Search any question & find its solution
Question:
Answered & Verified by Expert
Liquids A and B form an ideal solution. The vapour pressure of $A$ and $B$ are 50 and $32 \mathrm{~mm} \mathrm{Hg}$ respectively at $300 \mathrm{~K}$. One mole of liquid $\mathrm{A}$ is mixed with 1 mole of liquid B. What is the approximate mole fraction of $\mathrm{A}$ in vapour phase?
Options:
Solution:
1384 Upvotes
Verified Answer
The correct answer is:
0.61
$\begin{aligned} & \text { } \mathrm{P}_{\mathrm{A}}=50 ; \mathrm{P}_{\mathrm{B}}=32 \\ & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}}=\frac{50}{50+32}=\frac{50}{82}=0.609 \approx 0.61\end{aligned}$
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.