Question

calculate the number of Al ions present in 0.051g of aluminium oxide.​

Answer 1

Answer:

aluminium oxide = Al2O3

molecular mass of Al2O3 = 102 grams

now, 1 mole of Al2O3 contains 2 moles of Al

so, mass of Al in 1 mole of Al2O3 = 54g

Now, 102g aluminium oxide contains= 54g Al

so, 0.051g aluminium oxide contains 54/102 x 0.051g Al

= 0.027

the atomic mass of aluminium is 27u . This means that 1 mole of Al atoms has mass of 27g ,

which contains= 6.022x10^23 Al ions.

Now, 27g Al = 6.022×10^23

so, 0.027 g of Al = 6.022x10^23/27 x 0.027

= 6.022x10^20

Thus, the number of Al ions in 0.051g of aluminium oxide = 6.022x10^20