Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose (C12H22011) dissolved in 100.g of water?

Question

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
(C12H22011) dissolved in 100.g of water?

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Thành Đạt 5 months 2021-08-19T04:11:38+00:00 1 Answers 5 views 0

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    2021-08-19T04:12:39+00:00

    Answer:

    Freezing T° of solution = – 3.72°C

    Boiling T° of solution =  101.02°C

    Explanation:

    To solve this we apply colligative properties. Firstly, freezing point depression:

    ΔT = Kf . m . i

    ΔT = Freezing T° of pure solvent – Freezing T° of solution

    Kf = Cryoscopic constant, for water is 1.86 °C/m

    m = molality (moles of solute in 1kg of solvent)

    i = Ions dissolved in solution

    Our solute is sucrose, an organic compound so no ions are defined. i = 1.

    Let’s determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

    molality = 0.2 mol / 0.1kg of water = 2 m

    We replace data: ΔT = 1.86°C/m . 2m . 1

    Freezing T° of solution = – 3.72°C

    Now, we apply elevation of boiling point: ΔT = Kb . m . i

    ΔT = Boiling T° of solution – Boiling T° of  pure solvent

    Kf = Ebulloscopic constant, for water is 0.512 °C/m

    We replace:

    Boiling T° of solution – Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

    Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

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