AP exam review verified for 2027

AP Chem Unit 3 Review: Properties of Substances and Mixtures

Review AP Chem Unit 3 to understand how intermolecular forces, solid types, gas behavior, solution chemistry, and spectroscopy connect molecular structure to observable properties. This unit carries 18-22% of the exam and spans topics from London dispersion forces through Beer-Lambert law.

Use this hub to review all 13 topics, practice with available questions and FRQs, and estimate your score with the AP score calculator.

What is AP Chem unit 3?

Unit 3 asks you to explain properties of matter by reasoning from the particle level upward. Every macroscopic observation, whether a high boiling point, a gas that deviates from ideal behavior, or a colored solution that absorbs light, has a particle-level explanation rooted in forces, motion, or energy.

Unit 3 covers how intermolecular forces determine solid and liquid properties, how the ideal gas law and kinetic molecular theory describe gas behavior, how solutions are prepared and represented, how separation techniques exploit IMF differences, and how spectroscopy and Beer-Lambert law connect light absorption to molecular structure and concentration.

Intermolecular forces drive physical properties

London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole forces determine boiling point, melting point, vapor pressure, and solubility. Stronger forces mean higher boiling points and lower vapor pressures. Polarizability, molecular size, and the presence of N-H, O-H, or F-H bonds determine which forces are present and how strong they are.

Gas laws connect P, V, T, and n

PV = nRT describes ideal gas behavior. Dalton's law adds partial pressures for mixtures. Kinetic molecular theory explains pressure and temperature at the particle level using KE = 1/2 mv² and the Maxwell-Boltzmann distribution. Real gases deviate from ideal behavior at low temperature and high pressure due to intermolecular attractions and finite particle volume.

Solutions and spectroscopy require quantitative reasoning

Molarity M = n/L and dilution M1V1 = M2V2 are the core solution calculations. Solubility follows the like-dissolves-like principle. Chromatography and distillation separate mixtures using IMF differences. Beer-Lambert law A = εbc links absorbance to concentration, and photon energy E = hν links light frequency to electronic, vibrational, or rotational transitions.

Structure determines properties at every scale

Whether you are comparing boiling points of alkanes, explaining why a real gas compresses less than predicted, or reading a calibration curve, the reasoning chain is the same: identify the particles, identify the forces or energy relationships between them, and connect those to the observable property. Unit 3 builds this reasoning pattern across solids, liquids, gases, solutions, and light-matter interactions.

AP Chem unit 3 topics

3.1

Intermolecular and Interparticle Forces

Identify and rank London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole forces from molecular structure. Connect IMF strength to boiling point, melting point, and vapor pressure.

open guide
3.2

Properties of Solids

Classify solids as ionic, covalent network, metallic, or molecular. Explain melting point, hardness, conductivity, and brittleness from particle-level structure and force type.

open guide
3.3

Solids, Liquids, and Gases

Use particulate models to represent particle arrangement, motion, and spacing in each phase. Connect phase properties like compressibility and fluidity to particle behavior.

open guide
3.4

Ideal Gas Law

Apply PV = nRT to solve for any gas variable. Use Dalton's law and mole fractions to find partial pressures in gas mixtures. Interpret P-V-T-n graphs.

open guide
3.5

Kinetic Molecular Theory

Explain gas pressure and temperature using particle motion and elastic collisions. Interpret Maxwell-Boltzmann distribution curves for different temperatures and molecular masses.

open guide
3.6

Deviation from Ideal Gas Law

Explain why real gases deviate from ideal behavior at low temperature (intermolecular attractions) and high pressure (finite particle volume). Predict direction of deviation.

open guide
3.7

Solutions and Mixtures

Distinguish homogeneous from heterogeneous mixtures. Calculate molarity using M = n/L and apply the dilution equation M1V1 = M2V2.

open guide
3.8

Representations of Solutions

Draw and interpret particulate diagrams showing relative concentration and solute-solvent interactions, including correct ion orientation and dissociation for electrolytes.

open guide
3.9

Separation of Solutions and Mixtures

Explain chromatography and distillation results using IMF differences between components and the stationary or mobile phase. Interpret Rf values for relative polarity.

open guide
3.10

Solubility

Apply the like-dissolves-like principle to predict solubility of ionic and molecular compounds in polar and nonpolar solvents. Identify the specific IMFs that drive dissolution.

open guide
3.11

Spectroscopy and the Electromagnetic Spectrum

Match microwave, infrared, and UV-visible radiation to rotational, vibrational, and electronic transitions. Relate wavelength and frequency to photon energy.

open guide
3.12

Properties of Photons

Use c = λν and E = hν to calculate photon frequency, wavelength, and energy. Connect photon absorption or emission to quantized energy changes in atoms and molecules.

open guide
3.13

Beer-Lambert Law

Apply A = εbc to calculate absorbance, concentration, path length, or molar absorptivity. Use calibration curves to determine unknown concentrations from measured absorbance values.

open guide
practice snapshot

Hardest AP Chemistry unit 3 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

65%average MCQ accuracy

Across 22k multiple-choice practice attempts for this unit.

22kMCQ attempts

Practice activity included in this snapshot.

61%average FRQ score

Across 26 scored free-response attempts for this unit.

Hardest topics in unit 3

MCQ miss rate
3.12

Review Properties of Photons with attention to how the concept appears in AP-style source and evidence questions.

49%925 tries
3.6

Review Deviation from Ideal Gas Law with attention to how the concept appears in AP-style source and evidence questions.

38%2,203 tries
3.4

Review Ideal Gas Law with attention to how the concept appears in AP-style source and evidence questions.

38%918 tries
3.8

Review Representations of Solutions with attention to how the concept appears in AP-style source and evidence questions.

36%1,444 tries

Unit 3 review notes

3.1

Intermolecular Forces

Intermolecular forces (IMFs) are attractions between separate particles. Their strength determines physical properties like boiling point, melting point, and vapor pressure. For any comparison question, identify which forces are present, then rank their strength based on molecular structure.

  • London dispersion forces: Present in all molecules; arise from temporary fluctuating dipoles. Strength increases with molecular size, number of electrons, and contact surface area. Branched molecules have lower boiling points than straight-chain isomers because of reduced contact area.
  • Dipole-dipole interactions: Occur between polar molecules with permanent dipole moments. Stronger than London dispersion for small polar molecules. Orientation-dependent: positive end of one molecule attracts negative end of another.
  • Hydrogen bonding: A strong dipole-dipole interaction requiring a hydrogen atom bonded directly to N, O, or F interacting with a lone pair on another N, O, or F. Explains the unusually high boiling point of water and the properties of alcohols and amines.
  • Ion-dipole forces: Attractions between an ion and a polar molecule. The strongest IMF type; critical for explaining why ionic compounds dissolve in water (hydration of ions).
  • Polarizability: How easily an electron cloud distorts. Increases with more electrons and larger electron clouds; enhanced by pi bonding. Higher polarizability means stronger London dispersion forces.
Given two molecules, identify all IMFs present in each and predict which has the higher boiling point, explaining your reasoning in terms of force type and strength.
Force typeMolecules involvedRelative strengthStructural requirement
London dispersionAll moleculesWeakest to strongest (size-dependent)All molecules; increases with electron count and surface area
Dipole-dipolePolar moleculesModeratePermanent dipole moment (asymmetric polar bonds)
Hydrogen bondingMolecules with N-H, O-H, or F-HStrongH bonded to N, O, or F; lone pair acceptor nearby
Ion-dipoleIon + polar moleculeStrongest IMFIon present; polar solvent molecule
3.2

Solid Types and the Three Phases

Solids are classified by the particles present and the forces holding them together. Each type has a predictable set of macroscopic properties. The three phases differ in particle arrangement, freedom of motion, and compressibility.

  • Ionic solids: Held together by strong electrostatic attractions between oppositely charged ions in a crystal lattice. High melting points, low vapor pressure, brittle (like-charge repulsion when layers shift), conduct electricity only when molten or dissolved.
  • Covalent network solids: Atoms connected by covalent bonds throughout the entire structure (e.g., diamond, graphite, SiO2). Extremely high melting points. Diamond is hard and nonconducting; graphite conducts electricity due to delocalized electrons between layers.
  • Metallic solids: Metal cations surrounded by a sea of delocalized valence electrons. Conduct electricity and heat, are malleable and ductile. Alloys (substitutional or interstitial) alter properties by changing the lattice.
  • Molecular solids: Discrete molecules held together by IMFs (London dispersion, dipole-dipole, or hydrogen bonding). Generally low melting points and higher vapor pressures than ionic or network solids.
  • Phase particle models: Solids: particles in fixed positions, vibrational motion only. Liquids: particles in contact but mobile, translational motion. Gases: particles far apart, rapid random translational motion, highly compressible.
Given a substance's melting point, conductivity, and hardness, identify its solid type and explain the particle-level basis for each property.
Solid typeParticlesForcesMelting pointConductivity
IonicCations and anionsElectrostatic (Coulombic)HighOnly when molten or dissolved
Covalent networkAtomsCovalent bondsVery highUsually none (diamond); yes (graphite)
MetallicMetal cations + electron seaMetallic bondsVariable (generally high)Yes (solid and liquid)
MolecularMoleculesIMFs (LDF, dipole, H-bond)Low to moderateNo
3.4

Ideal Gas Law and Partial Pressures

The ideal gas law PV = nRT relates the four macroscopic properties of a gas sample. For mixtures, Dalton's law of partial pressures applies: each gas contributes independently to the total pressure based on its mole fraction.

  • PV = nRT: P is pressure (atm), V is volume (L), n is moles, R is 0.08206 L·atm·mol⁻¹·K⁻¹, T is temperature in Kelvin. Always convert Celsius to Kelvin before calculating.
  • Dalton's law of partial pressures: Ptotal = PA + PB + PC + ... Each gas in a mixture exerts pressure independently. Partial pressure PA = XA × Ptotal, where XA is the mole fraction of gas A.
  • Mole fraction: XA = moles of A / total moles. Dimensionless; used to find partial pressure from total pressure or to find composition from partial pressures.
  • Gas law graphs: P vs. 1/V is linear at constant T and n (Boyle's law). V vs. T is linear at constant P and n (Charles's law). P vs. T is linear at constant V and n (Gay-Lussac's law). V vs. n is linear at constant P and T (Avogadro's law).
A container holds 0.50 mol N2 and 0.25 mol O2 at a total pressure of 3.0 atm. Calculate the partial pressure of each gas and the mole fraction of N2.
3.5

Kinetic Molecular Theory and Real Gas Deviations

Kinetic molecular theory (KMT) explains ideal gas behavior at the particle level. Real gases deviate from ideal predictions when intermolecular attractions or finite particle volumes become significant.

  • KMT assumptions: Gas particles are in constant random motion, have negligible volume compared to the container, undergo elastic collisions, and have no intermolecular attractions. Average kinetic energy is proportional to Kelvin temperature.
  • KE = 1/2 mv²: Average kinetic energy of a particle depends on mass and velocity. At the same temperature, heavier molecules move more slowly than lighter ones because average KE is the same for all gases at a given T.
  • Maxwell-Boltzmann distribution: A graph showing the distribution of particle speeds or kinetic energies at a given temperature. At higher T, the curve shifts right and flattens (broader distribution, higher average speed). Heavier molecules have a curve shifted left compared to lighter molecules at the same T.
  • Real gas deviations: At low temperature, intermolecular attractions pull particles together, reducing pressure below ideal predictions. At very high pressure, finite particle volume becomes significant, increasing pressure above ideal predictions.
Sketch two Maxwell-Boltzmann curves: one for N2 at 300 K and one at 600 K. Describe how the peak position and curve shape change and explain why.
ConditionDominant effectDeviation from idealExplanation
Low temperatureIntermolecular attractionsPressure lower than idealParticles slow down and attract each other before hitting walls
High pressureFinite particle volumeVolume larger than idealParticle volume is no longer negligible relative to container volume
High temperature, low pressureNeither effect significantBehaves nearly ideallyParticles move fast and are far apart
3.7

Solutions, Mixtures, and Particulate Representa­tions

A solution is a homogeneous mixture with uniform macroscopic properties throughout. Molarity is the primary concentration unit. Particulate diagrams show both the relative number of solute particles (concentration) and how solute and solvent particles interact.

  • Molarity: M = moles of solute / liters of solution. Use total solution volume, not just solvent volume. Units are mol/L or M.
  • Dilution equation: M1V1 = M2V2. Moles of solute are conserved when a solution is diluted by adding solvent. Useful for preparing solutions from stock solutions.
  • Homogeneous vs. heterogeneous mixtures: Homogeneous mixtures (solutions) have uniform properties throughout. Heterogeneous mixtures have properties that vary by location. Solutions can be solid, liquid, or gas.
  • Particulate models of solutions: Diagrams must show correct relative particle counts for concentration comparisons and correct interactions: ion-dipole forces for ionic solutes in water, hydrogen bonding for polar molecular solutes, water molecules oriented with O toward cations and H toward anions.
Draw a particulate diagram showing NaCl dissolved in water. Label the ion-dipole interactions and show the correct orientation of water molecules around Na+ and Cl-.
3.9

Separation Techniques and Solubility

Separation of mixtures and solubility both depend on differences in intermolecular interactions. The like-dissolves-like principle governs which solutes dissolve in which solvents, and the same logic explains how chromatography and distillation separate components.

  • Like dissolves like: Polar and ionic solutes dissolve in polar solvents (e.g., NaCl in water) because solute-solvent interactions (ion-dipole or dipole-dipole) are comparable in strength to the forces being broken. Nonpolar solutes dissolve in nonpolar solvents via London dispersion forces.
  • Chromatography: Separates components based on differential IMF strength between each component and the stationary phase versus the mobile phase. More polar components interact more strongly with a polar stationary phase and travel more slowly (lower Rf). The chromatogram reveals relative polarities.
  • Distillation: Separates liquids based on differences in vapor pressure (and therefore boiling point), which reflect differences in IMF strength. The component with weaker IMFs has higher vapor pressure and distills first.
  • Filtration limitation: Filtration cannot separate the components of a true solution because solute particles are too small to be captured by a filter. Chromatography or distillation is required.
Two compounds are separated by paper chromatography. Compound A has a higher Rf than compound B. Which is more polar? Explain using IMF interactions with the stationary phase.
TechniqueBasis of separationIMF principle
ChromatographyDifferential interaction with stationary vs. mobile phaseMore polar compound interacts more strongly with polar stationary phase
DistillationDifferences in vapor pressure / boiling pointWeaker IMFs give higher vapor pressure; that component distills first
FiltrationParticle sizeDoes not separate true solutions; only removes suspended solids
3.11

Spectroscopy, Photon Energy, and the Electroma­gnetic Spectrum

Different regions of the electromagnetic spectrum interact with matter in different ways depending on the energy of the photons. The two key equations are c = λν (connecting wavelength and frequency) and E = hν (connecting frequency and photon energy).

  • Spectral region to transition type: Microwave radiation causes transitions in molecular rotational energy levels. Infrared radiation causes transitions in molecular vibrational energy levels. Ultraviolet/visible radiation causes transitions in electronic energy levels.
  • c = λν: Speed of light c = 3.00 × 10^8 m/s equals wavelength (λ, in meters) times frequency (ν, in Hz). Shorter wavelength means higher frequency.
  • E = hν: Photon energy equals Planck's constant h (6.626 × 10^-34 J·s) times frequency. Higher frequency (shorter wavelength) means higher energy per photon. UV photons carry more energy than IR photons.
  • Absorption and emission: When a photon is absorbed, the atom or molecule increases in energy by exactly the photon's energy. When a photon is emitted, the species decreases in energy by the same amount. Energy levels are quantized, so only specific photon energies are absorbed or emitted.
A photon has a wavelength of 500 nm. Calculate its frequency and energy. Identify which type of molecular transition this photon could cause.
Spectral regionTransition typeRelative photon energy
MicrowaveMolecular rotationLowest
InfraredMolecular vibrationModerate
Ultraviolet/VisibleElectronic transitionsHighest
3.13

Beer-Lambert Law

The Beer-Lambert law A = εbc quantitatively relates the absorbance of a solution to the concentration of the absorbing species, the path length of the light through the solution, and the molar absorptivity. It is the basis for using a spectrophotometer to determine unknown concentrations.

  • A = εbc: A is absorbance (unitless), ε is molar absorptivity (M⁻¹cm⁻¹, a property of the specific molecule at a specific wavelength), b is path length (cm), c is concentration (M). Absorbance is directly proportional to concentration when b and λ are held constant.
  • Calibration curve: A graph of absorbance (y-axis) vs. known concentration (x-axis) for standard solutions. The linear relationship allows you to read off the concentration of an unknown solution from its measured absorbance.
  • Wavelength of maximum absorbance: The spectrophotometer is set to λmax, the wavelength at which the species absorbs most strongly. This maximizes sensitivity and the linear range of the Beer-Lambert relationship.
  • Molar absorptivity: ε is a constant for a given chemical species at a given wavelength. A higher ε means the species absorbs light more intensely; a small amount of it produces a large absorbance signal.
A solution has an absorbance of 0.45 at λmax. The molar absorptivity is 150 M⁻¹cm⁻¹ and the path length is 1.00 cm. Calculate the concentration of the absorbing species.

Practice AP Chem unit 3 questions

Try stimulus-based AP practice questions and written prompts after you review the notes.

Example stimulus-based MCQs

open all practice
diagram

Stimulus-based practice question

Thin-layer chromatography is performed on a mixture of pentane (C5H12C_5H_{12}), 1-pentanol (C5H11OHC_5H_{11}OH), and pentanal (C4H9CHOC_4H_9CHO) using polar silica gel and nonpolar hexane. The chromatogram is shown.

Question

Which spot most likely represents 1-pentanol, and why?

Spot C, because it forms strong hydrogen bonds with the polar silica.

Spot A, because it forms strong hydrogen bonds with the polar silica.

Spot C, because it has the strongest dispersion forces with the solvent.

Spot A, because it has the strongest dispersion forces with the solvent.

MCQ

AP-style practice question

Question

At 20C20^\circ C, the solubility of sodium chloride (NaClNaCl) is 36.036.0 g per 100.0100.0 g of water. A student adds 50.050.0 g of NaClNaCl to 200.0200.0 g of water at 20C20^\circ C. How much additional NaClNaCl must be added to create a saturated solution?

14.0 g

22.0 g

36.0 g

72.0 g

Example FRQs

open all FRQs
FRQ

Intermolecular forces in carbon disulfide and carbonyl sulfide

3. Answer the following questions about carbon disulfide, CS₂, and carbonyl sulfide, COS.

Carbon disulfide, CS₂, and carbonyl sulfide, COS, are both liquids at room temperature. The molar mass of CS₂ is 76.14 g/mol, and the molar mass of COS is 60.07 g/mol. The Lewis electron-dot diagrams for both molecules are shown in Figure 1.

Figure 1. Lewis electron‑dot diagrams for CS₂ and COS (linear molecules)

Figure 1
A.
i.

Identify the types of intermolecular forces present in liquid CS₂ and liquid COS.

ii.

The boiling point of CS₂ is 319 K, and the boiling point of COS is 223 K. Explain this difference in terms of the intermolecular forces identified in part A(i).

A student places a 15.2 g sample of CS₂(l) in a rigid, evacuated 5.00 L container at 350 K. All of the liquid vaporizes.

B.

A 15.2 g sample of CS₂(l) is vaporized completely in a rigid 5.00 L container held at 350 K. (The molar mass of CS₂ is 76.14 g/mol.)

i.

Calculate the pressure, in atm, of the CS₂ gas in the container. (Assume R = 0.08206 L·atm/(mol·K))

ii.

The actual pressure of the CS₂ gas in the container is found to be slightly lower than the value calculated in part B(i). Explain this deviation using principles of non-ideal gas behavior.

The student compares the motion of particles in samples of CS₂(g) and COS(g) at 350 K.

C.
i.

Which gas particles, CS₂ or COS, have the higher average kinetic energy? Justify your answer.

ii.

Which gas particles, CS₂ or COS, have the higher root-mean-square speed? Justify your answer.

The student prepares a solution by dissolving iodine, I₂(s), in liquid CS₂.

D.

Explain why I₂ is more soluble in CS₂ than in water, H₂O, in terms of intermolecular forces.

The student uses a spectrophotometer to analyze the solution of I₂ in CS₂. A calibration curve is generated using standard solutions, as shown in Figure 2.

Figure 2. Beer’s Law calibration curve for iodine (I₂) dissolved in CS₂

Figure 2
E.

A sample of the I₂ solution has an absorbance of 0.60. Using Figure 2, calculate the concentration of I₂ in the sample.

F.

The student rinsed the cuvette with pure CS₂ solvent but failed to dry it before adding the sample solution from part E. Would the calculated concentration of I₂ be greater than, less than, or equal to the actual concentration in the sample solution? Justify your answer.

SAQ

Molecular properties and intermolecular forces in halogenated hydrocarbons

6. A scientist is investigating the physical properties of dichloromethane, CH2Cl2CH_2Cl_2, and carbon tetrachloride, CCl4CCl_4. The Lewis electron-dot diagrams for the molecules are shown in Figure 1.

Table 1. Physical properties of dichloromethane and carbon tetrachloride

Substance

Molar Mass (g/mol)

Boiling Point (°C)

Enthalpy of Vaporization (kJ/mol)

CH2Cl2CH_2Cl_2

84.93

39.6

28.0

CCl4CCl_4

153.82

76.7

30.0

Figure 1. Lewis electron‑dot diagrams for dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4).

Figure 1
A.

Identify the hybridization of the central carbon atom in the CH2Cl2CH_2Cl_2 molecule shown in Figure 1.

B.

The boiling point of CCl4CCl_4 is higher than the boiling point of CH2Cl2CH_2Cl_2, as shown in Table 1. Identify the types of intermolecular forces present in each substance and explain why the boiling point of CCl4CCl_4 is higher.

C.

A student introduces a 1.50 g sample of CH2Cl2(l)CH_2Cl_2(l) into a rigid, evacuated 2.00 L container at 315 K. The sample vaporizes completely. The student claims that the pressure of the gas is greater than 0.30 atm. Do you agree or disagree with the student's claim? Justify your answer with a calculation.

D.

The scientist wants to vaporize 50.0 g of the substance from Table 1 that has the stronger intermolecular forces. Calculate the thermal energy, in kJ, required for this process.

SAQ

Molecular geometry and intermolecular forces in halogenated methanes

5. A physical chemist is investigating the properties of fluoromethane (CH3FCH_3F) and chloromethane (CH3ClCH_3Cl). The Lewis diagrams for the molecules are shown in Figure 1, and selected physical property data are provided in Table 1.

Table 1. Physical properties of fluoromethane and chloromethane

Compound

Molar Mass (g/mol)

Dipole Moment (D)

Boiling Point (K)

CH3FCH_3F

34.03

1.85

195

CH3ClCH_3Cl

50.49

1.87

249

Figure 1. Lewis diagrams for fluoromethane (CH3F) and chloromethane (CH3Cl)

Figure 1
A.

Based on VSEPR theory, predict the molecular geometry around the carbon atom in the CH3ClCH_3Cl molecule shown in Figure 1.

B.

A student claims that CH3FCH_3F should have a higher boiling point than CH3ClCH_3Cl because the C–F bond is more polar than the C–Cl bond. Do you agree or disagree with the student's claim? Justify your answer using the data in Table 1.

C.

At 220 K, which compound has the higher vapor pressure? Justify your answer.

D.

A 2.50 g sample of CH3FCH_3F gas is pumped into a rigid, evacuated 1.25 L container at a temperature of 298 K. Calculate the pressure, in atm, of the gas in the container.

Key terms

TermDefinition
London dispersion forcesTemporary attractive forces between all molecules arising from instantaneous fluctuating dipoles. Strength increases with molecular size, electron count, and surface contact area.
Hydrogen bondingA strong IMF requiring a hydrogen atom bonded directly to N, O, or F interacting with a lone pair on another N, O, or F atom. Responsible for the high boiling point of water and many biological properties.
PolarizabilityThe ease with which an electron cloud distorts in response to an electric field. Larger molecules with more electrons and pi bonds have higher polarizability and stronger London dispersion forces.
PV = nRTThe ideal gas law relating pressure (atm), volume (L), moles, the gas constant R (0.08206 L·atm·mol⁻¹·K⁻¹), and absolute temperature (K).
Dalton's Law of Partial PressureThe total pressure of a gas mixture equals the sum of the partial pressures of each component. Each partial pressure equals the mole fraction of that gas times the total pressure.
Maxwell-Boltzmann distributionA graph showing the distribution of kinetic energies or speeds of gas particles at a given temperature. Higher temperature shifts the curve right and broadens it; heavier molecules shift it left.
MolarityConcentration expressed as moles of solute per liter of solution (M = n/L). The denominator is total solution volume, not solvent volume alone.
like dissolves likePolar and ionic solutes dissolve in polar solvents; nonpolar solutes dissolve in nonpolar solvents. Dissolution is favorable when solute-solvent IMFs are similar in type and strength to the forces being disrupted.
Stationary PhaseThe fixed medium in chromatography (e.g., silica gel, paper) with which solute components interact. More polar compounds interact more strongly with a polar stationary phase and travel more slowly.
DistillationA separation technique that exploits differences in vapor pressure (IMF strength) between liquid components. The component with weaker IMFs has higher vapor pressure and vaporizes first.
E = hνPlanck's equation relating photon energy (J) to frequency (Hz) via Planck's constant h = 6.626 × 10⁻³⁴ J·s. Higher frequency means higher energy per photon.
AbsorbanceA unitless measure of how much light a solution absorbs at a given wavelength. Directly proportional to concentration when path length and wavelength are held constant (Beer-Lambert law).
Calibration curveA graph of absorbance vs. known concentration for standard solutions. Used to determine the concentration of an unknown solution by comparing its measured absorbance to the linear trend.
particulate-level modelA representation of matter showing individual particles, their arrangement, spacing, and interactions. Used in AP Chem to explain macroscopic properties from molecular structure.

Common unit 3 mistakes

Using Celsius instead of Kelvin in gas law calculations

PV = nRT and KE = (3/2)kT both require absolute temperature. Always convert: K = °C + 273. Using 25 instead of 298 will give a wrong answer every time.

Confusing London dispersion forces with van der Waals forces

London dispersion forces are one specific type of intermolecular force. The term van der Waals forces is broader and includes dipole-dipole interactions as well. Do not use the two terms interchangeably on the exam.

Using solvent volume instead of solution volume for molarity

Molarity is moles of solute per liter of solution, not per liter of solvent. When a solid is dissolved in water to make 250 mL of solution, the denominator is 0.250 L of solution, not the volume of water added.

Assuming all gases deviate from ideal behavior in the same direction

At low temperature, attractions handle and pressure is lower than ideal. At very high pressure, particle volume dominates and the gas occupies more volume than ideal. The direction of deviation depends on which effect is larger under the given conditions.

Misreading chromatography polarity from Rf values

A higher Rf means the compound traveled farther with the mobile phase, indicating weaker interaction with the polar stationary phase, which means the compound is less polar. Students often reverse this relationship.

How this unit shows up on the AP exam

Particulate diagram tasks

AP Chem frequently asks you to draw or interpret particle-level diagrams for solutions, phases, or gas mixtures. For Unit 3, this means showing correct IMF interactions between solute and solvent, correct ion dissociation for electrolytes, correct relative particle counts for concentration comparisons, and correct phase representations with appropriate spacing and motion implied.

Justify and explain tasks using IMFs

Many Unit 3 questions ask you to justify a ranking of boiling points, vapor pressures, or solubilities by naming the specific IMF, explaining why it is present from molecular structure, and connecting its strength to the observed property. A complete response names the force, explains its structural basis, and links it to the macroscopic trend.

Quantitative calculation with conceptual follow-up

Gas law and Beer-Lambert law problems often pair a calculation with a conceptual question. For example, after solving PV = nRT for moles, you may be asked to explain why a real gas deviates from that value. After using A = εbc to find concentration, you may be asked to explain what happens to absorbance if path length doubles. Practicing both the calculation and the explanation together prepares you for multi-part questions.

Final unit 3 review checklist

  • Rank IMFs and connect to physical propertiesFor any molecule, identify all IMFs present (LDF, dipole-dipole, H-bonding, ion-dipole), rank their strength, and explain how that ranking predicts boiling point, vapor pressure, or solubility.
  • Classify solids and explain their propertiesGiven melting point, conductivity, hardness, or brittleness data, identify the solid type (ionic, covalent network, metallic, molecular) and explain each property from particle-level structure.
  • Solve ideal gas law and partial pressure problemsUse PV = nRT with T in Kelvin. Calculate partial pressures using PA = XA × Ptotal and verify that partial pressures sum to total pressure.
  • Interpret Maxwell-Boltzmann distributionsPredict how the distribution curve shifts with increasing temperature or increasing molecular mass. Explain what the curve shape means for average kinetic energy and particle speed.
  • Explain real gas deviations qualitativelyState whether a real gas will have higher or lower pressure than ideal at a given condition, and attribute the deviation to either intermolecular attractions (low T) or finite particle volume (high P).
  • Calculate molarity and use dilutionApply M = n/L using total solution volume. Use M1V1 = M2V2 for dilution problems. Draw or interpret particulate diagrams showing correct relative concentrations.
  • Apply Beer-Lambert law and read calibration curvesUse A = εbc to solve for any variable. Read an unknown concentration from a calibration curve by locating the measured absorbance on the y-axis and reading the corresponding concentration on the x-axis.

How to study unit 3

Step 1: Build your IMF and solid-type foundation (3.1-3.3)Read the topic guides for 3.1, 3.2, and 3.3. For each IMF type, write one example molecule and one property it explains. For each solid type, list the particles, forces, and two macroscopic properties. Practice drawing particulate models for each phase.
Step 2: Work through gas law problems (3.4-3.6)Review PV = nRT and Dalton's law with the topic guides for 3.4 and 3.5. Solve at least five ideal gas law problems, including mixture problems using mole fractions. Then read 3.6 and practice explaining real gas deviations in writing before checking your reasoning.
Step 3: Practice solution chemistry calculations and diagrams (3.7-3.8)Use the 3.7 topic guide to practice molarity and dilution calculations. Then use the 3.8 guide to practice drawing particulate diagrams for strong electrolytes, weak electrolytes, and nonelectrolytes. Check that your diagrams show correct ion orientation around water molecules.
Step 4: Connect IMFs to separation and solubility (3.9-3.10)Review chromatography and distillation using the 3.9 topic guide. For each technique, write a one-sentence explanation of why a specific component moves faster or distills first, naming the IMF involved. Then review 3.10 and practice predicting solubility using like-dissolves-like.
Step 5: Review spectroscopy and Beer-Lambert law (3.11-3.13)Use the topic guides for 3.11 and 3.12 to memorize the three spectral region-to-transition pairings and practice E = hν and c = λν calculations with unit tracking. Finish with 3.13: practice A = εbc calculations and reading calibration curves. Use available practice questions and FRQs to test all three topics together.

More ways to review

Topic study guides

Open the individual guides for Unit 3 when you want a closer review of one topic.

browse guides

FRQ practice

Practice free-response reasoning and compare your answer with scoring guidance.

practice FRQs

Cram archive videos

Watch past review streams filtered to Unit 3 when you want a video walkthrough.

open videos

Cheatsheets

Use unit cheatsheets for a quick visual review after you work through the notes.

open cheatsheets

Score calculator

Estimate your broader AP score goal after you review the course and exam format.

open calculator

Frequently Asked Questions

What topics are covered in AP Chem Unit 3?

AP Chem Unit 3 covers 13 topics across intermolecular forces, states of matter, and solution chemistry. Key topics include Intermolecular and Interparticle Forces, Properties of Solids, Ideal Gas Law, Kinetic Molecular Theory, Deviation from Ideal Gas Law, Solubility, Spectroscopy and the Electromagnetic Spectrum, Properties of Photons, and Beer-Lambert Law. Here's the full topic list: - 3.1 Intermolecular and Interparticle Forces - 3.2 Properties of Solids - 3.3 Solids, Liquids, and Gases - 3.4 Ideal Gas Law - 3.5 Kinetic Molecular Theory - 3.6 Deviation from Ideal Gas Law - 3.7 Solutions and Mixtures - 3.8 Representations of Solutions - 3.9 Separation of Solutions and Mixtures - 3.10 Solubility - 3.11 Spectroscopy and the Electromagnetic Spectrum - 3.12 Properties of Photons - 3.13 Beer-Lambert Law See AP Chem Unit 3 for matched practice on all of these.

How much of the AP Chem exam is Unit 3?

AP Chem Unit 3 makes up 18-22% of the AP exam, making it one of the heavier-weighted units. It covers solubility, intermolecular forces, gas laws, solutions and mixtures, and spectroscopy topics like the Beer-Lambert Law. Expect a solid chunk of both MCQ and FRQ questions drawn from this material.

What's on the AP Chem Unit 3 progress check (MCQ and FRQ)?

The AP Chem Unit 3 progress check in AP Classroom includes both MCQ and FRQ parts that pull from all 13 topics in this unit. The MCQ section tests conceptual understanding of intermolecular forces, gas laws, solubility, and spectroscopy. The FRQ part typically asks you to interpret data, apply the Beer-Lambert Law, or explain solution behavior at the particle level. The progress check is designed to mirror real exam difficulty, so it's worth treating it like a mini exam. For extra practice on the same topics, check out AP Chem Unit 3.

How do I practice AP Chem Unit 3 FRQs?

AP Chem Unit 3 FRQs most often target solubility, intermolecular forces, gas law calculations, and spectroscopy concepts like the Beer-Lambert Law. These questions ask you to explain phenomena at the particle level, interpret graphs or spectra, or set up and solve quantitative problems. The best way to practice is to write out full explanations, not just answers, because graders reward specific reasoning. Focus on topics 3.1, 3.4, 3.10, and 3.13 first since they show up most in free-response contexts. You can find practice sets and worked examples at AP Chem Unit 3.

Where can I find AP Chem Unit 3 practice questions?

The best place to find AP Chem Unit 3 practice questions, including MCQ and full practice test sets, is AP Chem Unit 3. That page has targeted multiple-choice questions covering solubility, intermolecular forces, the Ideal Gas Law, Kinetic Molecular Theory, and spectroscopy. When you're doing MCQ practice, pay close attention to questions on Deviation from Ideal Gas Law and Beer-Lambert Law since those tend to trip students up. Mixing topic-specific drills with timed full-unit practice tests is the most effective approach.

How should I study AP Chem Unit 3?

Start AP Chem Unit 3 by building a strong foundation in intermolecular forces (3.1), because those concepts carry through solubility, states of matter, and solution behavior later in the unit. Then work through the gas law topics (3.4, 3.5, 3.6) together since Kinetic Molecular Theory explains why ideal gas assumptions break down. Here's a practical study sequence: 1. Learn intermolecular forces and how they predict physical properties. 2. Work through Ideal Gas Law problems, then study where and why gases deviate. 3. Study Solutions and Mixtures (3.7-3.10), focusing on solubility rules and what affects them. 4. Finish with spectroscopy (3.11-3.13), including the Beer-Lambert Law and how photon energy relates to the electromagnetic spectrum. For each topic, practice explaining concepts in words before doing calculations. Unit 3 is 18-22% of the exam, so it rewards consistent review. Head to AP Chem Unit 3 for practice questions organized by topic.

Ready to review Unit 3?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.