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6 September, 11:13

Glycerol (C3H8O3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. Consider that you have an aqueous solution that contains 34.4 % glycerol by mass. If the vapor pressure of pure water is 23.8 torr at 25oC, what is the vapor pressure of the solution at 25oC?

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  1. 6 September, 14:12
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    The vapor pressure of the solution at 25°C is 26.01 Torr

    Explanation:

    This is a usual excersise of colligative properties. In this case we apply the vapor pressure lowering formula:

    ΔP = Xst. P°

    Where ΔP is the diferrence between Pressure of solution - Pressure of pure solvent.

    And Xst the molar fraction.

    P° is Pressure of pure solvent.

    So the formula will be:

    Pressure Solution - P° = Xst. P°

    Pressure Solution - 23.8 Tor = Xst. 23.8 Torr

    Xst : Mole fraction (Moles of solute or solvent / Total moles)

    34.4 % m/m means that in 100 g of solution I have 34.4 g of solute

    If I have 34.4 g of solute and the mass of 100 g in solution, I can know the mass of solvent, and finally the moles.

    100 g solution - 34.4 g solute = 65.6 g (mass of solvent)

    Molar mass of water : 18 g/m

    Moles of water: Mass of water / Molar mass

    65.6 g / 18g/m = 3.64 moles

    Moles of glycerol : Mass glycerol / Molar mass glycerol

    34.4 g / 92.1 g/m = 0.373 moles

    Total moles: moles of glycerol + moles of water

    0.373 m + 3.64 m = 4.01 m

    So Xst = 0.373 m / 4.01 m → 0.093

    Pressure Solution - 23.8 Tor = 0.093. 23.8 Torr

    Xst HAVE NO UNITS

    Pressure Solution = (0.093. 23.8 Torr) + 23.8 Tor

    Pressure Solution = 26.01 Torr
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