The rate of diffusion of a sample of ozonised oxygen is $0.98$ times more than that of pure oxygen. The percentage of ozone by volume in the ozone by volume in the ozonised sample is:

6.25 %

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7.25 %

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8.25 %

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10.4 %

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Solution

The correct option is C $10.4 %$
Molar mass of ozonized gas $= \frac{32}{0.98} = 33.65 g / m o l$ .
Let $x$ be the mole fraction of $O_{2}$ and $1 - x$ be the mole fraction of ozone.
$∴$ $x (32) + (1 - x) 48 = 33.65$
$x = 14.35 / 16 = 0.896$
% of $O_{3} = (1 - x) \times 100 = (1 - 0.896) \times 100 = 10.4$ %

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The rate of diffusion of a sample of ozonised oxygen is 0.98 times more than that of pure oxygen. The percentage of ozone by volume in the ozone by volume in the ozonised sample is:

The correct option is C 10.4%Molar mass of ozonized gas =320.98=33.65g/mol.Let x be the mole fraction of O2 and 1−x be the mole fraction of ozone.∴ x(32)+(1−x)48=33.65 x=14.35/16=0.896% of O3=(1−x)×100=(1−0.896)×100=10.4%

The rate of diffusion of a sample of ozonised oxygen is $0.98$ times more than that of pure oxygen. The percentage of ozone by volume in the ozone by volume in the ozonised sample is: