Question

volume of o2(g) and o3(g) obtained by 8 kg oxygen atoms at STP??

Answer 1

Mass of 1 oxygen atom = 16 g.

Mass of O₂ = 32 g.

Mass of O₃ = 48 g.

Now,

16 g of Oxygen atom combines to form 32 g of O₂.

∴ 1 g of O atom combines to forms 2 g of O₂.

∴ 8000 g of O atom combines to form 16000 g of O₂.

Similarly, Mass of O₃ (Ozone) formed = 3 ×  8000 g. = 24000 g.

Now,

1 mole of Oxygen at S.T.P. occupies the Volume of 22.4 liter.

∴ 32 g of Oxygen at S.T.P. occupies the volume of 22.4 liter.

∴ 1 g of Oxygen at S.T.P. occupies the volume of 22.4/32 liter.

∴ 16000 g of Oxygen at S.T.P. occupies the volume of 11200 liter.

Hence, the volume of oxygen O₂ at S.T.P. is 11200 liter.

1 mole of Ozone at S.T.P. occupies the volume of 22.4 liter. 

∴ 48 g of Ozone at S.T.P. occupies the volume of 22.4 liter. 

∴ 1 g of Ozone at S.T.P. occupies the volume of 22.4/48 liter. 

∴ 24000 g of Ozone at S.T.P. occupies the volume of 22.4/48 × 24000 liter. 
  = 11200 liter.

Hence, the volume of ozone formed is 11200 liter.