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

This problem involves aluminium ions. Please note that the mass of an aluminium ion is the same as that of an aluminium atom. 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.