Question

calculate the number of aluminium ions in 0.051g of Al(2)O(3).

Answer 1

Molecular mass of Al2O3 = 27*2 + 16*3 = 102g/mol 

so moles in 0.051 = mass/Mm = 0.051/102 = 0.0005 moles 

2 Al ions per molecule, so 0.001 moles of Al ions 

number of ions = moles x NA where NA is avogadros number = 6.022x10^23 

so number of ions = 6.022x10^23 x 0.001 = 6.022x10^20 ions

Answer 2

1 mole of Al2O3=mass of Al*2+mass of O*3
=27*2+16*3=102grams
now; one mole of Al2O3 contains 2 moles of Al
so; mass of Al=27*2=54grams
now; 102g aluminium contains=54g Al
so; 0. 051 g aluminium contains=54/102*0. 051g Al
=0. 027g Al
now; atomic mass of aluminium is 27 and it cintains 6. 022*10^23
so; 0. 027g of aluminium has ions=6. 022*10^23/27*0. 027
=6. 022*10^22