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
The vapour pressure of a solvent decrease by $2.5 \mathrm{~mm} \mathrm{Hg}$ by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is $250 \mathrm{~mm} \mathrm{Hg}$ )
ChemistrySolutionsMHT CETMHT CET 2021 (22 Sep Shift 1)
Options:
  • A $0.88$
  • B $0.01$
  • C $0.1$
  • D $0.99$
Solution:
2880 Upvotes Verified Answer
The correct answer is: $0.01$
$\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0}=\mathrm{X}_{\text {solute }} \\ & \mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}=2.5 \mathrm{~mm}, \mathrm{P}^{\circ}=250 \mathrm{~mm} \\ & \frac{2.5}{250}=\mathrm{X}_{\text {solute }} \\ & \mathrm{X}_{\text {solute }}=0.01\end{aligned}$

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.