Calculate the number of aluminium ions present in $0.051g$ of aluminium oxide.

The mass of an ion is the same as that of an atom of the same element. Atomic mass of $Al=27u$.

Moles of $A{l}_2{O}_3$=$\frac{0.051}{102}$

Number of molecules of $A{l}_2{O}_3=\frac{0.051}{102}\times 6.023\times {10}^{23}=3\times {10}^{20}$

1 molecule of $A{l}_2{O}_3$ contains 2 aluminium ions

so, $3\times {10}^{20}contains6\times {10}^{20}$ aluminium ions