Ostwald–Walker Method: Vapour Pressure Calculations
The Ostwald–Walker method, also known as the dynamic method, is an experimental technique used to measure the relative lowering of vapour pressure of solutions containing non-volatile solutes. For students preparing for Class 12 board examinations, JEE Main, NEET, and CUET, this topic connects Raoult’s law with laboratory measurements. Direct manometric measurement of minute vapour pressure drops is prone to large errors. The Ostwald–Walker method converts pressure differences into easily measurable mass changes to find solute molar masses.
Experimental Arrangement and Operating Principle
The apparatus consists of three sets of bulbs connected in series and kept in a thermostat at constant temperature:
- Solution Bulbs: A slow stream of dry air passes through bulbs containing the solution. The air becomes saturated with solvent vapour up to the solution vapour pressure p. The solution loses mass, and this loss w1 is directly proportional to p.
- Solvent Bulbs: The air stream, already carrying vapour at pressure p, enters bulbs containing pure solvent. Because pure solvent vapour pressure p° exceeds p, the air absorbs additional vapour until saturated at p°. The mass lost by the solvent bulbs, w2, is directly proportional to (p° − p).
- Absorption Tubes: Saturated air containing vapour at pressure p° passes through U-tubes packed with anhydrous calcium chloride. All solvent vapour is absorbed. The mass gained by the tubes, w3, equals w1 + w2 and is directly proportional to p°.
Ostwald–Walker Formulas and Proportionalities:
1. Mass loss of solution bulbs: w1 ∝ p
2. Mass loss of pure solvent bulbs: w2 ∝ (p° − p)
3. Mass gain of absorption tubes: w3 = w1 + w2 ∝ p°
4. Relative lowering of vapour pressure: (p° − p) / p° = w2 / (w1 + w2) = w2 / w3
5. Raoult’s law relation: (p° − p) / p = w2 / w1 = n2 / n1 = (W2 × M1) / (W1 × M2)
Where w1 = solution bulb loss (g); w2 = solvent bulb loss (g); w3 = tube gain (g); p° = pure solvent vapour pressure; p = solution vapour pressure; W1, W2 = solvent and solute masses (g); M1, M2 = solvent and solute molar masses (g/mol); n1, n2 = moles of solvent and solute.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 12, State Boards, JEE Main, NEET, CUET-UG
- Topic: Ostwald–Walker Method and Relative Lowering of Vapour Pressure
- Difficulty: School and entrance examination level
Step-by-Step Worked Examples
Question: A stream of dry air was passed through bulbs containing a solution of 12.00 g of an organic non-electrolyte in 100.00 g of water, and then through pure water bulbs at 298 K. The loss in mass of the solution bulbs was 2.500 g, and the loss in mass of the solvent bulbs was 0.050 g. Calculate the molar mass of the solute. Take the molar mass of water as 18.02 g/mol.
Solution:
Step 1: Record mass losses: w1 = 2.500 g (proportional to p) and w2 = 0.050 g (proportional to p° − p).
Step 2: Relate losses to vapour pressure ratio: (p° − p) / p = w2 / w1 = 0.050 / 2.500 = 0.0200.
Step 3: Connect to Raoult’s law: (p° − p) / p = n2 / n1.
Moles of water n1 = 100.00 g / 18.02 g/mol = 5.5494 mol.
Step 4: Moles of solute n2 = 0.0200 × 5.5494 mol = 0.11099 mol.
Step 5: Molar mass M2 = W2 / n2 = 12.00 g / 0.11099 mol = 108.12 g/mol.
Second route verification: Total loss w3 = w1 + w2 = 2.550 g. Relative lowering (p° − p) / p° = 0.050 / 2.550 = 1 / 51. Using n2 / (n1 + n2) = 1 / 51 gives n2 = n1 / 50 = 5.5494 / 50 = 0.11099 mol, confirming M2 = 12.00 / 0.11099 = 108.12 g/mol.
Answer: 108.12 g/mol
Question: Dry air was bubbled through a solution containing 15.00 g of a non-volatile compound in 250.00 g of ethanol (molar mass = 46.07 g/mol), and then through pure ethanol bulbs. The loss in mass of the solvent bulbs was 0.0450 g, and the total gain in mass in the absorption tubes was 1.8000 g. Find the relative lowering of vapour pressure and the solute molar mass.
Solution:
Step 1: Given: w2 = 0.0450 g, w3 = 1.8000 g, W1 = 250.00 g, M1 = 46.07 g/mol, W2 = 15.00 g.
Step 2: Relative lowering: (p° − p) / p° = w2 / w3 = 0.0450 / 1.8000 = 0.02500.
Step 3: Loss in solution bulbs: w1 = w3 − w2 = 1.8000 − 0.0450 = 1.7550 g.
Step 4: Ratio: (p° − p) / p = w2 / w1 = 0.0450 / 1.7550 = 1 / 39 ≈ 0.025641.
Step 5: Solvent moles: n1 = 250.00 / 46.07 = 5.4265 mol.
Step 6: Solute moles: n2 = 5.4265 / 39 = 0.13914 mol.
Step 7: Molar mass M2 = 15.00 / 0.13914 = 107.80 g/mol.
Second route verification: From (p° − p) / p° = n2 / (n1 + n2) = 0.02500, n2 = (0.02500 / 0.97500) × 5.4265 = 0.13914 mol, yielding M2 = 15.00 / 0.13914 = 107.80 g/mol.
Answer: Relative lowering = 0.02500; Molar mass = 107.80 g/mol
Common Mistakes Students Make
Keep these recurring pitfalls in mind:
- Confusing w2 with p°: The solvent bulb loss w2 is proportional to (p° − p), not p°. Incoming air is already saturated at pressure p; it only absorbs additional vapour up to p°.
- Misidentifying absorption tube gain: The drying tube absorbs all solvent vapour. Its mass gain w3 is proportional to p° and equals w1 + w2.
- Using approximate formulas carelessly: The relation (p° − p) / p = w2 / w1 = n2 / n1 is exact. Setting w2 / w3 = n2 / n1 is an approximation valid only for dilute solutions.
- Ignoring temperature constancy: Temperature variations between bulbs invalidate the proportionalities because vapour pressure depends on temperature.
Exam Relevance and Question Patterns
| Examination | Question Format | Key Concepts Tested |
|---|---|---|
| CBSE and State Boards | Numerical and short-answer problems | Apparatus layout, mass-loss relations, and calculating solute molar mass |
| JEE Main | MCQs and numerical value questions | Rapid calculation of w2/w1, exact vs dilute Raoult’s law |
| NEET | Conceptual MCQs and calculations | Identifying proportionalities for w1, w2, and w3 |
| CUET-UG | Objective questions | Direct formula substitution connecting bulb weight losses to vapor pressure |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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