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JEE Advanced 2026Paper 2
Exam
JEE Advanced
Year
2026
Paper
Paper 2
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2026 · Paper 2 · Chemistry · Solutions

Passage: Two volatile liquids A and B form an ideal solution....

Passage: Two volatile liquids A and B form an ideal solution. Consider a 55 molal solution of B in A inside a closed container having a total vapour pressure of 100100 mm Hg at 300300 K. The vapour pressure of pure A at 300300 K is 105105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.
Given:
The gas constant R=0.08R = 0.08 L atm K−1^{-1} mol−1^{-1}
Molar mass of A is 5050 g mol−1^{-1}
Molar mass of B is 5757 g mol−1^{-1}
Density of liquid B at 300300 K is 0.50.5 g/mL
11 atm =760= 760 mm Hg

Question: The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

Correct answer

0

Step-by-step solution

Molality of B in A is 55 m, which means 55 moles of B are dissolved in 11 kg (10001000 g) of solvent A.

Number of moles of A, nA=100050=20n_A = \dfrac{1000}{50} = 20 mol

Mole fraction of A in the liquid phase, xA=nAnA+nB=2020+5=0.8x_A = \dfrac{n_A}{n_A + n_B} = \dfrac{20}{20 + 5} = 0.8

Mole fraction of B in the liquid phase, xB=1−0.8=0.2x_B = 1 - 0.8 = 0.2

According to Raoult's law, the partial vapour pressure of A is:

PA=PA0xA=105×0.8=84P_A = P_A^0 x_A = 105 \times 0.8 = 84 mm Hg

Given total vapour pressure, PT=100P_T = 100 mm Hg

Partial vapour pressure of B, PB=PT−PA=100−84=16P_B = P_T - P_A = 100 - 84 = 16 mm Hg

Mole fraction of B in the vapour phase, yB=PBPT=16100=0.16y_B = \dfrac{P_B}{P_T} = \dfrac{16}{100} = 0.16

Answer: 0.160.16
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