Question

Calculate the number of aluminium ions $\left(A{l}^{3+}\right)$ present in 0.102 g of aluminium oxide $(A{l}_2{O}_3$).
[Molar mass of Al = 27 g/mol; Molar mass of O = 16 g/mol]

Answer 1

Molar mass of $A{l}_2{O}_3$ = 102 g/mol
Moles of $A{l}_2{O}_3$ = $\frac{Givenmass\left(m\right)}{Molarmass\left(M\right)}$ = ${10}^{-3}$ moles

As, 1 mole of $A{l}_2{O}_3$ contains 6.022 $\times$ ${10}^{23}$ molecules of $A{l}_2{O}_3$.
Therefore, ${10}^{-3}$ moles of $A{l}_2{O}_3$ contains ${10}^{-3}$ $\times$ 6.022 $\times$ ${10}^{23}$ molecules of $A{l}_2{O}_3$.

Also, single molecule of $A{l}_2{O}_3$ contains two $A{l}^{3+}$ ions.
Therefore, ${10}^{-3}$ $\times$ 6.022 $\times$ ${10}^{23}$ molecules of $A{l}_2{O}_3$ contains 2 $\times$ ${10}^{-3}$ $\times$ 6.022 $\times$ ${10}^{23}$ $A{l}^{3+}$ ions.

Hence, 0.102 g of aluminium oxide $(A{l}_2{O}_3$) contains 12.044 $\times$ ${10}^{20}$ number of $A{l}^{3+}$ ions.