The weight of 600 ml of a mixture of ozone and oxygen is 1 gm at STP. The volume of oxygen in the mixture is:

200 m l

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500 m l

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400 m l

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300 m l

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Solution

The correct option is B $400 m l$
Let us say volume of $O_{2} = x m o l$

Hence the volume of $O_{3} = (600 - x) m o l$

At STP, weight of $x m o l$ $O_{2} = \frac{22 \times x}{22400} g m$

Weight of $(600 - x) m o l O_{3} = \frac{48 \times (600 - x)}{22400} g m$

now, $\frac{32 x}{22400} + \frac{48 \times (600 - x)}{22400} = 1 32 x + 28800 - 48 x = 22400 x = 400 m o l$

So, the correct answer is option $C .$

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