The conversion of oxygen to ozone occurs to the extent of 15% the mass of ozone that can be prepared from 67.2 L of oxygen at 1 atm and 273 K will be:

14.4 gm

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96 gm

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640 gm

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64 gm

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Solution

The correct option is B 14.4 gm
$3 O_{2} \to 2 O_{3}$

As per stoichiometric equation, $3 \times 22.4 L$ of oxygen is converted to $2 \times 22.4 L$ of ozone. But given 15 % of oxygen is converted to ozone.

That means 67.2 L of oxygen used $\frac{15}{100} \times 67.2 \times \frac{2}{3} = 6.72 L$ ozone is produced.

The mass of ozone present in $22.4 L = 48 g$

$10.08 L$ of ozone means $= \frac{48}{22.4} \times 6.72 = 14.4 g$

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