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.

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Solution

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 = $32 g$ /mole / $0.9 8^{2}$ = $33.32 g / m o l e$

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

$X * 32 g / m o l e + (1 - X) * 48 g / m o l e = 33.32 g / m o l e$

Solve for X.

$(1 - X) * 100$ = % $O_{3}$ in the mixture

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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.

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.