Gaseous StateHard
Question
1120.0 mL of ozonized oxygen at S.T.P. weighs 1.76 g. Calculate the volume of oxygen in the ozonized oxygen.
Options
A.996 mL
B.896 mL
C.1120 mL
D.784 mL
Solution
n(O2 + O3) = 
mol = 0.05 mol.
& nO2 + nO3 = n(O2 + O3)⇒ nO2 + nO3 = 0.05 ........ (1)
32nO2 + 48 nO3 = 1.76
⇒ 2nO2 + 3nO3 = 0.11 ........ (2)
From (1) & (2) nO2 = 0.04 mol nO3 = 0.01 mol
Vol of O2 = 22400 ml × 0.04 = 896 ml
mol = 0.05 mol.& nO2 + nO3 = n(O2 + O3)⇒ nO2 + nO3 = 0.05 ........ (1)
32nO2 + 48 nO3 = 1.76
⇒ 2nO2 + 3nO3 = 0.11 ........ (2)
From (1) & (2) nO2 = 0.04 mol nO3 = 0.01 mol
Vol of O2 = 22400 ml × 0.04 = 896 ml
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