Relative Lowering of Vapour Pressure - UNSOLVED PRACTICE SET
Chapter: Solutions | Topic: Relative Lowering of Vapour Pressure
RELATIVE LOWERING OF VAPOUR PRESSURE - UNSOLVED PRACTICE SET
Topic: Relative Lowering of Vapour Pressure
Multiple Choice Questions
Q1. The relative lowering of vapour pressure of a solution is equal to:
- The mole fraction of the solvent
- The mole fraction of the solute
- The vapour pressure of the pure solvent
- The vapour pressure of the solution
Q2. For a dilute solution containing a non-volatile solute:
- (pĀ° p) / pĀ° = xā
- (pĀ° p) / p = xā
- (pĀ° p) / pĀ° = xā
- p / pĀ° = xā
Q3. The relative lowering of vapour pressure is a colligative property because it depends on:
- The nature of the solute
- The number of solute particles
- The colour of the solution
- The volume of the solution
Q4. If the vapour pressure of a pure solvent is 100 mmHg and that of the solution is 90 mmHg, the relative lowering is:
- 0.1
- 0.9
- 10
- 90
Q5. For determining molar mass using relative lowering of vapour pressure, the solute must be:
- Volatile
- Non-volatile
- Either volatile or non-volatile
- A gas
Q6. The molecular mass of a solute determined by relative lowering of vapour pressure will be:
- Higher than actual if the solute dissociates
- Lower than actual if the solute dissociates
- Accurate regardless of dissociation
- Higher than actual if the solute associates
Short Answer Questions
Q7. Derive the relationship between relative lowering of vapour pressure and mole fraction of solute for a dilute solution containing a non-volatile solute.
Q8. The vapour pressure of pure water at 25Ā°C is 23.8 mmHg. A solution of urea has a vapour pressure of 22.8 mmHg. Calculate the relative lowering of vapour pressure and the mole fraction of urea.
Q9. Why is the relative lowering of vapour pressure method less commonly used for determining molar mass compared to osmotic pressure or freezing point depression?
Q10. Explain how the relative lowering of vapour pressure is related to the elevation of boiling point and depression of freezing point.
Q11. Your grandmother covers water in a pot with a cloth during summer to keep it cool. Explain how this relates to vapour pressure and evaporation.
Q12. Why does adding salt to water lower its vapour pressure? Explain at the molecular level.
Long Answer Questions
Q13. Discuss relative lowering of vapour pressure in detail:
(a) Definition and mathematical expression
(b) Derivation from Raoult's law for non-volatile solutes
(c) Relationship between relative lowering and mole fraction of solute
(d) Determination of molar mass of non-volatile solutes
(e) Limitations and experimental methods (Ostwald-Walker dynamic method)
Q14. Explain the relationship between relative lowering of vapour pressure and other colligative properties:
(a) Derivation of elevation of boiling point from relative lowering of vapour pressure
(b) Derivation of depression of freezing point from relative lowering of vapour pressure
(c) The role of Clausius-Clapeyron equation in these derivations
(d) Why all colligative properties are fundamentally connected to vapour pressure lowering
Q15. Relative lowering of vapour pressure has practical significance in Indian contexts. Discuss:
(a) Why water in an earthen pot (matka) stays cool ā the role of evaporation and vapour pressure
(b) How salt is harvested from seawater in Gujarat and Tamil Nadu using solar evaporation
(c) Why sugar syrup (chashni) for Indian sweets is prepared by controlling boiling point elevation
(d) The design of cooling towers in Indian thermal power plants based on vapour pressure principles
Numerical / Application-Based Problems
Q16. The vapour pressure of pure benzene at 30Ā°C is 120 mmHg. When 7.8 g of a non-volatile solute is dissolved in 156 g of benzene, the vapour pressure of the solution becomes 115.2 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. A solution of 2.5 g of a protein in 100 mL of water at 25Ā°C has a vapour pressure of 23.65 mmHg. The vapour pressure of pure water at 25Ā°C is 23.76 mmHg.
(a) Calculate the relative lowering of vapour pressure.
(b) Calculate the molar mass of the protein.
(c) Why is this method less accurate for macromolecules compared to osmotic pressure measurement?
[Given: Density of water = 1 g/mL]
Q18. In a salt flat in Gujarat, seawater containing 3.5% NaCl by mass is allowed to evaporate.
(a) Calculate the vapour pressure of seawater at 25Ā°C if pure water has vapour pressure 23.76 mmHg.
(b) Calculate the relative lowering of vapour pressure.
(c) If the seawater is concentrated to 15% NaCl (brine), calculate the new vapour pressure. Explain why salt crystallises out when the solution becomes saturated.
[Given: Molar mass of NaCl = 58.5 g/mol, assume complete dissociation]