Question

$5\text{g}$ of a gaseous mixture of the $He$ and $Ar$ contained in a $10\text{L}$ vessel at $27°\text{C}$ exert a pressure of $1\text{atm}$ (absolute). Determine the mass percentage of $He$.
(Take $\text{R}=\frac{1}{12}{\text{L atm mol}}^{-1}{\text{K}}^{-1}$)

• A. $10.8\%$
• B. $14.2\%$
• C. $24.4\%$
• D. $74.6\%$

Answer 1

The correct option is C $24.4\%$

Let the mass of $He$ be $x\text{g}$, mass of $Ar$ will be $(5-x)\text{g}$
Moles $=\frac{\text{mass}}{\text{molas mass}}$
Moles of $He=\frac{x}{4}$
Moles of $Ar=\frac{5-x}{40}$
Total moles of gas $=\frac{x}{4}+\frac{5-x}{40}$
$PV=nRT\Rightarrow n=\frac{PV}{RT}=\frac{1\times 10}{1/12\times 300}=0.4\Rightarrow \frac{x}{4}+\frac{5-x}{40}=0.4\Rightarrow x=1.22\text{g}$
Mass percentage of Helium $=\frac{1.22}{5}\times 100=24.4\text{\%}$
Mass percentage of Argon $=75.6\text{\%}$