Question

2. An atom of gold has a mass of 3.271 x 10-22g. How many atoms of gold are in 3.00 cubic millimeters of gold?

Answer 1

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:

`\rho(Densisty)=(Mass)/(Volume)`
`\begin{gathered} Mass=Volume*\rho(Densisty) \\ Mass=3.00mm^3*19.3(g)/(mL) \end{gathered}`

We will use the following conversion factors:

1mL=1000mm^3

1atom=3.271x10^-22g

So, the atoms present in 3.00mm^2 will be:

`Atoms=Mass*(1atomAu)/(3.271*10^(-22)gAu)`
`Atomsau=3.00mm^3*(19.3g)/(mL)*(1mL)/(1000mm^3)*(1atomAu)/(3.271*10^(-22)gAu)`
`Atoms=1.77*10^(20)`

In 3.00mm^3 of gold, there are 1.77x10^20 atoms

Answer: 1.77x10^20 atoms