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12 May, 16:45

A 2.950*10-2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL. The density of water at 20.0∘C is 0.9982 g/mL. Part A Calculate the molality of the glycerol solution.

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  1. 12 May, 18:05
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    The molality of the glycerol solution is 2.960*10^-2 mol/kg

    Explanation:

    Number of moles of glycerol = Molarity * volume of solution = 2.950*10^-2 M * 1 L = 2.950*10^-2 moles

    Mass of water = density * volume = 0.9982 g/mL * 998.7 mL = 996.90 g = 996.90/1000 = 0.9969 kg

    Molality = number of moles of glycerol/mass of water in kg = 2.950*10^-2/0.9969 = 2.960*10^-2 mol/kg
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