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8 July, 04:18

Calculate the mass of ethylene glycol (C₂H₆O₂) that must be added to 1.00 kg of ethanol (C₂H₅OH) to reduce its vapor pressure by 11.0 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 1.00*10² torr.

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  1. 8 July, 04:32
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    123.5 g of C₂H₅OH are required

    Explanation:

    Let's think the colligative property of vapour pressure to solve this.

    ΔP = P°. Xm

    ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

    P° = Vapor pressure of pure solvent

    Xm = mole fraction of solute (moles of solute / (moles of solute + moles of solvent))

    11 Torr = 100 Torr. Xm

    11 Torr / 100 Torr = 0.11 → Xm

    0.11 = moles of solute / moles of solute + moles of solvent

    Let's determine the moles of solvent.

    We have a mass of 1 kg, which is the same as 1000 g

    Molar mass ethanol = 46 g/mol

    1000g / 46g/mol = 21.7 moles

    0.11 moles = moles of solute / moles of solute + 21.7 moles

    0.11 moles (moles of solute + 21.7 moles) = moles of solute

    0.11 moles of solute + 2.39 moles = moles of solute

    2.39 moles = 1 - 0.11 moles of solute

    2.39 moles = 0.89 moles of solute

    2.39 / 0.89 = moles of solute → 2.68 moles

    Let's convert the moles to mass

    2.68 moles. 46 g/mol = 123.5 g
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