# Question a351e

Dec 4, 2017

$3.299 \cdot {10}^{22}$ ${\text{atoms cm}}^{- 3}$

#### Explanation:

The trick here is to recognize that a unified atomic mass unit, or $\text{u}$, is equivalent to ${\text{1 g mol}}^{- 1}$. This means that you have

${\text{207.2 u" = "207.2 g mol}}^{- 1}$

which tells you that $1$ mole of lead has a mass of $\text{207.2 g}$. Since you know that lead has a density of ${\text{11.35 g cm}}^{- 3}$, you can say that ${\text{1 cm}}^{3}$ 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 mole Pb" = 6.022 * 10^(23)color(white)(.)"atoms Pb}}}} \to$ Avogadro's constant

to find the number of atoms present in ${\text{1 cm}}^{3}$ 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

$3.299 \cdot {10}^{22} \textcolor{w h i t e}{.} {\text{atoms cm}}^{- 3}$

The answer is rounded to four sig figs.