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4 May, 23:44

An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.75 g/cm3. Calculate the molality, molarity, and mole fraction of the ethylene glycol.

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  1. 5 May, 02:30
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    Xm ethylene glycol = 0.16

    Molality = 10.75 m

    Molarity = 11.3M

    Explanation:

    40.0% by mass. This data says that 40 g of ethyelen glycol are contained in 100 g of solution

    Therefore: Mass of solute is 40g

    Mass of solvent = 60 g

    Mass of solution = 100 g

    We convert the mass to moles → 40 g. 1mol / 62 g = 0.645 moles

    We convert the solvent mass to moles → 60 g. 1mol / 18g = 3.33 moles

    Xm ethylene glycol = Moles of ethylene glycol / Total moles

    0.645 / (0.645 + 3.33) = 0.16

    Molality are the moles of solute in 1kg of solvent.

    We convert the mass of solvent to kg → 60 g. 1kg/1000g = 0.060kg

    Molality (mol/kg) = 0.645 mol / 0.060kg = 10.75 m

    To determine molarity we use the density of solution in order to find out the volume: density = mass / volume

    1.75 g/cm³ = 100 g / volume → 100 g / 1.75 g/cm³ = 57.1 cm³ (1cm³ = 1ml)

    We convert the solution's volume to L → 57.1 mL. 1L/1000mL = 0.0571L

    Molarity → moles of solute in 1L of solution

    Molarity (mol/L) → 0.645 mol / 0.0571L = 11.3M
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