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

1 Answer
Dec 7, 2015

Answer:

#"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)#