40 ml gaseous mixture of CO, CH_(4) and Ne was exploded with 10 ml of oxygen. On cooling, the gases occupied 36.5 ml. After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol, the volume further reduced, percentage of CH_(4) in the original mixture is

40 ml gaseous mixture of CO, CH_(4) and Ne was exploded with 10 ml of

1.	40 ml gaseous mixture of CO, CH_(4) and Ne was exploded with 10 ml of oxygen. On cooling, the gases occupied 36.5 ml. After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol, the volume further reduced, percentage of CH_(4) in the original mixture is
Answer» <html><body><p>a. `22.4`<br/><a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>. `77.5`<br/>c. `7.5`<br/>d. `15`</p>Solution :`underset(x)(CO)+(1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)underset((x)/(2))(O_(2))rarrunderset(x)(CO_(2))` <br/> `underset(y)(CH_(4))+underset(2y)(2O_(2))rarrunderset(y)(CO_(2))+2H_(2)O` <br/> On passing through KOH, volume decrease due to `CO_(2)` absorption <br/> `impliesx+y=9….(1)` <br/> Final volume after reaction = <br/> `CO_(2)+Ne+"left over "O_(2)` <br/> `=(x+y)+(40-x-y)+(10-x//2-2y)` <br/> `36.5=50-(x)/(2)-2y` <br/> `x+4y=27....(2)` <br/> solving, `y=6,x=3` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> %" of "CH_(4)=(6)/(40)xx100=15%`</body></html>