Question

A gaseous mixture of helium and oxygen is found to have a density of $0.5{\text{g dm}}^{-3}$ at $25{}^{\circ }\text{C}$ and $0.662atm$. What is the percentage by mass of helium in this mixture?
(Given: $R=0.08Latmmo{l}^{-1}{K}^{-1},\frac{5960}{331}=18$)

Answer 1

Let the mean molar mass of the mixture be $=M$.
From ideal gas law,
we know that, $PM=dRT$
Where, $d=\text{density of the gas mixture}$

$\therefore 0.662\times M=0.5\times 0.08\times 298$
$\Rightarrow M=18\text{g}$
Let the mole fraction of $He$ in the mixture be $\alpha$.
So, ${\chi }_{He}=\alpha ,{\chi }_{O_2}=(1-\alpha )$
$\text{Average molar mass}$
$M=\alpha \times M_{He}+(1-\alpha )M_{O_2}$
$\Rightarrow 18=\alpha \times 4+(1-\alpha )32$
$\Rightarrow \alpha =\frac{14}{28}=0.5$
$\therefore 1-\alpha =0.5$

$\text{\% by mass of}He=\frac{\text{mass of helium}}{\text{Total mass of gases}}\times 100$
$=\frac{0.5\times 4}{0.5\times 4+0.5\times 32}\times 100$
$=11.11\%$