Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone

Question

Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone

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Gia Bảo 6 months 2021-07-31T03:53:19+00:00 1 Answers 16 views 0

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    2021-07-31T03:54:42+00:00

    Answer:

    Molarity = 0.809 M

    mole fraction = 0.047

    Explanation:

    The complete question is

    Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

    Solution –

    Solution for molarity:

    1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

    1)  

    Mass of 1.09 mole of acetone

    = 1.09  mol x 58.0794 g/mol = 63.306 g

    Density of acetone = 0.788 g/cm3  

    Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

    For ethanol

    1000 g divided by 0.789 g/cm3 = 1267.427 cm3

    Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

    a) Molarity:

    1.09 mol / 1.347 L = 0.809 M

    Mole Fraction  

    a) moles of ethanol:

    1000 g / 46.0684 g/mol = 21.71 mol

    b) moles of acetone:

    1.09 / (1.09 + 21.71) = 0.047

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