# . Calculate the number of aluminium ions present in 0.051 g 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 = 27 u)

Ans. 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.