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10 June, 08:43

When a specific amount of acetone (C3H6O) is added to 100.0 g of pure water at 65°C, the vapor pressure of water over the solution is lowered by 1.576 kPa. Given the vapor pressure of water at 65°C is 25.022 kPa, what is the mass of acetone added?

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  1. 10 June, 12:02
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    mass of acetone added:

    w acetone = 21.676 g

    Explanation:

    decrease in vapor pressure:

    ΔP = - (P*a) (Xb, l)

    ∴ ΔP = - 1.576 KPa = Pa - P*a

    ∴ a: water (solvent)

    ∴ b: C3H6O (solute)

    ∴ P*a (65°C) = 25.022 KPa

    ⇒ ΔP / ( - P*a) = Xb, l

    ⇒ Xb. l = - 1.576 KPa / ( - 25.022 KPa) = 0.063

    ∴ Xb = nb/nt = nb / (na + nb) = (wb/Mb) / [ (wa/Ma) + (wb/Mb) ]

    ∴ Mb = 58.08 g/mol (molar mass acetone)

    ∴ Ma = 18.015 g/mol (molar mass water)

    ∴ wa = 100 g

    ⇒ 0.063 = (wb/58.08) / [ (100/18.015) + (wb/58.08) ]

    ⇒ (0.063) [ (5.55 + (wb/58.08) ] = wb/58.08

    ⇒ 0.3497 + 1.085 E-3wb = wb/58.08

    ⇒ 20.3106 + 0.063wb = wb

    ⇒ 20.3106 = wb - 0.063wb

    ⇒ 20.3106 = 0.937wb

    ⇒ 20.3106/0.937 = wb

    ⇒ 21.676 g = wb
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