Relative Lowering Of Vapour Pressure
The vapour pressure of pure water at 25°C is 23.8 mmHg. When 18 g of glucose (C₆H₁₂O₆, molar mass = 180 g mol⁻¹) is dissolved in 178.2 g of water, the vapour pressure of the solution is approximately:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- relative lowering of vapour pressure
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
23.56 mmHg
Moles of glucose = 18 / 180 = 0.1 mol. Moles of water = 178.2 / 18 = 9.9 mol. Mole fraction of solvent (water): x₁ = 9.9 / (9.9 + 0.1) = 9.9 / 10.0 = 0.99. By Raoult's law for the solvent: P_solution = x₁ × P°_water = 0.99 × 23.8 = 23.562 mmHg ≈ 23.56 mmHg. The relative lowering of vapour pressure (P° − P)/P° = x₂ = 0.1/10 = 0.01, meaning a 1% lowering. This result illustrates that for dilute solutions the vapour pressure lowering is small but measurable, and is proportional to the mole fraction of the non-volatile solute (glucose does not contribute to the vapour phase).
This hard difficulty chemistry question is from the chapter solutions, covering the topic of relative lowering of vapour pressure. It appeared in the 2025 exam.
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