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5 May, 20:29

A solution contains 22.4 g glucose (C6H12O6) dissolved in 0.500 L of water. What is the molality of the solution

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  1. 5 May, 20:40
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    Molality is 0.25 m

    Explanation:

    Molality → Moles of solute / kg of solvent

    We need the moles of solute → 0.124 moles

    22.4 g. 1 mol / 180 g = 0.124 moles

    We need the mass of solvent in kg. We determine the mass of solvent with density.

    Density = Mass / Volume

    Mass = Density. volume → 1 g/mL. 500 mL = 500 g

    If we convert the mass in g to kg → 500 g. 1kg / 1000 g = 0.5 kg

    In conclussion, molality → 0.124 mol / 0.5 kg = 0.25 m
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