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 %

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)