Question

On electrical decomposition of $150$ml dry pure $O_2$, $10\%$ of $O_2$ gets changed to $O$, then the volume of gaseous mixture after reaction and volume of remaining gas left after passing in turpentine oil will be:

• A. $145$ ml
• B. $149$ ml
• C. $128$ ml
• D. $105$ ml

Answer 1

Answer: (D)

$105$ ml

Let us assume $1$ mole of $O_2=150ml$

$O_2\rightleftharpoons 20$

$0.1$ $0.2$

Number of moles of $O_2$ remaining= $1-0.1=0.9$

Number of moles of $O$ produced= $0.2$

$O_2+O\rightleftharpoons O_3$ $\Rightarrow$ Initial volume $\Rightarrow (0.7+0.2)\times 150=145ml$

$0.9$ $0.2$ $0$

$0.7$ $0$ $0.2$

$\Rightarrow$ Entire $O_3$ is absorbed by turpentine oil.

$\Rightarrow 0.7$ mole of $O_2$ is left $\Rightarrow \left(0.7\right)\times 150=105ml$