$600$ mL of a mixture of $O_{3}$ and $O_{2}$ weighs $1$ g at NTP. Calculate the volume of ozone in the mixture.

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Solution

Let $V m l$ be the volume of ozo0ne
volume of oxygen $(600 - v)$
Mass of ozone $= 22400 m l$
Hence the mass of $V m l$ of ozone $= 48$
$= \frac{V}{22400}$
$\frac{32 \times (600 - V)}{22400}$
$48 = \frac{32 \times (600 - v)}{22400}$
$48 \times V + 32 \times (600 - V) = 22400$
$V = 200 m l$

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