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12 October, 22:45

Determine the molarity and mole fraction of a 1.15 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the final volume equals the sum of the volumes of acetone and ethanol.

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  1. 12 October, 23:02
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    The molarity is 0.85M

    The mole fraction = 0.05

    Explanation:

    Step 1: Data given

    Molarity of acetone = 1.15 m

    Density of acetone = 0.788 g/cm³

    Density of ethanol = 0.789 g/cm³

    Molar mass of acetone = 58.08 g/mol

    Step 2: Calculate number of moles acetone

    Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)

    1.15 m = 1.15 moles acetone / 1 kg ethanol

    Step 3: Calculate mass of acetone

    1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone

    Step 4: Calculate volume of acetone

    66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone

    Step 5: Calculate volume of ethanol

    1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol

    Step 6: Calculate total volume solution

    Total solution volume = 84.76 + 1267.43 = 1352.2 mL = 1.3522L

    Step 7: Calculate molarity

    Molarity of acetone = moles acetone / volume solution = 1.15moles / 1.3522L

    Molarity = 0.85 M

    Step 8: Calculate moles ethanol

    moles ethanol = mass / molar mass = 1000g / 46.0g/mol = 21.74 moles

    Step 9: Calculate mole fraction

    mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05
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