many liters of seawater must be processed to recover 1 troy ounce of gold, assuming \(81.5 \%\) efficiency \((d\) of seawater \(=1.025 \mathrm{~g} / \mathrm{mL} ; 1\) troy ounce \(=31.1 \mathrm{~g}) ?\)

The term ‘parts per billion’ (ppb) is a unit of measurement used to describe the concentration of a substance in a mixture. When we say that gold is found in seawater at a concentration of 1.1 x 10^{-2} ppb, it means that for every billion parts of seawater, there are 1.1 x 10^{-2} parts of gold.
Imagine you have a billion tiny droplets of seawater, only 1.1 x 10^{-2} of those are gold. It’s a very tiny amount, but when you have vast oceans, it can add up.
In this exercise, the gold concentration is first given in ppb, which needs to be converted into grams per liter (g/L), a more practical unit for calculations.
This requires a series of mathematical conversions, first translating ppb into grams per milliliter (g/mL) and then to grams per liter (g/L) because 1 ppb is equivalent to 1 x 10^{-9} grams per milliliter.

A troy ounce is a unit of measure used for precious metals like gold, silver, and platinum. It is not the same as a regular ounce. One troy ounce is approximately 31.1 grams. This is important when calculating gold recovery because the final target amount of gold to be recovered is given in troy ounces.
In our exercise, we need to recover 1 troy ounce of gold. To do efficient calculations, we convert this amount to grams, giving us the value 31.1 grams.
Understanding this conversion is crucial because all subsequent calculations will use this gram value instead of troy ounces.

Efficiency in this context refers to how effectively the process can extract gold from seawater. Given an efficiency of 81.5%, it means that only 81.5% of the target amount of gold can actually be recovered from seawater.
To calculate how much gold we actually need to aim for, considering inefficiency, we use the formula:
\[ \text {Recovered Gold (grams)} = \frac{\text{Target Gold (grams)}}{\text{Efficiency}} \]
Here, the initial target is 31.1 grams. Given an 81.5% efficiency, we calculate:
\[ \frac{31.1 \text{ grams}}{0.815} = 38.16 \text{ grams} \]
Hence, we need to aim for 38.16 grams of gold to ensure that 31.1 grams will be effectively recovered. This takes into account the losses in the process.

Unit conversion is at the heart of this exercise. We have to align different measurement units to make accurate calculations. Initially, gold concentration is given in parts per billion (ppb), which we convert to grams per liter (g/L) to match our needs.
Let's break down the steps:
1. Convert ppb to grams per milliliter (g/mL):
\[ 1 \text{ ppb} = 1 \times 10^{-9} \text{g/mL} \]
So, 1.1 x 10^{-2} ppb is: \[1.1 \times 10^{-2} \times 10^{-9} \text{ g/mL} = 1.1 \times 10^{-11} \text{g/mL} \]
2. Convert grams per milliliter to grams per liter (g/L):
Knowing that 1 mL = 0.001 L: \[1.1 \times 10^{-11} \text{ g/mL} = 1.1 \times 10^{-8} \text{ g/L}\]
This unit alignment ultimately helps us conclude that for 38.16 grams of gold at a concentration of 1.1 x 10^{-8} grams per liter, we need a substantial volume of seawater.
The volume calculation then follows as:
\[ V = \frac{38.16 \text { grams}}{1.1 \times 10^{-8} \text { g/L}} = 3.47 \times 10^{9} \text{ L} \]
So, to recover 1 troy ounce of gold, assuming 81.5% efficiency, we need to process about 3.47 x 10^9 liters of seawater.