Question

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

Answer 1

`"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 * 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) 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 `"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 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 = color(green)("390 cm"^3)`