What is the molality of a solution prepared with 92.2 g of ethyl propanoate and 579.6 mL of acetone? The density of acetone is 0.784 g/mL.

Answer 1

Ethyl propanoate = C₂H₅O₂CCH₂CH₃ molar mass = 102 g / mol

molality = moles solute / kg solvent

moles solute = 92.2 g C₂H₅O₂CCH₂CH₃ x 1 mol / 102 g = 0.9039 mols C₂H₅O₂CCH₂CH₃

kg solvent = 579.6 ml acetone x 0.784 g / ml = 454.4 g x 1 kg / 1000 g = 0.4544 kg

Molality (m) = 0.9039 mols / 0.4544 kg = 0.503 m (3 sig. figs.)