Vapour Pressure of Solutions - Raoult's Law - UNSOLVED PRACTICE SET

Q7. State Raoult's law for solutions containing:

(a) A non-volatile solute in a volatile solvent

(b) Two volatile liquids

Q8. Explain why the vapour pressure of a solvent decreases when a non-volatile solute is added to it.

Q9. The vapour pressure of pure water at 25°C is 23.8 mmHg. Calculate the vapour pressure of a solution containing 18 g of glucose in 178.2 g of water.

[Given: Molar masses: glucose = 180 g/mol, water = 18 g/mol]

Q10. What is the difference between an ideal solution and a non-ideal solution? Give one example of each.

Q11. Your mother adds salt to water while boiling potatoes. Why does the water boil at a higher temperature? Explain using Raoult's law.

Q12. Why does a solution of ethanol and cyclohexane show positive deviation from Raoult's law?

Q13. Discuss Raoult's law in detail:

(a) Statement and mathematical expression for non-volatile solutes

(b) Statement and mathematical expression for volatile liquid mixtures

(c) Derivation of the relationship between relative lowering of vapour pressure and mole fraction of solute

(d) Graphical representation of Raoult's law

(e) Limitations and conditions for validity

Q14. Explain deviations from Raoult's law:

(a) Ideal solutions — characteristics, examples (benzene + toluene, n-hexane + n-heptane)

(b) Positive deviation — causes (weaker A–B interactions), examples (ethanol + cyclohexane), graphical representation

(c) Negative deviation — causes (stronger A–B interactions), examples (acetone + chloroform), graphical representation

(d) Azeotropes — minimum boiling azeotropes and maximum boiling azeotropes

Q15. Raoult's law has practical applications in Indian industry and daily life. Discuss:

(a) How fractional distillation in Indian petroleum refineries separates crude oil components based on differences in vapour pressure

(b) Why adding salt to ice helps make ice cream faster (freezing point depression related to vapour pressure lowering)

(c) The use of pressure cookers in Indian kitchens — how increased pressure raises boiling point

(d) Why desalination of seawater is energy-intensive — the vapour pressure of saline water is lower than pure water

Q16. The vapour pressure of pure benzene at 30°C is 120 mmHg. When 6 g of a non-volatile solute is dissolved in 234 g of benzene, the vapour pressure of the solution becomes 118.8 mmHg.

(a) Calculate the relative lowering of vapour pressure.

(b) Calculate the mole fraction of the solute.

(c) Calculate the molar mass of the solute.

[Given: Molar mass of benzene = 78 g/mol]

Q17. Two volatile liquids A and B have pure vapour pressures of 450 mmHg and 700 mmHg respectively at 25°C. A solution contains 2 moles of A and 3 moles of B.

(a) Calculate the total vapour pressure of the solution assuming ideal behaviour.

(b) Calculate the mole fraction of A in the vapour phase.

(c) If the solution shows positive deviation and the actual total vapour pressure is 620 mmHg, calculate the deviation from ideal behaviour.

Q18. A sugar factory in Uttar Pradesh produces syrup by evaporating water from sugarcane juice. The juice contains 15% sugar by mass.

(a) If the vapour pressure of pure water at the boiling temperature is 760 mmHg, calculate the vapour pressure of the juice.

(b) Calculate the boiling point elevation of the juice (Kb for water = 0.512 K·kg/mol).

(c) If the factory processes 10,000 L of juice per hour, calculate the mass of water that must be evaporated to concentrate the syrup to 60% sugar. Explain why this requires significant energy input.

[Given: Molar mass of sugar (sucrose) = 342 g/mol, density of juice ≈ 1.05 g/mL]