In order to solve this problem, first of all we have to find out the mass of aluminium atoms in 0.051 g of aluminium oxide (which will give us the mass of aluminium ions) This can be done as follows:

1 mole of Al₂O₃ = Formula mass of Al₂O₃ in grams

= Mass of Al × 2 + Mass of O × 3

= 27 × 2 + 16 × 3

= 54 + 48

= 102 grams

Now, 1 mole of Al₂O₃ contains 2 moles of Al.

So, Mass of Al in 1 mole of Al₂O₃ = Mass of Al × 2

= 27 × 2

= 54 grams

Now, 102 g aluminium oxide contains = 54 g Al

So, 0.051 g aluminium oxide contains = 54/102 × 0.051 g Al

= 0.027 g Al

The atomic mass of aluminium is given to be 27 u. This means that 1 mole of aluminium atoms (or aluminium ions) has a mass of 27 grams, and it contains 6.022 × 10²³ aluminium ions.

Now, 27 g of aluminium has ions = 6.022 × 10²³

So, 0.027 g of aluminium has ions = 6.022 × 10²³/27 × 0.027

= 6.022 × 10²⁰

Thus, the number of aluminium ions (Al³⁺) in 0.051 gram of aluminium oxide is 6.022 × 10²⁰.