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10 October, 21:58

Calculate the vapor pressure at 25°C of a solution containing 55.3 g ethylene glycol (HOCH2CH2OH) and 285.2 g water. The vapor pressure of pure water at 25°C is 23.8 torr.

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  1. 10 October, 22:14
    0
    Vapor pressure of solution = 23.9 Torr

    Explanation:

    Let's apply the colligative poperty of vapor pressure to solve this:

    ΔP = P°. Xm

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

    We have solvent and solute mass, so let's find out the moles of each.

    55.3 g / 62 g/mol = 0.89 moles

    285.2 g / 18 g/mol = 15.84 moles

    Let's determine the mole fraction of ethylene glycol.

    Mole fraction = Moles of ethylene glyco / Total moles

    0.89 moles / (0.89 + 15.84) = 0.053

    25.3 Torr - Vapor pressure of solution = 25.3 Torr. 0.053

    Vapor pressure of solution = 25.3 Torr. 0.053 - 25.3 Torr

    Vapor pressure of solution = 23.9 Torr
  2. 10 October, 23:32
    0
    P (25°C) = 22.523 torr

    Explanation:

    low volatile solute and diluted solution:

    ΔP = P - P*a = - Xb. P*a

    ∴ a: water

    ∴ b: C2H6O2

    ∴ P*a (25°C) = 23.8 torr

    Xb = (wb/Mb) / (wa/Ma + wb/Mb)

    ∴ wb = 55.3 g

    ∴ Mb = 62.07 g/mol

    ∴ wa = 285.2 g

    ∴ Ma = 18.015 g/mol

    ⇒ Xb = (55.3/62.07) / ((285.2/18.015) + (55.3/62.07))

    ⇒ Xb = 0.0536

    ⇒ P (25°C) = P*a - Xb. P*a

    ⇒ P (25°C) = 23.8 torr - ((0.0536) (23.8 torr))

    ⇒ P (25°C) = 22.523 torr
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