Pure glacial acetic acid, HC₂H₃O₂, has a freezing point of 16.62°C. Its freezing point depression constant is: K_f = 3.57 °C m^-1. A solution was made by taking 9.755 g of an unknown nonelectrolyte and dissolving it in 90.50 g of glacial acetic acid. The measured freezing point of the solution was 8.64°C. Calculate the molecular weight of the unknown substance.Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Question

Quizplus · Accepted Answer

The freezing point depression is calculated as ΔT = 16.62°C - 8.64°C = 7.98°C. Using the formula ΔT = Kf * m, where m is molality, we find m = 7.98°C / 3.57°C m^-1 = 2.235 m. Molality is moles of solute per kg of solvent, so moles of solute = 2.235 m * 0.0905 kg = 0.2023 moles. The molar mass is then 9.755 g / 0.2023 moles = 48.2 g mol^-1.

Pure Glacial Acetic Acid, HC2H3O2, Has a Freezing Point of 16.62°C

Pure Glacial Acetic Acid, HC₂H₃O₂, Has a Freezing Point of 16.62°C