Calculate the number of aluminium ions present in 0.051g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g

i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃

Then, 0.051 g of Al₂O₃ contains = {6.022×10^{23}\over102} × 0.051 molecules

= 3.011 × 10²⁰ molecules of Al₂O₃

The number of aluminium ions (Al³⁺ ) present in one molecules of aluminium oxide is 2.

Therefore, The number of aluminium ions (Al³⁺ ) present in 3.11 × 10²⁰ molecules (0.051g) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰

= 6.022 × 10²⁰