Question

When $100ml$ of a $O_2-O_3$ mixture was passed through turpentine, there was reduction of volume by $20mL$. If $100ml$ of such a mixture is heated, what will be the increase in volume?

• A. 10%
• B. 20%
• C. 40%
• D. 30%

Answer 1

The correct option is A 10%

Given,

Total volume of $O_2-O_3$ mixture$=100mL$

Reduction in volume$=20mL$

We know that $O_3$ gets absorbed in turpentine oil, therefore

Volume of $O_3$ that gets absorbed $=20mL$

Volume of $O_2=(100-80)mL=20mL$

On heating, the following reaction takes place between $O_3$ and $O_2$

${20}_3\rightleftharpoons 3O_2$

Since volume is proportional to moles, therefore

${20}_3\rightleftharpoons 3O_2$

$2mol$ $3mol$

$20mL$ $30mL$

Therefore, increase in volume will be $=(30-20)mL=10mL$

Hence, percentage increase in volume will be = Increase in volume/Total volume *100

$\Rightarrow \frac{10mL}{100mL}\times 100$

$\Rightarrow 10$%

The correct answer will be option A.