$600 m$ of a mixture of $O_{3}$ and $O_{2}$ weight $1 g$ at $N T P$ . Calculate the volume of ozone in the mixture.

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Solution

The molecular mass of oxygen= 32
The molecular mass of ozone=48
Therefore,
Let volume of $O_{2} = x m l$
volume of $O_{3} = 600 - x m l$
At STP, wt of x ml of $O_{2} = \frac{32 \times x}{22400} g m$
wt of $600 - x m l O_{3} = \frac{48 \times (600 - x)}{22400} g m$
Hence, $\frac{32 \times x}{22400} + \frac{48 \times (600 - x)}{22400} = 1$
Therefore, volume of ozone in the mixture is $400 m l$

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