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

Mass of $A{l}_2{O}_3=0.051g$

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

$=5\times {10}^{-4}$ moles

$A{l}_2{O}_3⟶2A{l}^{3+}+3O^{2-}$

As one molecule of aluminium oxide contains 2 Aluminium ions.

Therefore, No. of Al ions in $A{l}_2{O}_3$ $=2\times 5\times {10}^{-4}\times 6.02\times {10}^{23}$

$=6.02\times {10}^{20}ions$