Search any question & find its solution
Question:
Answered & Verified by Expert
At $\mathrm{T}(\mathrm{K})$, a gaseous mixture containing $\mathrm{H}_2, \mathrm{He}$ and $\mathrm{O}_2$ exerted a pressure of 1 bar. The weight percentage of $\mathrm{H}_2$ and $\mathrm{He}$ is 20 and 16 respectively. The partial pressure (in bar) of $\mathrm{H}_2, \mathrm{He}$ and $\mathrm{O}_2$ is respectively
Options:
Solution:
1420 Upvotes
Verified Answer
The correct answer is:
$0.625,0.250,0.125$
$P_{\text {total }}=1 \mathrm{bar}$, Let total mass $=100 \mathrm{~g}$
$\begin{aligned} & \Rightarrow \mathrm{W}\left(\mathrm{H}_2\right)=20 \mathrm{~g}, \mathrm{~W}(\mathrm{He})=16 \mathrm{~g}, \mathrm{~W}\left(\mathrm{O}_2\right)=64 \mathrm{~g} \\ & \Rightarrow \mathrm{n}\left(\mathrm{H}_2\right)=\frac{20}{2}=10 \mathrm{~mol} \\ & \mathrm{n}(\mathrm{He})=\frac{16}{4}=4 \mathrm{~mol} \\ & \mathrm{n}\left(\mathrm{O}_2\right)=\frac{64}{32}=2 \mathrm{~mol} \\ & \Rightarrow \mathrm{x}\left(\mathrm{H}_2\right)=\frac{10}{10+4+2}=\frac{10}{16}=0.625 \\ & \mathrm{x}(\mathrm{He})=\frac{4}{16}=0.250\end{aligned}$
$\begin{aligned} & x\left(\mathrm{O}_2\right)=\frac{2}{16}=0.125 \\ & \Rightarrow P\left(\mathrm{H}_2\right)=\mathrm{x}_{\mathrm{H}_2} \mathrm{P}_{\text {total }}=0.625 \times 1=0.625 \\ & \mathrm{P}(\mathrm{He})=0.25 \text { and } \mathrm{P}\left(\mathrm{O}_2\right)=0.125\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{W}\left(\mathrm{H}_2\right)=20 \mathrm{~g}, \mathrm{~W}(\mathrm{He})=16 \mathrm{~g}, \mathrm{~W}\left(\mathrm{O}_2\right)=64 \mathrm{~g} \\ & \Rightarrow \mathrm{n}\left(\mathrm{H}_2\right)=\frac{20}{2}=10 \mathrm{~mol} \\ & \mathrm{n}(\mathrm{He})=\frac{16}{4}=4 \mathrm{~mol} \\ & \mathrm{n}\left(\mathrm{O}_2\right)=\frac{64}{32}=2 \mathrm{~mol} \\ & \Rightarrow \mathrm{x}\left(\mathrm{H}_2\right)=\frac{10}{10+4+2}=\frac{10}{16}=0.625 \\ & \mathrm{x}(\mathrm{He})=\frac{4}{16}=0.250\end{aligned}$
$\begin{aligned} & x\left(\mathrm{O}_2\right)=\frac{2}{16}=0.125 \\ & \Rightarrow P\left(\mathrm{H}_2\right)=\mathrm{x}_{\mathrm{H}_2} \mathrm{P}_{\text {total }}=0.625 \times 1=0.625 \\ & \mathrm{P}(\mathrm{He})=0.25 \text { and } \mathrm{P}\left(\mathrm{O}_2\right)=0.125\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.