Question

Question #a351e

Answer 1

`3.299 * 10^(22)` `"atoms cm"^(-3)`

Explanation:

The trick here is to recognize that a unified atomic mass unit, or `"u"`, is equivalent to `"1 g mol"^(-1)`. This means that you have

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

which tells you that `1` mole of lead has a mass of `"207.2 g"`. Since you know that lead has a density of `"11.35 g cm"^(-3)`, you can say that `"1 cm"^3` of lead will contain

`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 `"1 cm"^3` of lead.

`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.