Theory — Properties That Depend on the Number of Particles
A colligative property depends on the number of dissolved solute particles in a solution and not on their chemical identity. Adding a nonvolatile solute lowers the tendency of the solvent to escape, which raises the temperature needed to boil it and lowers the temperature at which it freezes.
Boiling-point elevation and freezing-point depression
Both effects are proportional to the molality of the solution, the moles of solute per kilogram of solvent, through constants that belong to the solvent. The boiling point goes up and the freezing point goes down.
Freezing-point depression: change in Tf = i × kf × m
Here kb is the molal boiling-point elevation constant and kf is the molal freezing-point depression constant of the solvent, m is the molality, and i is the van't Hoff factor.
The van't Hoff factor
The van't Hoff factor i is the number of particles a formula unit produces in solution. A nonelectrolyte such as glucose stays whole, so i is one. An ionic solute breaks into ions, so it contributes more particles: sodium chloride gives two, calcium chloride and sodium sulfate give three. Because colligative effects count particles, an ionic solute lowers the freezing point more than a nonelectrolyte at the same molality.
NaCl, KCl, KI, HCl: i = 2
CaCl₂, Na₂SO₄: i = 3
Finding a molar mass
Because the freezing-point depression depends on molality, measuring it for a known mass of an unknown nonelectrolyte gives the number of moles present, and therefore the molar mass. First find the molality from the measured depression, then the moles in the solvent, then divide the mass by the moles.
moles of solute = m × (kilograms of solvent)
molar mass M = (mass of solute) / (moles of solute)
Count particles
Colligative properties depend on how many particles dissolve, not on what they are, which is why the van't Hoff factor matters.
Freezing drops
The freezing point always falls, which is why salt is spread on icy roads and antifreeze protects an engine.
Mass from depression
A measured freezing-point depression, with the solvent constant, reveals the molar mass of an unknown solute.
| Quantity | Relationship | Notes |
|---|---|---|
| Boiling-point elevation | change in Tb = i kb m | boiling point rises |
| Freezing-point depression | change in Tf = i kf m | freezing point falls |
| van't Hoff factor i | particles per formula unit | 1 for nonelectrolytes |
| Molar mass | M = kf · mass · 1000 / (change in Tf · grams solvent) | nonelectrolyte |
Apparatus
The equipment a real colligative-properties experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.
Instructions — Running the Virtual Experiment
This is a record, calculate, and compare lab. In each part you set up the solution, calculate the requested quantity by hand, enter your value, and only then does the simulation reveal its result so you can compare. Record every value in your worksheet.
Simulation — The Colligative Bench
| Solvent | Solute (i) | m | Target | Your value (°C) | Actual (°C) |
|---|---|---|---|---|---|
| No rows yet — set up the solution, calculate the point, enter it, and check. | |||||
Build the solution
| Solute | m | Your i | i actual | Your fp (°C) | fp actual (°C) |
|---|---|---|---|---|---|
| No rows yet — choose a solute, give i and the freezing point, and check. | |||||
Electrolyte in water
| Solvent | Mass solute (g) | Mass solvent (g) | ΔTf (°C) | Your M (g/mol) | M actual (g/mol) |
|---|---|---|---|---|---|
| No rows yet — read ΔTf, calculate the molar mass, enter it, and check. | |||||
Unknown nonelectrolyte
Team Questions
Example Lab Report
A worked example showing the expected format and the record, calculate, and compare workflow.
Colligative Properties
Chemistry | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Objective
To calculate the boiling-point elevation and freezing-point depression of solutions, to account for the van't Hoff factor of electrolytes, and to determine an unknown molar mass from a measured freezing-point depression, comparing every calculated value with the simulation.
Part A — Boiling and Freezing Points (worked example)
For 1.00 m NaCl in water (kf = 1.86 °C/m, i = 2): change in Tf = i kf m = 2 × 1.86 × 1.00 = 3.72 °C, so the freezing point is 0.00 minus 3.72 = minus 3.72 °C. For the boiling point, change in Tb = 2 × 0.512 × 1.00 = 1.02 °C, so the solution boils at 101.02 °C. Both matched the simulation.
Part B — van't Hoff Factor (worked example)
For 0.50 m CaCl₂ in water, i = 3, so change in Tf = 3 × 1.86 × 0.50 = 2.79 °C and the freezing point is minus 2.79 °C. At the same 0.50 m, glucose (i = 1) gives only minus 0.93 °C, confirming that more ions lower the freezing point more.
Part C — Molar Mass (worked example)
Dissolving 1.50 g of an unknown nonelectrolyte in 25.0 g of benzene (kf = 5.12 °C/m) gave a measured depression of 2.40 °C. M = kf × mass × 1000 / (change in Tf × grams solvent) = 5.12 × 1.50 × 1000 / (2.40 × 25.0) = 128 g/mol, matching the simulation.
Discussion and Conclusion
Every calculated value agreed with the simulation. The freezing point fell and the boiling point rose in proportion to the molality, electrolytes shifted the points further because of their extra particles, and the measured freezing-point depression recovered the molar mass of the unknown, confirming that colligative properties count particles.