Answer to Question #180699 in Chemistry for Kyle

Question

The density of a 15.00% by mass aqueous solution of acetic acid, CH3COOH, is 1.087 g/mL. What is the molarity, the molality, and the mole fraction of each component? Show your solution.

Answer 1

It is possible that an incorrect value of the density of a 15.00% by mass aqueous solution of acetic acid is indicated in the problem (1.087 g/mL). Since this value should have been 1.0187 g/mL.

However, the solution below is given for a density of 1.087 g/mL.

Solution:

acetic acid - CH₃COOH = C₂H₄O₂

The molar mass of C₂H₄O₂ is 60.052 g/mol.

The molar mass of H₂O is 18.015 g/mol.

 

Let's assume that the volume of the solution is 1 L (1000 mL).

Thus,

Mass of 1 L C₂H₄O₂ solution = Solution volume × Density = 1000 mL × 1.087 g/mL = 1087 g

Mass of C₂H₄O₂ = (%C₂H₄O₂ × Mass of solution) / 100% = (15% × 1087 g) / 100% = 163.05 g C₂H₄O₂

Mass of H₂O = Mass of solution - Mass of C₂H₄O₂ = 1087 g - 163.05 g = 923.95 g H₂O

Moles of C₂H₄O₂ = Mass of C₂H₄O₂ / Molar mass of C₂H₄O₂ = 163.05 g / 60.052 g mol^-1 = 2.7151 mol

Moles of H₂O = Mass of H₂O / Molar mass of H₂O = 923.95 g / 18.015 g mol^-1 = 51.2878 mol

 

Mole Fraction (χ):

Mole fraction (χ) is the ratio of moles of one substance in a mixture to the total number of moles of all substances.

Total number of moles = Moles of C₂H₄O₂ + Moles of H₂O = 2.7151 mol + 51.2878 mol = 54.0029 mol

χ_C2H4O2 = Moles of C₂H₄O₂ / Total number of moles = 2.7151 mol / 54.0029 mol = 0.0503 = 0.05

χ_C2H4O2 = 0.05

χ_H2O = Moles of H₂O / Total number of moles = 51.2878 mol / 54.0029 mol = 0.9497 = 0.95

χ_H2O = 0.95

 

Molarity (M):

Molarity or molar concentration is the number of moles of solute per liter of solution.

Molarity of solution = Moles of C₂H₄O₂ / Volume of solution

Molarity of solution = 2.7151 mol / 1 L = 2.7151 mol/L = 2.7151 M

Molarity of solution = 2.7151 M

 

Molality (m):

Molality of a solution is the moles of solute divided by the kilograms of solvent.

Mass of solvent (H₂O) = 923.95 g = 0.92395 kg

Molality of solution = Moles of C₂H₄O₂ / Kilograms of solvent

Molality of solution = 2.7151 mol / 0.92395 kg = 2.93858 mol/kg = 2.9386 m

Molality of solution = 2.9386 m

 

Answers:

χ_C2H4O2 = 0.05

χ_H2O = 0.95

Molarity of solution = 2.7151 M

Molality of solution = 2.9386 m