Colligative Properties

(a) Ethanol and water form a minimum boiling azeotrope at about 95.6% ethanol and 4.4% water by mass. - This azeotropic mixture boils at a constant temperature. - Since both liquid and vapour have the same composition at this point, further separation by fractional distillation is not possible. (b) Why does a mixture of chloroform and acetone show deviation from ideal behaviour?
Solution:
(b) Chloroform and acetone exhibit strong intermolecular hydrogen bonding between them: - Acetone has a polar carbonyl group (C=O) - Chloroform has acidic hydrogen (C–H O hydrogen bonding) These new solute–solvent interactions are stronger than the individual ones. Hence, the mixture shows negative deviation from Raoult’s law. (c)(i) The vapour pressure of pure benzene at a certain temperature is 1.25 atm. When 1.2 g of non-volatile, non-electrolyte solute is added to 60 g of benzene (M = 78 g/mol), the vapour pressure of the solution becomes 1.237 atm. Calculate the molar mass of the non-volatile solute.
Solution:
(c)(i) Given: - , - , - , OR (c)(ii) The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute.
Solution:
(c)(ii) Boiling point elevation: Molality:

Step 1: Molality Step 2: van’t Hoff factor (i) MgSO dissociates completely → Step 3: Boiling point elevation