The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given that (1) the amount of the reparations was \(\$ 28.8\) billion dollars, (2) the value of gold at the time was about \(\$ 21.25\) per troy ounce ( \(1 \text { troy ounce }=31.103 \mathrm{g}),\) and (3) gold occurs in seawater to the extent of \(4.67 \times 10^{17}\) atoms per ton of seawater \((1 \text { ton }=2000\) lb), how many cubic kilometers of seawater would have had to be processed to obtain the required amount of gold? Assume that the density of seawater is \(1.03 \mathrm{g} / \mathrm{cm}^{3}\) (Haber's scheme proved to be commercially infeasible, and the reparations were never fully paid.)

Step by step solution

01

Determine the total weight of gold required

First, find out the total weight of gold required to pay off the reparations. This can be done by dividing the total reparations by the price of gold per troy ounce, then converting that to grams: \[ Weight = \frac{\$28.8 billion}{\$21.25 per troy ounce}\times 31.103g per troy ounce \]

02

Find the number of atoms of gold required

Next, use Avogadro's Law to convert the weight from grams to atoms. Avogadro's number (\(6.022 \times 10^{23}\)) represents the number of atoms in 1 mole, which weighs the atomic weight in grams: \[ Atoms = Weight(g) \times \frac{1 mole}{197.0g} \times \frac{6.022 \times 10^{23} atoms}{1 mole} = Weight \times \frac{6.022 \times 10^{23}}{197.0} \] We use \(197.0 g\) as it is the atomic weight of gold.

03

Calculate the total tons of seawater required

To find the total tons of seawater required, divide the total number of atoms required by the number of atoms per ton of seawater: \[ Tons = \frac{Atoms}{4.67 \times 10^{17} atoms per ton} \]

04

Convert tons of seawater to cubic kilometers

Finally, convert tons to kilograms, then kilograms to cubic centimeters (using the density of seawater), and finally cubic centimeters to cubic kilometers: \[ Cubic \, kilometers = \, Tons \times \frac{1000 \, kg}{1 ton} \times \frac{1 \, cm^3}{1.03 g} \times \frac{1 \, m^3}{10^6 \, cm^3} \times \frac{1 \, km^3}{10^9 \, m^3} \]

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