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8 June, 13:11

What is the molality, m, of an aqueous solution of ammonia that is 12.83 M NH3 (17.03 g/mol) ? This solution has a density of 0.9102 g/mL.

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  1. 8 June, 15:27
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    Molality = 18.5 m

    Explanation:

    Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)

    12.83 M means molarity → mol of solute in 1L of solution

    Density refers always to solution → Mass of solution / Volume of solution

    1L = 1000 mL

    We can determine the mass of solution with density

    0.9102 g/mL = Mass of solution / 1000 mL

    Mass of solution = 0.9102 g/mL. 1000 mL → 910.2 g

    Let's convert the moles of solute (NH₃) to mass

    12.83 mol. 17.03 g / 1 mol = 218.5 g

    We can apply this knowledge:

    Mass of solution = Mass of solvent + Mass of solute

    910.2 g = Mass of solvent + 218.5 g

    910.2 g - 218.5 g = 691.7 g → Mass of solvent.

    Let's convert the mass in g to kg

    691.7 g. 1kg / 1000 g = 0.6917kg

    We can determine molalilty now → 12.83 mol / 0.6917kg

    Molality = 18.5 m
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