Calculate the freezing point of a nonionizing antifreeze solution containing 388 g ethylene glycol, C2H6O2, and 510 g of water.

Question

Calculate the freezing point of a nonionizing antifreeze solution containing 388 g ethylene glycol, C2H6O2, and 510 g of water.

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Kim Chi 6 months 2021-08-29T01:31:40+00:00 1 Answers 2 views 0

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    2021-08-29T01:33:13+00:00

    Answer:

    T° freezing solution = -22.8°C

    Explanation:

    To solve this problem we apply, the Freezing Point Depression. This is a colligative property which its formula is:

    T° freezing pure solvent – T° freezing solution = Kf . molality . i

    i = Van’t Hoff factor.

    We have been informed is a nonionizing solute, so i = 1

    Our solute is ethylene glycol, so le’ts determine the moles to get molality

    388 g . 1 mol / 62.07 g = 6.25 moles

    molality (m) = moles of solute /kg of solvent

    We convert mass of solvent, water, to kg → 510 g . 1kg /1000g = 0.510 kg

    6.25 mol /0.510kg = 12.25 m

    We replace at formula → 0°C – T° freezing solution = 1.86°C/m . 12.25 m . 1

    T° freezing solution = -22.8°C

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