Question #a351e

1 Answer
Stefan V.
Dec 4, 2017

#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.

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