Question

$90mL$ of pure dry $O_2$ is subjected to silent electric discharge. If only $10$ of it is converted to $O_3$, volume of the mixture of gases ($O_2$ and $O_3$) after the reaction will be ------ and after passing through turpentine oil will be .....

• A. $84mL$ and $78mL$
• B. $81mL$ and $87mL$
• C. $78mL$ and $84mL$
• D. $87mL$ and $81mL$

Answer 1

Answer: (B)

$81mL$ and $87mL$

$3O_2⟶2O_3$

Given volume of $O_2$ which is converting into $O_3$ is 10%
Now we should convert 90ml in 10%=$\frac{90\times 10}{100}=9ml$.

After reaction(90-9)$=81ml$.

$\frac{2}{3}\times 9=6ml$ of $O_3$ volume of the mixture of the gases ( $O_2$ and $O_3)=81+6=87$ ml.
Turpentine oil volume will be $=87$ mL

hence the option (B) is correct.