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NEET-XII-Chemistry

02: Solutions

with Solutions - page 4
Qstn# B-34 Prvs-QstnNext-Qstn
  • #34
    Vapour
    pressure of water at 293 Kis 17.535 mm Hg. Calculate the
    vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in
    450 g of water.
    Ans : Vapour pressure of
    water, =
    17.535 mm of Hg
    Mass of glucose, w2 = 25 g
    Mass of water, w1 = 450 g
    We know that,
    Molar mass of glucose
    (C6H12O6), M2 = 6
    × 12 + 12 × 1 + 6 × 16
    = 180 g mol-1
    Molar mass of water, M1 = 18 g mol-1
    Then, number of moles
    of glucose,
    = 0.139 mol
    And, number of moles of
    water,
    = 25 mol
    We know that,

    ⇒ 17.535 - p1 = 0.097
    ⇒ p1 = 17.44 mm of Hg
    Hence, the vapour
    pressure of water is 17.44 mm of Hg.