AP Chemistry, AP exams, 18–22%

Unit 3: Properties of Substances and Mixtures

Unit 3 carries the largest published weighting in AP Chemistry. It spans intermolecular forces, the properties of solids, liquids, and gases, the ideal gas law and kinetic molecular theory, solutions and separations, and the spectroscopy topics including the Beer-Lambert law. Because it is both wide and heavily weighted, it deserves the most scheduled review time of any unit.

Position: 3 of 9 units
Published weighting: 18–22% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Chemistry. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Intermolecular and interparticle forces: London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole ranked and applied to physical properties.
Properties of solids, liquids, and gases: How particle-level arrangement and motion explain the macroscopic behavior of each phase.
Ideal gas law and kinetic molecular theory: The gas law relationships, the assumptions behind them, and Maxwell-Boltzmann speed distributions.
Deviation from ideal behavior: When high pressure or low temperature makes real gases depart from ideal predictions, and why.
Solutions, mixtures, and separations: Molarity, particulate representations of solutions, solubility, and separations including chromatography and distillation.
Spectroscopy and the Beer-Lambert law: The electromagnetic spectrum, photon energy, and using absorbance to determine concentration.

Study tips

How to practice Substances and Mixtures.

Section-specific guidance that follows from how this material is usually tested.

Rank intermolecular forces for any pair of substances in one step, then justify the ranking in a sentence; that is a standard free-response task.
Practice gas law problems with unit tracking rather than formula matching, especially when moles are the unknown.
Give this unit the most review sessions in your plan; its published weighting is roughly double that of most other units.

Where this fits

Unit 3: Properties of Substances and Mixtures is 3 of 9 units in the AP Chemistry guide, and it carries a published weighting of 18–22% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Chemistry guide Official exam information

Sample Substances and Mixtures questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the strongest intermolecular force acting between two water molecules?
1. ALondon dispersion forces
2. BIon-dipole forces
3. CCovalent bonding
4. DHydrogen bonding
Show answer
Answer: D. Hydrogen bonding
Water has hydrogen atoms bonded directly to a highly electronegative oxygen, which is the requirement for hydrogen bonding, the strongest intermolecular force present in pure water. London dispersion forces are also present but are much weaker in a molecule this small. Covalent bonding is the tempting distractor, but it holds atoms together inside a molecule rather than between molecules.
Key insight: Hydrogen bonding requires H bonded to N, O, or F, which is why water boils far above what its molar mass alone would predict.
Question 2Foundational
True or false: at the same temperature, a sample of O2 gas has a lower average kinetic energy than a sample of He gas because oxygen molecules are heavier.
1. ATrue
2. BFalse
Show answer
Answer: B. False
Average kinetic energy of a gas depends only on absolute temperature, so at the same temperature every gas has the same average kinetic energy per particle. Because kinetic energy combines mass and speed, the heavier oxygen molecules must move more slowly on average than helium atoms to hold the same energy. Confusing average speed with average kinetic energy is the classic error in reading Maxwell-Boltzmann distributions.
Key insight: Same temperature means same average kinetic energy; heavier particles simply move more slowly.
Question 3Intermediate
A rigid 2.00 L container holds 0.500 mol of an ideal gas at 300. K. Using R = 0.0821 L atm / (mol K), what is the pressure of the gas in atmospheres?
1. A3.08 atm
2. B6.16 atm
3. C12.3 atm
4. D24.6 atm
Show answer
Answer: B. 6.16 atm
Rearranging PV = nRT gives P = nRT/V, so P equals 0.500 mol times 0.0821 times 300 kelvin, divided by 2.00 L, which is 12.3 divided by 2.00, or 6.16 atm. The value 12.3 atm is the tempting distractor because it is nRT before dividing by the volume. Carrying units through the calculation confirms that liters cancel and atmospheres remain.
Key insight: Solve the ideal gas law symbolically first, then substitute, so no term gets dropped.
Question 4Intermediate
The boiling points of the halogens increase steadily from F2 to Cl2 to Br2 to I2. What explains this trend?
1. ALarger, more polarizable electron clouds create stronger London dispersion forces
2. BHydrogen bonding becomes possible in the heavier halogens
3. CThe molecules become increasingly polar down the group
4. DThe covalent bond inside each molecule becomes stronger down the group
Show answer
Answer: A. Larger, more polarizable electron clouds create stronger London dispersion forces
All four are nonpolar diatomic molecules, so dispersion forces are the only intermolecular attraction available. More electrons make the electron cloud easier to distort, producing larger temporary dipoles and stronger attractions, so more energy is needed to separate the molecules. The claim about covalent bond strength is the tempting distractor, but boiling separates molecules from one another and does not break bonds within them.
Key insight: Boiling point reflects intermolecular forces, never the strength of the bonds inside the molecule.
Question 5Advanced
A colored solution has an absorbance of 0.600 in a cell with a path length of 1.00 cm, and the molar absorptivity of the solute is 1.50 x 10^3 L / (mol cm). What is the concentration of the solution?
1. A4.00 x 10^-4 mol/L
2. B4.00 x 10^-3 mol/L
3. C9.00 x 10^2 mol/L
4. D2.50 x 10^3 mol/L
Show answer
Answer: A. 4.00 x 10^-4 mol/L
The Beer-Lambert law states that absorbance equals molar absorptivity times path length times concentration, so concentration equals 0.600 divided by the product of 1.50 x 10^3 and 1.00, which is 4.00 x 10^-4 mol/L. The value 9.00 x 10^2 comes from multiplying absorbance by molar absorptivity instead of dividing. Absorbance is unitless, so the surviving units are mol/L.
Key insight: Absorbance rises with concentration, so a small absorbance in a standard cell means a dilute solution.

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Unit 3: Properties of Substances and Mixtures