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 gets 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,
Volume of $CO=x$ mL
Volume of $C{H}_4=y$ mL
Volume of $Ne=40-x-y$ mL
Again,
$CO+\frac{1}{2}{O}_2\to C{O}_2$
volume of $C{O}_2$ formed from $CO=x$ mL
Loss of $O_2=\frac{1}{2}x$ mL
$C{H}_4+2O_2\to C{O}_2+2H_{2}O$
volume of $C{O}_2$ formed from $C{H}_4=y$ mL

Volume after cooling,
$46.5=(40-x-y)+(20-\frac{x}{2}-2y)+(x+y)$
$\Rightarrow 46.5=60-\frac{x}{2}-2y$
$\Rightarrow x+4y=27\dots \dots \left(i\right)$

Volume loss due to treatement with $KOH$
$x+y=9mL\dots \dots \left(ii\right)$
Subtracting equation (ii) from (i),
$3y=18\Rightarrow y=6\text{mL}$
$\therefore x=3\text{ml}$
$\%$ of $C{H}_4=\frac{6}{40}\times 100=15\%$