Answer:
[tex]6.022x10^{23}atoms \ Al[/tex]
Explanation:
Hello,
In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:
[tex]27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al[/tex]
Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.
Best regards.