Question

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

Answer 1

• The chemical formula of Aluminium oxide is ${\mathrm{Al}}_2{\mathrm{O}}_3$.

$$\begin{gathered} {\text{1 mole of Al}}_2{\mathrm{O}}_3=2\times 27+3\times 16 \\ =102\mathrm{g} \\ \mathrm{Moles}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3=\frac{0.051}{102} \\ =5\times {10}^{-4}\mathrm{moles} \\ 1\mathrm{mole}\mathit{=}6.02\times {10}^{23}\mathrm{molecules} \\ \therefore 5\times {10}^{-4}\mathrm{g}\text{of}{\mathrm{Al}}_2{\mathrm{O}}_3=5\times {10}^{-4}\times 6.02\times {10}^{23} \\ =3.011\times {10}^{20}\text{molecules} \end{gathered}$$

Number of ${\mathrm{Al}}^{3+}$ ions in 1 molecule of ${\mathrm{Al}}_2{\mathrm{O}}_3$ is 2.

Thus, the number of Aluminium ions in 0.051 g of ${\mathrm{Al}}_2{\mathrm{O}}_3$ will be,

$2\times 3.011\times {10}^{20}=6.022\times {10}^{20}$