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17 July, 16:00

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.824 g/cm^3 (cubed)

a.) Calculate the concentration of the solution in molality.

b.) Calculate the concentration of the solution in mole fraction of LiBr

c.) Calculate the concentration of the solution in mass percentage of CH3CN

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  1. 17 July, 16:12
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    a. [LiBr] = 2.70 m

    b. Xm for LiBr = 0.1

    c. 81% by mass CH₃CN

    Explanation:

    Solvent → Acetonitrile (CH₃CN)

    Solute → LiBr, lithium bromide

    We convert the moles of solute to mass → 1.80 mol. 86.84 g/1 mol = 156.3 g

    This mass of solute is contained in 1L of solution

    1 L = 1000 mL → 1mL = 1cm³

    We determine solution mass by density

    Solution density = Solution mass / Solution volume

    Solution density. Solution volume = solution mass

    0.824 g/cm³. 1000 cm³ = 824 g

    Mass of solution = 824 g (solvent + solute)

    Mass of solute = 156.3 g

    Mass of solvent = 824 g - 156.3 g = 667.7 g

    Molality → Moles of solute in 1kg of solvent

    We convert the mass of solvent from g to kg → 667.7 g. 1kg / 1000g = 0.667 kg

    Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

    Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

    Moles of solvent → 667.7 g. 1mol / 41g = 16.3 moles

    Total moles = 16.3 + 1.8 = 18.1

    Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

    Mass percentage → (Mass of solvent, in this case / Total mass). 100

    We were asked for the acetonitrile → (667.7 g / 824 g). 100 = 81%
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