Question

In one experiment $100mL$ of ozonised oxygen was reduced to $60mL$ when treated with turpentine oil. What would be increase in volume if the original sample was heated until no further change occured [All volumes are measured under identical condition].

• A. $10mL$
• B. $20mL$
• C. $30mL$
• D. $40mL$

Answer 1

The correct option is A $10mL$

Given : In one experiment 100 mL100 mL of ozonised oxygen was reduced to 60 mL60 mL when treated with turpentine oil. What would be increase in volume if the original sample was heated until no further change occured [All volumes are measured under identical condition].

Solution :

100 ml of mixture contain $O_3$ and $O_2$.

Turpentine oil absorb only $O_3$.

Since reduction in volume is 20 mL indicates that mixture contain 20 mL of $O_3$. It means 100 mL of mixture contains 20 ml $O_3$ and 80 ml $O_2$. .

On Heating $O_3$ decomposed as follows :

2$O_3$ $\to$​ 3 $O_2$

2 ml $O_3$ gives 3 ml $O_2$.

Therefore 20 mL of would give 30 mL of $O_2$.

​Total volume of mixture after heat$O_3$ng

​80 mL of $O_2$. + 30 mL of $O_2$. = 110 mL

​The total increase in volume = 110mL-100 mL

​ = 10 mL

The Correct Opt = A