Question

calculate the number of aluminium ions present in 0.051 g of aluminium oxide(atomic mass of aluminium=27u)

Answer 1

1 mole of aluminium oxide (Al2O3) = 2*27 + 3 * 16

= 102 g

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

Then, 0.051 g of Al2O3 contains =

= 3.011 * 1020 molecules of Al2O3

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

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

= 6.022 * 1020

Answer 2

Let us compute the molar mass of the molecule first

Mass of Al = 27 AMU
Mass of O = 16 AMU
So mass of Al2O3 "molecule" = 2x27 + 3x16 = 102 AMU

So 1 mole of  ALUMINIUM OXIDE will weigh 102 g
1 mole of any pure substance contains Avogrado's number of molecules, that is 6.023x10^23.
Thus 0.051g of ALUMINIUM OXIDE will contain 6.023x(10^23)x.051/102 = 6.023x(10^20)/2 "molecules "

Each molecule contains two ions of aluminium.

So total no. Of ions in the given amount of substance =  2x6.023x(10^20)/2 = 6.023x10^20