Answer:
a. 1.78x10⁻⁵
b. 9.92x10⁻⁴ mol/kg
c. 17.8x10⁻³ g % mass
Step-by-step 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)