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28 August, 21:07

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 12.0 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00*102 torr.

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  1. 28 August, 23:15
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    183.7 g of ethylene glycol

    Explanation:

    This a case of colligative property, about vapour pressure lowering.

    ΔPv = P°. Xst

    ΔPv = 12 Torr

    P° = Vapour pressure of pure solvent → 100 Torr

    12 Torr = 100 Torr. Xst

    12 Torr / 100 Torr = 0.12 → Xst

    Xst is mole fraction for solute which means moles of solute / Total moles and, the total moles are the sum of moles of solvent + moles of solute.

    We can make this equation:

    Moles of solute / (Moles of solute + Moles of solvent) = 0.12

    We don't know the moles of solute, but we do know the moles of solvent by the mass.

    Mass / Molar mass = Mol

    1000 g / 46 g/m = 21.7 moles

    Moles of solute / Moles of solute + 21.7 moles = 0.12

    Moles of solute = 0.12 (Moles of solute + 21.7 moles)

    Moles of solute - 0.12 moles of solute = 2.60 moles

    0.88 moles of solute = 2.60 moles

    Moles of solute = 2.60 moles / 0.88 → 2.96 moles of solute

    Mol. molar mass = Mass

    2.96 m. 62g/m = 183.7 g
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