What is the number of aluminium ions present in 0.051 g of aluminium oxide $\left(A{l}_2{O}_3\right)$?

The correct option is C $6.022\times {10}^{20}$
Molar mass of aluminium oxide $\left(A{l}_2{O}_3\right)=2\times 27+3\times 16=102g$
Number of molecules of $A{l}_2{O}_3$ in
0.051 g sample$=\frac{\mathrm{Given}\mathrm{mass}\times \mathrm{Avogadro}\text{'}s\mathrm{number}}{\mathrm{Molar}\mathrm{mass}}$$=\frac{0.051\times 6.022\times {10}^{23}}{102}=3.011\times {10}^{20}$

The number of aluminium ions $\left(A{l}^{3+}\right)$ present in one molecule 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 $\left(A{l}_2{O}_3\right)=2\times 3.011\times {10}^{20}$
$=6.022\times {10}^{20}$