Calculate the number of aluminium ions present in 0.051 g of aluminium oxide (Al₂O₃). (Atomic masses: Al = 27 u; O = 16 u)

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{aluminium}\mathrm{oxide}=2\times 27+3\times 16 \\ =54+48 \\ =102 \end{gathered}$$

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{aluminium}\mathrm{oxide}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}} \\ =\frac{0.051}{102} \\ =0.0005\mathrm{mol} \end{gathered}$$

As 1 mole of aluminium oxide contains 2 moles of aluminium.

So, 0.0005 mole of aluminium oxide contains = 0.0005 ⅹ 2

0.0005 mole of aluminium oxide contains = 0.001 mole of Aluminium

As 1 mole contains 6.022 ⅹ 10²³ ions
Therefore, 0.001 moles of aluminium oxide = 6.022 ⅹ 10²³ ⅹ 0.001 ions
= 6.022 ⅹ 10²⁰ ions of aluminium

Hence, 0.051 g of aluminium oxide contains 6.022ⅹ10²⁰ ions of aluminium.