Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A closed container contains mixture of non-reacting gases A and B. Partial pressure of $A$ and $B$ are 4.5 bar and 5.5 . bar respectively. Find mole fraction of $\mathrm{A}$ and $\mathrm{B}$ respectively.
ChemistryStates of MatterMHT CETMHT CET 2023 (13 May Shift 2)
Options:
  • A 0.035 and 0.065
  • B 0.055 and 0.045
  • C 0.45 and 0.55
  • D 0.55 and 0.45
Solution:
1556 Upvotes Verified Answer
The correct answer is: 0.45 and 0.55
$\begin{aligned} \mathrm{P}_{\text {Total }}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}} & =4.5+5.5 \\ & =10.0 \mathrm{bar}\end{aligned}$
$\mathrm{P}_{\mathrm{A}}=x_{\mathrm{A}} \times \mathrm{P}_{\text {Total }}$
$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {Total }}}=\frac{4.5}{10}=0.45$
$x_{\mathrm{B}}=\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\text {Total }}}=\frac{5.5}{10}=0.55$

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.