# Question #a351e

##### 1 Answer

#### Explanation:

The trick here is to recognize that a **unified atomic mass unit**, or

#"207.2 u" = "207.2 g mol"^(-1)#

which tells you that **mole** of lead has a mass of

#11.35 color(red)(cancel(color(black)("g"))) * "1 mole Pb"/(207.2color(red)(cancel(color(black)("g")))) = "0.0548 moles Pb"#

Now all you have to do is to use the fact that you have

#color(blue)(ul(color(black)("1 mole Pb" = 6.022 * 10^(23)color(white)(.)"atoms Pb"))) -># Avogadro's constant

to find the number of atoms present in

#0.05478 color(red)(cancel(color(black)("moles Pb"))) * (6.022 * 10^(23)color(white)(.)"atoms Pb")/(1color(red)(cancel(color(black)("mole Pb")))) = 3.299 * 10^(22)color(white)(.)"atoms Pb"#

Therefore, you can say that the density of lead in *atoms per cubic centimeter* is equal to

#3.299 * 10^(22)color(white)(.)"atoms cm"^(-3)#

The answer is rounded to four **sig figs**.