Question

$40mL$ gaseous mixture of $CO,C{H}_4$ and Ne was exploded with $20mL$ of oxygen. On cooling, the gases occupied $46.5mL$. After treatment with $KOH$, the volume reduced by $9mL$ and again on treatment with alkaline pyrogallol, the volume further reduced. Percentage of $C{H}_4$ in the original mixture is:

• A. 22.5 %
• B. 77.5 %
• C. 7.5 %
• D. 15 %

Answer 1

The correct option is D 15 %

Let, the volume of $CO,C{H}_4$ and Ne be x, y and z respectively
$\underset{x}{CO}+\underset{x/2}{\frac{1}{2}{O}_2}\to \underset{x}{C{O}_2}\underset{y}{C{H}_4}+\underset{2y}{2O_2}\to \underset{y}{C{O}_2}+2H_{2}O\left(l\right)$
Remaining volume of $O_2=20-\frac{x}{2}-2y$
Volume after reaction : $x+y+20-\frac{x}{2}-2y+z=46.5$
$\frac{x}{2}-y+z=26.5$ mL ...(i)
$x+y+z=40$ (given) ...(ii)
$KOH$ absorbs $C{O}_2$ gas only

$2\underset{\left[Base\right]}{KOH}+\underset{\left[Acid\right]}{C{O}_2}\to K_{2}C{O}_3+H_{2}O$

So, $x+y=9mL$ ...(iii)
from equation (ii) and (iii),
$Z=31mL$ (volume of $Ne$)
By putting the value of Z in equation (i),
$\frac{x}{2}-y=-4.5\Rightarrow x-2y=-9$ ...(iv)
from equation (iii) and (iv),
$x-2y=-9$
$x+y=9$
___________
$-3y=-18\Rightarrow y=6mL$ (volume of $C{H}_4$)
So, from equation (iii) by putting the value of y
x = 3 mL (volume of CO)
Total volume of gaseous mixture=40 mL
So, % of $C{H}_4=6/40\times 100\Rightarrow 15\%$