A sample of ozonised oxygen contains 90% by volume of oxygen. What will be the volume of the resulting gases when 120ml of such a sample is heated and then cooled to the original conditions of temperature and pressure ?

Volume of Ozonized oxygen = 120ml

Volume percentage of Oxygen = 90%

Volume of Ozonized oxygen = $\left(120\times \frac{90}{100}\right)=108g$

Vol. of Ozone=120-108=12g

 When heated and cooled all the Ozone converts to oxygen

$$\begin{gathered} 2{\mathrm{O}}_3\left(\mathrm{g}\right)\rightleftharpoons 3{\mathrm{O}}_2\left(\mathrm{g}\right) \\ 2\mathrm{v} 3\mathrm{v} \\ 12\mathrm{ml} ? \\ {\mathrm{V}}_{{\mathrm{O}}_2} \mathrm{formed}=12\times \frac{3}{2}=18\mathrm{ml} \end{gathered}$$

Volume of oxygen formed due to the decomposition of Ozone = 18ml
After the process only oxygen gas is left after the process= (108 + 18)ml
Volume of resulting gases (Oxygen) =126ml