Question

Number of Al atoms present in 255g of Al2O3

Answer 1

No of Al2O3 atom = (mass / mol mass) *Avogadro no.

(Mol wt of Al2O3=27*2+16*3=102)

= (255/102)*(6.023*10^23)

No of Al2O3 atoms = 1.505*10^24

Now, No of Al2 atoms =1.505*10^24 Therefore, No of Al atom =7.52*10^23