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At $\mathrm{T}(\mathrm{K})$, a gaseous mixture of $\mathrm{H}_2$ and $\mathrm{O}_2$ containing $20 \%$ (weight/weight) of $\mathrm{H}_2$ exerts a total pressure of 2 bar. What is the partial pressure of $\mathrm{O}_2$ (in bar)?
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The correct answer is:
0.4
$\mathrm{P}_{\text {total }}=2$ bar, $\mathrm{P}_{\mathrm{O}_2}=\mathrm{x}_{\mathrm{O}_2} \mathrm{P}_{\text {total }}$
For $100 \mathrm{~g}$ of this gaseous mixture, $\mathrm{W}\left(\mathrm{H}_2\right)=20 \mathrm{~g}$, W( $\left.\mathrm{O}_2\right)$ $=80 \mathrm{~g}$.
$\Rightarrow \mathrm{n}_{\mathrm{H}_2}=\frac{20}{2}=10 \mathrm{~mol}, \mathrm{n}_{\mathrm{O}_2}=\frac{80}{32}=2.5 \mathrm{~mol}$
$\Rightarrow \mathrm{x}_{\mathrm{O}_2}=\frac{2.5}{2.5+10}=0.2$
$\Rightarrow \mathrm{P}_{\mathrm{O}_2}=0.2 \times 2=0.4 \mathrm{bar}$
For $100 \mathrm{~g}$ of this gaseous mixture, $\mathrm{W}\left(\mathrm{H}_2\right)=20 \mathrm{~g}$, W( $\left.\mathrm{O}_2\right)$ $=80 \mathrm{~g}$.
$\Rightarrow \mathrm{n}_{\mathrm{H}_2}=\frac{20}{2}=10 \mathrm{~mol}, \mathrm{n}_{\mathrm{O}_2}=\frac{80}{32}=2.5 \mathrm{~mol}$
$\Rightarrow \mathrm{x}_{\mathrm{O}_2}=\frac{2.5}{2.5+10}=0.2$
$\Rightarrow \mathrm{P}_{\mathrm{O}_2}=0.2 \times 2=0.4 \mathrm{bar}$
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