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14 March, 10:12

An IV bag is labeled as 0.800% w/w sodium chloride in water solution with a density of 1.036g/mL. What is the concentration of sodium chloride in the solution expressed in molarity, molality, and mole fraction?

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  1. 14 March, 11:24
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    [NaCl] = 0.14 M

    [NaCl] = 0.14 m

    Mole fraction of NaCl → 2.48*10⁻³

    Explanation:

    This is a problem of concentration.

    0.8% w/w means that in 100 g of solution, we have 0.8 g of solute, in this case NaCl.

    Let's determine the volume with the density.

    Solution density = Solution mass / Solution volume

    1.036 g/mL = 100 g / Solution volume

    Solution volume = 100 g / 1.036 g/mL → 96.5 mL

    Let's calculate molarity (mol/L)

    We convert the mass to moles (mass / molar mass)

    0.8 g / 58.45 g/mol = 0.0137 moles

    We must convert the volume of solution to L.

    Molarity is mol/L, moles of solute in L of solution.

    96.5 mL. 1L/1000 mL = 0.0965 L

    0.0137 mol / 0.0965 L = 0.14 M

    Let's determine molality and mole fraction.

    Molality are moles of solute in 1 kg of solvent (mol/kg)

    Mass of solution = Mass of solute + Mass of solvent

    100 g = 0.8 g + Mass of solvent

    100 g - 0.8 g = Mass of solvent → 99.2 g

    Then, we must convert the mass of solvent to kg

    99.2 g. 1kg / 1000 g = 0.0992 kg

    Molality: 0.0137 mol / 0.0992 kg → 0.14 m

    Mole fraction → moles of solute / moles of solute + moles of solvent

    Let's find out the moles of solvent (mass / molar mass)

    99.2g / 18 g/mol = 5.511 mol

    Total moles = 5.511 + 0.0137 → 5.5247 moles

    Mole fraction = 0.0137 / 5.5247 → 2.48*10⁻³
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