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30 January, 09:40

How many grams of ethylene glycol (C2H6O2) must be added to 1.15 kg of water to produce a solution that freezes at - 4.46°C?

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  1. 30 January, 13:05
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    232.5 g C2H6O2

    Explanation:

    The equation you need to use here is ΔTf = i Kf m

    Since pure water freezes at 0 C, your ΔTf is just 4.46 C

    i = 1 (ethylene glycol is a weak electrolyte)

    Kf = molal freezing constant, which for water is 1.86 C/m

    m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)

    Than,

    4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)

    Solve for x, you should get x = 2.75 mol C2H6O2

    3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2
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