600ml of a mixture of $O_{3}$ and $O_{2}$ weighs 1g at NTP. Calculate the volume of ozone in the mixture. 105ml of pure water $(4^{\circ} C)$ is saturated with $N H_{3}$ gas, producing a solution of density 0.9g/ml.

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Solution

Let us consider the volume of $O_{2} = x$ ml
At $S T P$ , weight of $x$ ml $O_{2} = \frac{32 \times x}{22400} g m$
Weight of $(600 - x)$ ml of $O_{3} = \frac{48 \times (600 - x)}{22400} g m$
$∴$ $\frac{32 x}{22400} + \frac{48 \times (600 - x)}{22400} = 1$
$\Rightarrow 32 x + 28800 - 48 x = 22400$
$\Rightarrow x = 400$ mL
$∴$ Hence the volume of $O_{3} = 600 - 400 = 200$ ml

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