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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are- [JEE Main 8-4-2019 Morning]

    A) 500 mmHg, 0.5, 0.5

    B) 450 mmHg, 0.4, 0.6

    C) 450 mmHg, 0.5, 0.5

    D) 500 mmHg, 0.4, 0.6

    Correct Answer: D

    Solution :

    \[{{P}_{total}}={{X}_{A}}.P_{A}^{0}+{{X}_{B}}.P_{B}^{0}=0.5\times 400+0.5\times 600\]           \[=500mmHg\] Now, mole fraction of A in vapour, \[{{Y}_{A}}=\frac{{{P}_{A}}}{{{P}_{total}}}=\frac{0.5\times 400}{500}=0.4\] and mole fraction of B in vapour, \[{{Y}_{B}}=1-0.4=0.6\] Correct option :


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