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At $300 \mathrm{~K}$ and 760 torr pressure, the density of a mixture of $\mathrm{He}$ and $\mathrm{O}_2$ gases is $0.543 \mathrm{gL}^{-1}$.
The mass percent of oxygen approximately is $\left(\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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The mass percent of oxygen approximately is $\left(\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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Verified Answer
The correct answer is:
$80$
$\mathrm{PV}=\mathrm{nRT}$ or $\mathrm{PV}=\mathrm{RT}$ (for $1 \mathrm{~mole}$ )
$\Rightarrow \mathrm{V}=\frac{\mathrm{RT}}{\mathrm{P}}=\frac{(0.0821)(300)}{1}(760 \text { torr }=1 \mathrm{~atm})$
$=24.63 \mathrm{~L}$ (of the mixture)
$\Rightarrow$ Total mass of the mixture :
$\begin{aligned}
& \mathrm{M}=\mathrm{V} \times \mathrm{d}=24.63 \mathrm{~L} \times 0.543 \mathrm{gL}^{-1} \\
& =13.37 \mathrm{~g}
\end{aligned}$
Mole fraction of $\mathrm{O}_2=\mathrm{x}$
Mole fraction of $\mathrm{He}=1-\mathrm{x}$.
Average molecular weight
$\begin{aligned}
& =\left[\left(\mathrm{x} \times \mathrm{M}_{\mathrm{O}_2}\right)+(1-\mathrm{x}) \mathrm{M}_{\mathrm{He}}\right] \\
& =[32 \mathrm{x}+(1-\mathrm{x}) 4] \\
& =13.37 \\
& \Rightarrow \mathrm{x}=0.334 \\
& \Rightarrow \text { Mass of } \mathrm{O}_2=0.334 \times 32=10.688 \\
& \Rightarrow \text { Mass } \% \text { of } \mathrm{O}_2=\frac{10.688}{13.37} \times 100 \\
& \cong 80 \%
\end{aligned}$
$\Rightarrow \mathrm{V}=\frac{\mathrm{RT}}{\mathrm{P}}=\frac{(0.0821)(300)}{1}(760 \text { torr }=1 \mathrm{~atm})$
$=24.63 \mathrm{~L}$ (of the mixture)
$\Rightarrow$ Total mass of the mixture :
$\begin{aligned}
& \mathrm{M}=\mathrm{V} \times \mathrm{d}=24.63 \mathrm{~L} \times 0.543 \mathrm{gL}^{-1} \\
& =13.37 \mathrm{~g}
\end{aligned}$
Mole fraction of $\mathrm{O}_2=\mathrm{x}$
Mole fraction of $\mathrm{He}=1-\mathrm{x}$.
Average molecular weight
$\begin{aligned}
& =\left[\left(\mathrm{x} \times \mathrm{M}_{\mathrm{O}_2}\right)+(1-\mathrm{x}) \mathrm{M}_{\mathrm{He}}\right] \\
& =[32 \mathrm{x}+(1-\mathrm{x}) 4] \\
& =13.37 \\
& \Rightarrow \mathrm{x}=0.334 \\
& \Rightarrow \text { Mass of } \mathrm{O}_2=0.334 \times 32=10.688 \\
& \Rightarrow \text { Mass } \% \text { of } \mathrm{O}_2=\frac{10.688}{13.37} \times 100 \\
& \cong 80 \%
\end{aligned}$
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