Chemistry · Solutions and Colligative Properties
JEE Main 2026 — 23 January, Evening Shift — Question 63
Two liquids A and B form an ideal solution. At 320 K , the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg . Vapour pressure (in mm Hg ) of B in the pure state is . (Nearest integer)
Answer: 200
Numerical answer — enter this value.
Step-by-step solution
3 \mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}=2000 \end{gathered}$$ Now 1 moles of A is further added so $\mathrm{n}_{\mathrm{A}}=4$ mole, $\mathrm{n}_{\mathrm{B}}=1$ mole $\mathrm{X}_{\mathrm{A}}^{\prime}=\frac{4}{5}, \mathrm{X}_{\mathrm{B}}^{\prime}=\frac{1}{5}$ $\mathrm{P}_{\mathrm{s}}=520=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \times \frac{4}{5}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \times \frac{1}{5}$ $$\begin{gathered} 4 \mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}=2600 \end{gathered}$$ By equation (2) - equation (1) $\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}=600 \mathrm{~mm} \mathrm{Hg}$ $\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}=200 \mathrm{~mm} \mathrm{Hg}$
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- Exam
- JEE Main 2026
- Subject
- Chemistry
- Chapter
- Solutions and Colligative Properties
- Topic
- Liquid in Liquid Solutions (Raoult's Law)