Question

how many atoms of aluminium are present in 0.051 gram of aluminium oxide

Answer 1

Molecular weight of Aluminium oxide (Al₂O₃) = (2*27) + (3*16) = 54 + 48
                                                                                                = 102
102 grams of Al2O3 contains 6.02*10²³ number of molecules
0.051 grams Al2O3 contains ‘X’ number of molecules
X = (0.051*6.02*10²³)/102
  = 0.003 *10²³ = 3*10²⁰
0.05 grams of Al2O3 contains 3*10²⁰ number of molecules. 
Each 1 molecule of Al2O3 contains – 2 Al⁺³ ions
3*10²⁰ number of molecules contains – 2 * 3*10²⁰ number of Al⁺³ ions
6*10²⁰ number of Al⁺³ ions.
Therefore 0.051 grams of Al2O3 contains 6*10²⁰ number of Al⁺³ ions.

Answer 2

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