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NEET-XII-Chemistry

02: Solutions

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  • #8
    The
    vapour pressure of pure liquids A and B are 450 and 700 mm Hg
    respectively, at 350 K. Find out the composition of the liquid
    mixture if total vapour pressure is 600 mm Hg. Also find the
    composition of the vapour phase.
    Ans : It is
    given that:
    =
    450 mm of Hg
    =
    700 mm of Hg
    ptotal = 600 mm of Hg
    From
    Raoult’s law, we have:

    Therefore,
    total pressure,

    Therefore,
    = 1 -
    0.4
    = 0.6
    Now,

    = 450 ×
    0.4
    = 180 mm
    of Hg

    = 700 ×
    0.6
    = 420 mm
    of Hg
    Now, in
    the vapour phase:
    Mole
    fraction of liquid A

    = 0.30
    And,
    mole fraction of liquid B = 1 - 0.30
    =
    0.70
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