Colligative Properties

Why salt the roads?

After a snowstorm, road crews spread salt and the ice melts, because salt water doesn’t freeze at 0 °C. The antifreeze in a car’s radiator, a solution of ethylene glycol in water, keeps the engine from freezing in winter. Yet a spoonful of salt in a pot of pasta water hardly changes its boiling point at all. These effects have something in common: their size depends only on how many solute particles are dissolved, not on what the solute is.

This lesson is optional in this course. US honors chemistry covers these calculations; Chinese high-school chemistry does not.

When a nonvolatile solute dissolves in a solvent, some properties of a dilute solution depend only on the concentration of solute particles, not on their identity. These are colligative properties依数性:

Vapor pressure lowering蒸气压下降: vapor pressure is the pressure of the vapor once molecules leave and return to the liquid’s surface at equal rates (lesson 2.8.5). Solute particles take up some of the surface, so fewer solvent molecules escape into the gas each second, and the solution’s vapor pressure is lower than the pure solvent’s. (This is a simplified picture; the full explanation uses entropy.)
Boiling point elevation沸点升高: with a lower vapor pressure, the solution must be heated further before its vapor pressure reaches the air pressure and it boils.
Freezing point depression凝固点降低: solute particles get in the way of solvent molecules lining up into a crystal, so the solution must be colder to freeze.
Osmotic pressure渗透压: if a membrane that lets only water through separates a dilute solution from a concentrated one, water moves into the concentrated side. Saline IV fluid (0.9% NaCl, as in 1.7.4) has the same osmotic pressure as blood, so red blood cells neither swell and burst nor shrivel.

Molality

Boiling and freezing point changes are calculated with molality质量摩尔浓度, not molarity. Molality is the moles of solute per kilogram of solvent, with the symbol m and units of mol/kg:

m = \frac{\text{mol solute}}{\text{kg solvent}}

Why switch? The volume of a solution changes with temperature, but the mass of solvent doesn’t. Colligative properties are all about temperature changes, so mass of solvent is the better measure. For very dilute aqueous solutions, molality and molarity are nearly equal.

ΔT = i·K·m

The boiling point elevation and freezing point depression of a dilute solution are proportional to the molality of solute particles:

\Delta T_\text{f} = i K_\text{f} m \qquad \Delta T_\text{b} = i K_\text{b} m

K_f and K_b depend only on the solvent. For water, K_f = 1.86 °C·kg/mol and K_b = 0.512 °C·kg/mol.
i is the van ’t Hoff factor范特霍夫因子: the number of particles one formula unit becomes in solution. For nonelectrolytes such as glucose, sucrose, and ethylene glycol, i = 1. NaCl splits into Na⁺ and Cl⁻, so i = 2; for CaCl₂, i = 3.
In real solutions some cations and anions stay paired up, so the measured i is a little smaller than the ideal value, and more so at higher concentration.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, O 16.00, Na 22.99, Cl 35.45, Ca 40.08).

Try different solutes at the same molality and compare how far the freezing point drops. Then work out how much salt it takes to melt ice:

Try it: freezing point depression and boiling point elevation

Solute

Mass of solutegMass of waterg

Schematic: pale blue dots are water molecules and colored dots are solute particles (real solutions have far fewer solute particles).

How Chinese textbooks differ

Chinese textbooks (following the Chinese national standard) write molality as b, which avoids confusing it with mass m. Its name, 质量摩尔浓度, means “mass-based molar concentration.”
Chinese high-school chemistry doesn’t do these calculations. They first appear in university general chemistry, so a student from China may never have used ΔT_f = iK_fm before.

Example 1: freezing and boiling points of salt water

You dissolve 5.844 g of NaCl in 500.0 g of water. Estimate the freezing point and the boiling point of the solution at 1 atm.

Step-by-step solution

The formulas. ΔT_f = iK_fm and ΔT_b = iK_bm. For water, K_f = 1.86 °C·kg/mol and K_b = 0.512 °C·kg/mol.
Moles of solute.
$\begin{aligned} n(\ce{NaCl}) &= \frac{5.844\ \text{g}}{58.44\ \text{g/mol}} \\ &= 0.1000\ \text{mol} \end{aligned}$
Molality. Divide by the mass of solvent in kilograms, not the volume of solution:
$\begin{aligned} m &= \frac{0.1000\ \text{mol}}{0.5000\ \text{kg}} \\ &= 0.2000\ \text{mol/kg} \end{aligned}$
Van ’t Hoff factor. NaCl splits into Na⁺ and Cl⁻, doubling the number of particles, so i = 2.
Freezing point.
$\begin{aligned} \Delta T_\text{f} &= 2 \times 1.86 \times 0.2000 \\ &= 0.744\ \text{°C} \end{aligned}$
The solution freezes at about 0 − 0.744 = −0.744 °C.
Boiling point.
$\begin{aligned} \Delta T_\text{b} &= 2 \times 0.512 \times 0.2000 \\ &= 0.205\ \text{°C} \end{aligned}$
It boils at about 100 + 0.205 = 100.205 °C.
A common mistake. Forgetting i = 2, which gives −0.372 °C. The measured drop is actually a little less than 0.744 °C, because oppositely charged ions attract each other, so the solution acts as if it had slightly fewer than twice as many particles.

Example 2: how much antifreeze?

What is the minimum mass of ethylene glycol (C₂H₆O₂, a nonelectrolyte) you must add to 1.00 kg of water to lower its freezing point to −5.00 °C?

Step-by-step solution

The plan. Work backward from the temperature drop: ΔT_f → molality → moles → grams.
Molality needed. ΔT_f = 0 − (−5.00) = 5.00 °C, and ethylene glycol is a nonelectrolyte, so i = 1:
$\begin{aligned} m &= \frac{\Delta T_\text{f}}{i K_\text{f}} \\ &= \frac{5.00}{1 \times 1.86} \\ &= 2.688\ \text{mol/kg} \end{aligned}$
Moles of ethylene glycol. 2.688 mol/kg × 1.00 kg = 2.688 mol.
Mass. The molar mass of C₂H₆O₂ is 2(12.01) + 6(1.008) + 2(16.00) = 62.07 g/mol:
$2.688 \times 62.07 = \boxed{167\ \text{g}}$
Check. 167 g ÷ 62.07 g/mol ÷ 1.00 kg = 2.69 mol/kg, and 2.69 × 1.86 = 5.00 °C, as required.
Think about it. This solution is no longer very dilute, so the answer is only an estimate. Car antifreeze contains much more ethylene glycol and protects down to well below −30 °C.

Practice: what is the molality of solute particles, i·m?

0.1 m glucose
0.1 m NaCl
0.1 m CaCl₂
0.2 m KBr
0.2 m MgCl₂
0.3 m sucrose
0.1 m Na₂SO₄
0.2 m ethylene glycol
0.1 m Al(NO₃)₃

Common mistakes

Mistake 1: “Salting pasta water makes it boil faster and hotter.” A spoonful of salt (about 10 g) in 2 kg of water gives m ≈ 0.085 mol/kg, which raises the boiling point by only about 2 × 0.512 × 0.085 ≈ 0.09 °C. And a higher boiling point means the water actually takes slightly longer to boil.

Mistake 2: “Colligative properties depend on what the solute is.” They depend only on the number of particles. At the same molality, NaCl gives two particles per formula unit, so it lowers the freezing point twice as much as glucose. It isn’t because NaCl is “saltier.”

Mistake 3: “Molality and molarity are the same thing.” Molality divides by kilograms of solvent; molarity divides by liters of solution. For very dilute aqueous solutions the numbers are close, but for concentrated solutions they differ a lot.

Quick check

Which of these has the lowest freezing point?

0.500 mol of sucrose is dissolved in 1.00 kg of water. What is the approximate freezing point of the solution?

Summary

Colligative properties (vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure) depend only on the number of solute particles, not on what they are.
Molality m = mol solute / kg solvent. ΔT_f = iK_fm and ΔT_b = iK_bm; for water, K_f = 1.86 and K_b = 0.512 °C·kg/mol.
For an electrolyte, the ideal i is the number of ions per formula unit, so CaCl₂ has a bigger effect than NaCl or glucose at the same molality.

On to AP (for later)

The AP Course and Exam Description lists colligative properties and molality calculations as content that is not assessed on the exam. AP Unit 3 still asks you to explain vapor pressure, boiling point, and solubility from particles and intermolecular forces. University general chemistry goes further, using freezing point depression and osmotic pressure to find molar masses.

Data sources: K_f and K_b for water from OpenStax Chemistry 2e, Table 11.2.

Colligative Properties依数性

Learning goals

Before you start

Why salt the roads?