Library / Solution chemistry, equilibria and kinetics / SCH-46
Solution chemistry, equilibria and kinetics
Solvent vapour pressure above an ideal solution
Partial pressure of the solvent over an ideal solution of a non-volatile solute, from the solvent mole fraction and the pure-solvent vapour pressure at the same temperature (Raoult’s law).
- Ref
- SCH-46
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- p [kPa]
Raoult’s law says the partial pressure of a component over an ideal liquid mixture is its mole fraction times its pure-component vapour pressure at the same temperature. For a solvent with a non-volatile solute, p = x p*. Dissolving a solute therefore lowers the vapour pressure by p* (1 − x). The law is accurate when the solvent fraction is near 1 and the molecules are similar. If other volatile components are present, this gives only the solvent’s partial pressure.
Partial pressure of the solvent over an ideal solution of a non-volatile solute, from the solvent mole fraction and the pure-solvent vapour pressure at the same temperature (Raoult’s law).
Calculator
- Vapour-pressure lowering, p* − p1 kPa
- Relative lowering, 1 − x0.05
- Lowering per 0.01 of solute mole fraction0.2 kPa
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- With 95 mol % solvent the vapour pressure is 19 kPa, 1 kPa below the pure solvent.
- The fractional lowering (5 %) equals the solute mole fraction, which is the basis of the colligative effects on boiling and freezing points.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kPa) | Local step | Change in result |
|---|---|---|---|---|---|---|
| p* Pure-solvent vapour pressure | 0.01 to 1,000 kPa | 0.0095 | 950 | 950 | +10% | +10 % |
| x Mole fraction of solvent | 0.01 to 1 | 0.2 | 20 | 19.8 | +10% | — |
Sweep of Mole fraction of solvent
| Mole fraction of solvent | Solvent vapour pressure (kPa) |
|---|---|
| 0.01 | 0.2 |
| 0.1338 | 2.675 |
| 0.2575 | 5.15 |
| 0.3812 | 7.625 |
| 0.505 | 10.1 |
| 0.6288 | 12.58 |
| 0.7525 | 15.05 |
| 0.8762 | 17.52 |
| 1 | 20 |
Sweep one variable
Variables and how each one contributes
Solvent vapour pressure
What the equation gives you. Everything else feeds it.
Mole fraction of solvent
The mole fraction of the solvent, between 0 and 1. Raises the vapour pressure in direct proportion. Raising it by 10 % (to 1.045) gives no valid result at the other default values.
Pure-solvent vapour pressure
The pure-solvent vapour pressure at the same temperature, in kPa. Raises the vapour pressure in direct proportion. Raising it by 10 % (to 22) moves the result by +10 %, from 19 to 20.9 kPa.
Worked example with the values above
- Pure-solvent pressure: p* = 20 kPa.
- Multiply by the mole fraction 0.95.
- p = 19 kPa.
Assumptions and limits
- Dilute or ideal solutions with a non-volatile solute, unless a second volatile component is stated.
- Constant temperature (or constant pressure for boiling-point elevation) and a membrane that passes only the solvent for osmosis.
Common mistakes
- Using molarity where molality is needed for freezing and boiling points.
- Forgetting the van’t Hoff factor for salts.
- Adding the elevation or depression to the temperature in the wrong direction.
Statistics and uncertainty
Colligative formulas assume ideal behaviour, which holds only for dilute solutions. The effective particle factor i of an electrolyte is below its ideal value, by 5 % to 30 % even at 0.1 mol/kg.
Where it is used
- Estimating antifreeze, de-icing and desalination behaviour.
- Finding molar masses and reading simple phase diagrams.
Maths you need
- Raoult’s law: partial pressure = mole fraction × pure vapour pressure
- Colligative properties: properties that depend on particle number
- Molality: moles of solute per kilogram of solvent