Commercial nail polish remover usually consists of a mixture of organic liquids and water. One such formulation contains ethanol

Question

Commercial nail polish remover usually consists of a mixture of organic liquids and water. One such formulation contains ethanol (C2H6O, density = 0.789 g/mL), acetone (C3H6O, density = 0.784 g/mL), and water (H2O, density = 1.000 g/mL). A bottle having a total solution volume of 250 mL can be prepared by mixing 44.0 mL of ethanol, 150 mL of acetone, and 63.0 mL of water. Calculate the molarity, molality, and mole fraction of ethanol in this mixture. Note: as the major component of the solution, the solvent in this mixture is acetone.

Answer 1

Molarity of ethanol = moles ethanol / liter solution

moles ethanol = 44.0 ml x 0.789 g / ml x 1 mol ethanol/ 46.1 g = 0.753 mols ethanol

Volume = 250 ml x 1 L/1000 ml = 0.250 L

Molarity = 0.753 mol / 0.250 L = 3.01 M

Molality of ethanol = moles ethanol / kg acetone

moles ethanol = 0.753 mols (see calculation above)

kg acetone = 150 ml x 0.784 g / ml x 1 kg / 1000 g = 0.1176 kg

Molality = 0.784 mol / 0.1176 kg = 6.67 m

Mole fraction of ethanol = mols ethanol / total moles of solution

mols ethanol = 0.753 mols (see calculation above)

mols acetone = 150 ml x 0.784 g/ml x 1 mol acetone / 58.1 g = 2.02 mols acetone

mols H₂O = 63.0 mls x 1.00 g/ml x 1 mol / 18 g = 3.50 mols H₂O

Total moles = 0.753 + 2.02 + 3.50 = 6.27 mols

Mole fraction ethanol = X_ethanol = 0.753/6.27 = 0.120