Rate of diffusion of pure oxygen was found to be 0.95 times the mixture of ionosed oxygen. Calculate the %of ozone in the mixture

Dear Student


Rate of Diffusion of the mixture of ozonized oxygen = x

So, Rate of Diffusion of Pure Oxygen will be 0.95x


Now, the % of ozone in the mixture = ?


We know, 

rate of diffusion is inversely proportional to the square root of the molar mass

Since, ozonised oxygen diffuses at a slower rate than that of oxygen, therefore the former one will have a higher molar mass

We know, 
No. of Moles = $\frac{Mass}{Molar Mass}$

So, Molar Mass of Ozonised gas = $\frac{32 g/mole}{0.95 x 2}$ = 33.32 g/mol
Molar Mass of O₂ = 32g/mol

Molar Mass of Ozone, O₃ = 48 g/mol



Since, we have a mixture of O₂ and O₃ , therefore we will equate it with that of the mixture of ozonised oxygen

So, the Mole Fraction of O₂ be A, and that of O₃ be 1 - A

Thus, 

(A x 32 g/mol) + [(1 - A) x 48 g/mol] = 33.32 g/mol

​32A g/mol + 48 g/mol - 48A g/mol = 33.32 g/mol

32A g/mol - 48A g/mol = (33.32 - 48) g/mol

-16A g/mol = - 14.68 g/mol

​or, 16A  = 14.68 

​A = $\frac{14.68}{16}$ = 0.9175

Therefore, % of A = 0.9175 x 100% = 91.75 %



Regards