56.8 grams of glucose is dissolved in acetic acid. The boiling point of the acetic acid raises to 126.6°C. What is the molal concentration

Question

56.8 grams of glucose is dissolved in acetic acid. The boiling point of the acetic acid raises to 126.6°C. What is the molal concentration of the solution?

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Thành Công 3 years 2021-09-02T12:38:14+00:00 1 Answers 7 views 0

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  1. King
    0
    2021-09-02T12:39:32+00:00
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    Answer:

    Molal concentration of glucose = 2.64 m

    Explanation:

    The excersise can be solved by the formula for colligative property, elevation of boiling point.

    ΔT = Kb . m . i

    As glucose (C₆H₁₂O₆) is an organic compound, i = 1. (No ions generated)

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

    Kb = Ebulloscopic constant.

    We need data from acetic acid.

    Boiling T° of pure solvent: 118.1°C

    Kb = 3.22 °C/m

    We replace data:

    126.6°C – 118.1°C = 3.22°C/m . m . 1

    (126.6°C – 118.1°C) / 3.22 m/°C = m

    Molality = 2.64 mol/kg

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