Answer:
[tex]2.19 \times 10^{-4} \mathrm{Rh}[/tex]
Explanation:
Atomic mass of Nickel = 58.69 g/mol
Mass of 1 mole of nickel atom = 58.69 gm
Now, mass of 1 nickel atom = Gram atomic mass/Avogadro number
=[tex]\frac{58.69}{6.023\times10^{22}} =9.744\times10^{23} g/atom[/tex]
Now, price of Rhodium =$1500 per troy ounce
Price of nickel in market  = 0.577$/oz
No. of rhodium atoms needed to buy 1 nickel atom is
[tex]9.744 \times 10^{-23} \mathrm{g} \times \frac{1 \mathrm{oz}}{31.1035 \mathrm{g}} \times \frac{0.577 \mathrm{s}}{1 \mathrm{oz}} \times \frac{1 \mathrm{oz}}{1500 \mathrm{s}}[/tex]
[tex]\quad \times \frac{31.1035 \mathrm{g}}{1 \mathrm{oz}} \times \frac{6.023 \times 10^{23} \mathrm{Rh} \text { atoms }}{102.9 \mathrm{g}}=2.19 \times 10^{-4} \mathrm{Rh} \text { atoms }[/tex]
