# 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 = $\frac{Mass}{MolarMass}$

So, Molar Mass of Ozonised gas = $\frac{32g/mole}{0.95x2}$ = 33.32 g/mol

Molar Mass of O

_{2}= 32g/mol

Molar Mass of Ozone, O

_{3}= 48 g/mol

Since, we have a mixture of O

_{2}and O

_{3}, therefore we will equate it with that of the mixture of ozonised oxygen

So, the Mole Fraction of O

_{2}be A, and that of O

_{3}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

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