Answer :
To solve this problem we are going to assume that the temperature at which the gold is found is room temperature.
We are going to use a property of materials that relates mass to volume, this is density. At this temperature, the density of gold is 19.3g/mL, the equation of density is:
[tex]\rho(Densisty)=\frac{Mass}{Volume}[/tex][tex]\begin{gathered} Mass=Volume\times\rho(Densisty) \\ Mass=3.00mm^3\times19.3\frac{g}{mL} \end{gathered}[/tex]We will use the following conversion factors:
1mL=1000mm^3
1atom=3.271x10^-22g
So, the atoms present in 3.00mm^2 will be:
[tex]Atoms=Mass\times\frac{1atomAu}{3.271\times10^{-22}gAu}[/tex][tex]Atomsau=3.00mm^3\times\frac{19.3g}{mL}\times\frac{1mL}{1000mm^3}\times\frac{1atomAu}{3.271\times10^{-22}gAu}[/tex][tex]Atoms=1.77\times10^{20}[/tex]In 3.00mm^3 of gold, there are 1.77x10^20 atoms
Answer: 1.77x10^20 atoms