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2 April, 08:25

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/ml; = 62.07 g/mol) and water (d = 1.00 g/ml) at 20°c. the density of the mixture is 1.070 g/ml. express the concentration of ethylene glycol as:

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  1. 2 April, 09:18
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    a) Volume percent

    Formula: % v/v = [volume solute / volume solution] * 100

    Just to make it easy take a base of 50 volume parts of ethylen glycol and 50 volume parts of water to make 100 volumes of mixture (this assumpion will be valid for all the questions):

    % v/v = [ 50 ml ethyleneglycol] / [100 ml mixture] * 100 = 50%

    Answer: 50% v/v

    b) Mass percent

    % m/m = [mass ethylene glycol / mass solution] * 100

    mass ethylene glycol = 50 ml * 1.114 g / ml = 55.7 g

    mass of mixture = 100 ml * 1.07 g/ml = 107 g

    % m/m = [55.7 / 107 g] * 100 = 52.06 %

    Answer: 52.06%

    c) Molarity

    M = number of moles of solute / liters of solution

    number of moles of solute = mass in grams / molar mass

    number of moles of ehtylene glycol = 55.7 g / 62.07 g/mol = 0.8974 mol

    liters of solution = 0.1 liter

    M = 0.8974 mol / 0.1 liter = 8.974 M

    Answer: 8.974 M

    d) Molality

    m = number of moles of solute / kg of solvent

    number of moles of ethylen glycol = 0.8974 mol

    mass of water = 50 ml * 1 g/ml = 50 g = 0.05 kg

    m = 0.8974 mol / 0.05 kg = 17.95 m

    Answer: 17.95 m

    e) mole fraction

    mole fraction = [number of moles of solute] / [number of moles of mixture] * 100

    number of moles of ethylen glycol = 0.8974 mol

    number of moles of water = 50 g / 18.01 g / mol = 2.776 mol

    mole fraction = 0.8974 mol / [0.8974 mol + 2.776 mol] = 0.244

    Answer: 0.244
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