What is the volume of #2.3 * 10^25# atoms of gold if its density is #19.3##g##/##cm^3#?
1 Answer
Answer:
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
So, if you need
#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
#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)#