Question

The cause for deviation from Raoult’s law in the colligative properties of non-ideal solutions lies in the nature of interactions at the molecular level. These properties show deviations from Raoult’s law due to difference in interactions between solute–solvent, solute–solute and solvent–solvent. Some liquids on mixing form azeotropes which are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes called minimum boiling azeotrope and maximum boiling azeotrope. (a) Pure ethanol cannot be prepared by fractional distillation of ethanol–water mixture. Comment.

Hint:

Deviations from Raoult’s law arise due to unequal solute–solvent interactions. Azeotropes boil at constant composition, limiting distillation.

Answer 1

(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. \[ \boxed{\text{Ethanol–water forms an azeotrope that cannot be separated by distillation.}} \] (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$\cdots$O hydrogen bonding) These new solute–solvent interactions are stronger than the individual ones. Hence, the mixture shows negative deviation from Raoult’s law. \[ \boxed{\text{Strong intermolecular attraction → negative deviation from ideality.}} \] (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) \[ \text{Relative lowering of vapour pressure: } \frac{P^0 - P_s}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \] Given: - \( P^0 = 1.25\ \text{atm} \), \( P_s = 1.237\ \text{atm} \) - \( w_1 = 1.2\ \text{g (solute)} \), \( w_2 = 60\ \text{g (solvent)} \) - \( M_2 = 78\ \text{g/mol (benzene)} \), \( M_1 = ? \) \[ \frac{1.25 - 1.237}{1.25} = \frac{1.2/M_1}{60/78} \Rightarrow \frac{0.013}{1.25} = \frac{1.2 \times 78}{M_1 \times 60} \Rightarrow 0.0104 = \frac{93.6}{60 M_1} \Rightarrow M_1 = \frac{93.6}{60 \times 0.0104} \Rightarrow M_1 \approx \boxed{150\ \text{g/mol}} \] 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. \( K_b = 2.53\ \text{K kg mol}^{-1} \)
Solution:
(c)(ii) Boiling point elevation: \[ \Delta T_b = T_{\text{solution}} - T_{\text{pure}} = 354.11 - 353.23 = 0.88\ \text{K} \] Molality: \[ \Delta T_b = K_b \cdot m = \frac{K_b \cdot w_1 \cdot 1000}{M_1 \cdot w_2} \Rightarrow 0.88 = \frac{2.53 \cdot 1.8 \cdot 1000}{M_1 \cdot 90} \Rightarrow M_1 = \frac{2.53 \cdot 1.8 \cdot 1000}{0.88 \cdot 90} \Rightarrow M_1 = \frac{4554}{79.2} \approx \boxed{57.5\ \text{g/mol}} \]