Calculate the number of aluminium ions present in 0.051 g of aluminium oxide (Atomic mass of Al = 27u).

1 mole of aluminium oxide $\left(A{l}_2{O}_3\right)=2\times 27+3\times 16$
= 102 g
i.e., 102 g of $A{l}_2{O}_3=6.022\times {10}^{23}$ molecules of $A{l}_2{O}_3$
Then, 0.051 g of $A{l}_2{O}_3$ contains =
$\frac{6.022\times {10}^{23}}{102}\times 0.051$ molecules
$=3.011\times {10}^{20}$ molecules of $A{l}_2{O}_3$
The number of aluminium ions $\left(A{l}^{3+}\right)$ present in one moleucle of aluminium oxide is 2.
Therefore, the number of aluminium ions $\left(A{l}^{3+}\right)$ present in $3.011\times {10}^{20}$ molecules (0.051 g) of aluminium oxide $\left(A{l}_2{O}_3\right)=2\times 3.011\times {10}^{20}$
$=6.022\times {10}^{20}$