What is the volume of 2.3⋅1025 atoms of gold if its density is 19.3g/cm3?
Answers
convert the number of atoms of gold to moles of gold by using Avogadro's number
use gold's molar mass to convert the number of moles to number of grams
use gold's density to determine what volume would contain that many grams
As you know, one mole of any element contains exactly
6.022
⋅
10
23
atoms of that element - this is known as Avogadro's number.
So, if you need
6.022
⋅
10
23
atoms of gold to have one mole of gold, then the number of atoms you have will be equivalent to
2.3
⋅
10
25
atoms Au
⋅
1 mole Au
6.022
⋅
10
23
atoms Au
=
38.2 moles Au
Now, gold's molar mass will tell you what the mass of one mole of gold is. In this case, that many moles would correspond to a mass of
38.2
moles Au
⋅
196.97 g
1
mole Au
=
7524.3 g
Finally, a density of
19.3 g/cm
3
tells you that
1 cm
3
of gold will have a mass of
19.3
grams. In your case, the volume that would have a mass of
7524
grams is
7524.3
g
⋅
1 cm
3
19.3
g
=
389.9 cm
3
Rounded to two sig figs, the number of sig figs you have for the number of atoms of gold, the answer will be
V
=
390 cm
3