Question

672 ml of ozonized oxygen $(0_3+0_2)$ were found to weight 1g Calculate volume of oronized oxygen.

Answer 1

56 ml

Answer 2

Let $x$ be the volume (in ml) of ozone $(O_3)$ in the mixture. Then, the volume of oxygen $(O_2)$ is $672 - x$ ml.

At STP, $22.4\,L = 22400\,ml$ is the molar volume.

• Molar mass of $O_3 = 48\,g$
• Molar mass of $O_2 = 32\,g$

The mass contributed by ozone is:

$$\text{Mass of } O_3 = \frac{x \times 48}{22400}\,g$$

The mass contributed by oxygen is:

$$\text{Mass of } O_2 = \frac{(672 - x) \times 32}{22400}\,g$$

The total mass is given as 1 g, so:

$$\frac{48x}{22400} + \frac{32(672 - x)}{22400} = 1$$

Multiply through by $22400$:

$$48x + 32(672 - x) = 22400$$

Simplify:

$$48x + 21504 - 32x = 22400$$ $$16x + 21504 = 22400$$ $$16x = 22400 - 21504 = 896$$ $$x = \frac{896}{16} = 56 \, ml$$

Thus, the volume of ozone in the ozonized oxygen is 56 ml.