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 `Al_(2)O_(3) = "Formula mass of" Al_(2)O_(3) "in grams"`
`= "Mass of Al" xx 2 + "Mass of O" xx 3`
`= 27 xx 2 + 16 xx 3`
`= 102` grams
Now, 1 mole of `Al_(2)O_(3)` contains 2 moles of Al.
So, Mas of `Al` in 1 mole of `Al_(2)O_(3) =` Mass of `Al xx 2`
` = 27 xx 2`
`= 54` grams
Now, `102 g` aluminimum oxide contains `= 54 g Al`
So, `0.051` g aluminium oxide contains `= (54)/(102) xx 0.051 g Al`
The atomic mass of aluminium is given to be `27 u`. This mass that 1 mole of aluminium atoms (or alunimum ions) has a mass of ` 27` grams, and it contains `6.022 xx 10^(23)` aluminium ions.
Now, `27 g` of aluminium has ions ` = 6.022 xx 10^(23)`
So, `0.027 g` of aluminium has ions ` = (6.022 xx 10^(23))/(27) xx 0.027`
`= 6.022 xx 10^(20)`
thus, the number of aluminium ions `(Al^(3+))` in `0.051` grams of aluminium oxide is `6.022 xx 10^(20)`.
