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JEE Main Chemistry Solutions 2026 JEE Main 2026 (24 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m ^ -2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8 ?

Options

  1. A. 0.340
  2. B. 0.663
  3. C. 0.5217
  4. D. 0.480

Answer

C. 0.5217

Step-by-step solution

For an ideal solution with Raoult's law, the partial pressure of A in vapour is P_A = P_A^0 y_A where y_A = 0.8 is the mole fraction in vapour phase. Thus P_A = 55 0.8 = 44 kNm^ -2 . Using the equilibrium condition y_A = P_A^0 _A P_ total where _A is mole fraction in liquid and P_ total = P_A^0 _A + P_B^0(1- _A): 0.8 = 55 _A 55 _A + 15(1- _A) 0.8(55 _A + 15 - 15 _A) = 55 _A 44 _A + 12 - 12 _A = 55 _A 12 = 23 _A _A = 0.5217

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