HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Chemistry

02: Solutions

with Solutions - page 4
Qstn# B-36 Prvs-QstnNext-Qstn
  • #36
    100
    g of liquid A (molar mass 140 g mol-1)
    was dissolved in 1000 g of liquid B (molar mass 180 g mol-1).
    The vapour pressure of pure liquid B was found to be 500 torr.
    Calculate the vapour pressure of pure liquid A and its vapour
    pressure in the solution if the total vapour pressure of the solution
    is 475 Torr.
    Ans : Number of moles of
    liquid A,
    = 0.714 mol
    Number of moles of
    liquid B,
    = 5.556 mol
    Then, mole fraction of
    A,

    = 0.114
    And, mole fraction of
    B, xB = 1 - 0.114
    = 0.886
    Vapour pressure of pure
    liquid B, =
    500 torr
    Therefore, vapour
    pressure of liquid B in the solution,

    = 500 × 0.886
    = 443 torr
    Total vapour pressure
    of the solution, ptotal = 475 torr
    Vapour
    pressure of liquid A in the solution,
    pA = ptotal - pB
    = 475 - 443
    = 32 torr
    Now,


    = 280.7 torr
    Hence, the vapour
    pressure of pure liquid A is 280.7 torr.
    Page No 62: