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4 March, 04:20

A 0.944 mM solution of glucose, C6H12O6, in water has a density of 1.0021 g/mL at 20.0 oC. What is the concentration of this solution in the following units. Show ALL your work including units for credit

a. mole fraction

b. molality

c. Mass %

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Answers (1)
  1. 4 March, 04:45
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    a. 1.78x10⁻⁵

    b. 9.92x10⁻⁴ mol/kg

    c. 17.8x10⁻³ g % mass

    Explanation:

    Concentration in mM, means that x milimoles are contained in 1 L of solution. In this case 0.994 mmoles

    If we want to convert to Molarity we should / 1000.

    9.94x10⁻⁴ M (moles of glucose that are contained in 1L of solution)

    C₆H₁₂O₆ (glucose) → Molar mass = 180.15 g/m

    Let's apply density to know the total mass of solution:

    Density = Solution mass / Solution volume

    1.0021 g/mL = Solution mass / 1000mL

    (Remember that molarity has a volume of 1L, which can also be written as 1000 mL)

    1.0021 g/mL. 1000 mL = 1002.1 g (Solution mass)

    Mole of solute. Molar mass = Solute mass

    9.94x10⁻⁴ m. 180.15 g/m = 0.179 g (Mass of glucose, in solution)

    Total mass = Mass of solute + Mass of solvent

    1002.1 g = 0.179 g + Mass of solvent

    1002.1 g - 0.179 g = 1001.921 g (Mass of solvent)

    Mole of solvent = Mass / Molar mass of solvent

    1001.921 g / 18 g/m = 55.66 mole

    a. Mole fraction = (Mole of solute / Total mole)

    9.94x10⁻⁴ m / 9.94x10⁻⁴ m + 55.6 m = 1.78x10⁻⁵

    Mole fraction has no units

    b. Molality is mole of solute in 1kg of solvent

    9.94x10⁻⁴ mole of solute are in 1001.921 g of water

    Let's convert 1001.921 g to kg

    1.001921 kg = 1001.921 g

    1.001921 kg contain 9.94x10⁻⁴ mole of solute

    1 kg of solute contains (1kg. 9.94x10⁻⁴m / 1.001921 kg) = 9.92x10⁻⁴ m

    c. Mass % (means mass of solute in 100g of solution)

    1002.1 g contain 0.179 g of glucose

    In 100 g of solution (0.179 g. 100g) / 1002.1 g = 0.0178 g (17.8x10⁻³ g % mass)
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