Question

Ratio of effusion of a sample of ozonized oxygen is $0.95$ times the ratio of pure oxygen. Determine the percentage of ozone in the ozonized sample.

Answer 1

Diffusion rates are inversely proportional to the square root of the molar masses.

Since the ozonised gas diffuses slower than $O_2$, it must have a higher molar mass which is reasonable since ozone is $O_3$

MW ozonised gas = $32g$/mole /$0.98^2$ =$33.32g/mole$

You have a mixture of $O_2$ with a molar mass of 32 g/mole and ozone with a molar mass of 48 g/mole

Let X be the mole fraction that is O2. 1 - X is the mole fraction that is ozone

$X\ast 32g/mole+(1-X)\ast 48g/mole=33.32g/mole$

Solve for X.

$(1-X)\ast 100$ = % $O_3$ in the mixture