Question

The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. $H_{2}S{O}_4$ are used to absorb the selective gases like $O_2,O_3,C{O}_2\text{and}{H}_{2}O\left(v\right)$ respectively. A $2.0$ mole mixture of $H_2\left(g\right),O_2\left(g\right)\text{and}He\left(g\right)$ are placed together in a closed container at pressure equal to $50\text{atm}$. An electric spark is passed and pressure is noted as $12.5\text{atm}$ after cooling the content to original temperature (room temp). Oxygen gas is introduced for pressure to change to $25\text{atm}$, keeping volume and temperature constant. Again electrical spark is passed and pressure drops to $10\text{atm}$ under original temperature conditions. The moles of $He\left(g\right)$ present in the mixture sample is:

• A. $0.2$
• B. $0.1$
• C. $0.5$
• D. $1.4$

Answer 1

The correct option is B $0.1$

Let the number of moles of $H_2,O_2$ and $\text{He are x, y}$ and $\text{z}$ respectively
$\therefore x+y+x=2.0$
Pressure exerted by 2 moles of gaseous mixture is $50\text{atm}.$
After the first electric spark decrease in pressure = $50–12.5=37.5\text{atm}$
$\therefore$ decrease in no. of moles of gaseous mixture is $1.5$
But from the given information limiting reagent in first step is $O_2$
$\therefore \text{from}{H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$
$\therefore H_2\text{and}{O}_2$ reacted are in the ratio of $2:1$
$\therefore 0.5\text{moles of}{O}_2\text{should be reacted with}1.0\text{mole of}{H}_2$
$\therefore \text{No. of mole of}{O}_2\text{in the mixture (y)}=0.5$
When again $O_2$ is passed, pressure increases from 12.5 to 25.0 $\text{atm}$
$\therefore$ Change in pressure = 12.5 $\text{atm}$
$\therefore$ No. of moles of $O_2$ added = $\frac{12.5\times 2}{50}=0.5\text{mole}$
Now, $H_2$ will be completely reacted
No. of moles $H_2$ actually left after first electric spark = $x-1$
$H_2+\frac{1}{2}{O}_2⟶H_{2}O$
$(x-1)\left(\frac{x-1}{2}\right)$
Now change in pressure in $25-10=15\text{atm}$
$\therefore$ change in no of moles is $\frac{15\times 2}{50}=0.6$
$\therefore (x-1)+\left(\frac{x-1}{2}\right)=0.6$
$\Rightarrow x=1.4$
$\therefore$ no. of moles of $H_2=1.4$
$\therefore$ no. of moles of $He=2.0-(1.4+0.5)=0.1$