# What is the volume of 2.3 * 10^25 atoms of gold if its density is 19.3g/cm^3?

Dec 7, 2015

${\text{390 cm}}^{3}$

#### Explanation:

Your strategy here will be to

• 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 $6.022 \cdot {10}^{23}$ atoms of that element - this is known as Avogadro's number.

So, if you need $6.022 \cdot {10}^{23}$ atoms of gold to have one mole of gold, then the number of atoms you have will be equivalent to

2.3 * 10^(25) color(red)(cancel(color(black)("atoms Au"))) * "1 mole Au"/(6.022 * 10^(23) color(red)(cancel(color(black)("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 color(red)(cancel(color(black)("moles Au"))) * "196.97 g"/(1 color(red)(cancel(color(black)("mole Au")))) = "7524.3 g"

Finally, a density of ${\text{19.3 g/cm}}^{3}$ tells you that ${\text{1 cm}}^{3}$ 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 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(19.3 color(red)(cancel(color(black)("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

$V = \textcolor{g r e e n}{{\text{390 cm}}^{3}}$