When 100 ml of a mixture of $O_2$ and $O_3$ was passed through turpentine, there was a reduction of volume by 20 ml. When 100 ml of such a mixture is heated, the volume increases to $(\text{22}\times \text{n})\text{ml}$. Find the value of n.

Ozone is highly soluble in turpentine. Thus, the decrease in volume of 20 ml is due to the solubility of ozone. Thus 100ml of a mixture contains 20 ml of ozone.

 

On heating, 20 ml of ozone decomposes to form 30 ml of oxygen. The net increase in volume will be (30−20)=10ml.

When 100 ml of such a mixture is heated, the volume increases to $(\text{22}\times 5)\text{ml}$. the value of n=5.