How many aluminum atoms in a #0.45*mol# quantity of this metal?

1 Answer
anor277
Dec 14, 2017

#0.45*molxx"Avogadro's number"#

Explanation:

And #"Avogadro's number"-=6.022xx10^23*mol^-1#

And so there are.....

#6.022xx10^23*mol^-1xx0.45*mol~=3.0xx10^23# #"individual aluminum atoms."#

What is the mass of this number of atoms?

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