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600 ml of a mixture of $$\text{O}_2$$ and $$\text{O}_3$$ weighs 1 gram at NTP. The volume of ozone (in ml) in the mixture _______.
Correct Answer: 200
Let the volume of ozone be $$x$$ mL. One mole occupies 22400 mL at NTP, so $$\frac{32(600 - x) + 48x}{22400} = 1$$. This gives $$19200 + 16x = 22400$$ and $$x = 200\text{ mL}$$.
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