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Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 28 January, Morning Shift — Question 46

At T(K),2\mathrm{T}(\mathrm{K}), 2 moles of liquid A and 3 moles of liquid BB are mixed. The vapour pressure of ideal solution formed is 320 mm Hg . At this stage, one mole of A and one mole of BB are added to the solution. The vapour pressure is now measured as 328.6 mm Hg . The vapour pressure (in mm Hg ) of A and B are respectively:

  1. Option A:

    300, 200

  2. Option B:

    600, 400

  3. Option C:

    400, 300

  4. Option D:

    500, 200

    Correct

Answer: D

Step-by-step solution

2 moles of A+3\mathrm{A}+3 moles of B

XA=2/5,XB=3/5\mathrm{X}_{\mathrm{A}}=2 / 5, \mathrm{X}_{\mathrm{B}}=3 / 5

PS=XAPA∘+XBPB∘P_{S}=X_{A} P_{A}^{\circ}+X_{B} P_{B}^{\circ} 320=PA∘(25)+PB∘(35)320=\mathrm{P}_{\mathrm{A}}^{\circ}\left(\frac{2}{5}\right)+\mathrm{P}_{\mathrm{B}}^{\circ}\left(\frac{3}{5}\right)

2 \mathrm{P}_{\mathrm{A}}^{\circ}+3 \mathrm{P}_{\mathrm{B}}^{\circ}=1600 \end{gathered}$$ Now 1 mole of $A$ & 1 mole of $B$ is added $\mathrm{X}_{\mathrm{A}}^{\prime}=\frac{3}{7}, \mathrm{X}_{\mathrm{B}}^{\prime}=\frac{4}{7}$ $\mathrm{P}_{\mathrm{S}}^{\prime}=328.6=\mathrm{P}_{\mathrm{A}}^{\circ}\left(\frac{3}{7}\right)+\mathrm{P}_{\mathrm{B}}^{\circ}\left(\frac{4}{7}\right)$ $$\begin{gathered} 3 \mathrm{P}_{\mathrm{A}}^{\circ}+4 \mathrm{P}_{\mathrm{B}}^{\circ}=2300.2 \end{gathered}$$ Now eq (I) $\times 3-$ eq (II) $\times 2$ $6 \mathrm{P}_{\mathrm{A}}^{\circ}+9 \mathrm{P}_{\mathrm{B}}^{\circ}=4800$ $6 \mathrm{P}_{\mathrm{A}}^{\circ}+8 \mathrm{P}_{\mathrm{B}}^{\circ}=4600.4$ $\mathrm{P}_{\mathrm{B}}^{\circ} \simeq 200 \mathrm{~mm}$ of Hg $\mathrm{P}_{\mathrm{A}}^{\circ} \simeq 500 \mathrm{~mm}$ of Hg

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Liquid in Liquid Solutions (Raoult's Law)