Calculate the number of aluminium ions (Al3+) in 0.056 g of alumina (Al2O3).

Molecular mass of alumina (Al₂0₃) = 2 × Al³⁺ + 3 × 0^2-

= 2 × 27 u + 3 × 16u

= 102 u

Gram molecular mass = 102 g

1 mol of alumina (Al₂0₃) = 102 g

102 g of Al₂O₃ = 1 mol

∴ 0.056 g of Al₂O₃ = ( 1 × 0.056/102) mol

= 5.49 x 10^-4 mol

We know that one mol of alumina contains 2 mol of Al³⁺ ions.

∴ 5.49 x 10^-4 mol of Al₂O₃ contains 2 x 5.49 x 10^-4 mol of Al³⁺ ions

∴ Number of Al³⁺ ions in 0.056 g = 2 × 5.49 × 10^-4 x 6.022 x 10²³

                                                          = 6.613 x 10²⁰ ions of AI³⁺