A #2.72*g# mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

1 Answer
Feb 10, 2016

Answer:

Avogadro's number of #Al# atoms has a mass of #26.98# #g#. So you tell us.

Explanation:

If there are #6.022# #xx# #10^23# individual aluminum atoms, there is a mass of #26.98# #g# of aluminum. #6.022# #xx# #10^23# individual aluminum atoms represents #1# #mol# of aluminum.

So, #(2.72*cancelg)/(26.98*cancelg*mol^-1)# #=# #??# #mol#?

Whatever this molar quantity is, we multiply this by Avogadro's number, #6.022# #xx# #10^23# #mol^-1#, to get the number of aluminum atoms.