AP Chemistry / Topic

Intermolecular Forces and Properties

Unit 3 connects particle-level forces to bulk properties: London dispersion, dipole-dipole, and hydrogen bonding, plus the gas laws, kinetic molecular theory, solutions and molarity, and spectroscopy. It is one of the most-weighted units on the exam.

54 practice questions2 FRQs42 skill areas

What the exam asks

  • Rank substances by boiling point, vapor pressure, viscosity, or surface tension using IMF strength
  • Identify when hydrogen bonding applies (H bonded directly to N, O, or F)
  • Apply the ideal gas law and Dalton's law of partial pressures
  • Explain non-ideal gas behavior at high pressure and low temperature
  • Work with molarity, dilutions, and particle diagrams of solutions
  • Connect photon energy and wavelength to spectroscopy regions (Beer-Lambert law for absorbance)

Key formulas and rules

Ideal gas law

PV = nRT

Dalton's law

P(total) = P1 + P2 + ...; Pi = Xi·P(total)

Molarity and dilution

M = mol/L; M1V1 = M2V2

Kinetic energy of gases

average KE ∝ temperature (K); lighter gases move faster at the same T

Beer-Lambert law

A = εbc — absorbance is proportional to concentration

IMF ranking

H-bonding > dipole-dipole > London dispersion (but dispersion grows with molar mass and surface area)

Question bank breakdown

Multiple choice

54

Free response

2

Difficulty mix

19 easy · 22 medium · 13 hard

Skills covered

Ideal gas lawTypes of intermolecular forcesPhysical properties and IMFs - Boiling pointPhysical properties and IMFs - Vapor pressurePhysical properties and IMFs - ViscosityPhysical properties and IMFs - Surface tensionDalton's law of partial pressuresKinetic molecular theoryDeviations from ideal gas behaviorSolutions and solubilityColligative propertiesPhase diagramsHeating curves and phase changesIntermolecular Force IdentificationHydrogen Bonding RequirementsLondon Dispersion Forces and PolarizabilityBoiling Point Trends and Molecular ShapeHydrogen Bonding and Boiling PointTypes of SolidsCovalent Network SolidsMetallic SolidsIdeal Gas Law CalculationsCombined Gas LawGas Density and Molar Mass

Every question in the bank comes with a per-choice explanation, so you learn why each wrong answer is wrong — not just the key.

Sample question

From the free tier of the ScoreMint AP Chemistry bank — try it, then check the answer.

A sealed, rigid container holds 2.00 mol of an ideal gas at 300 K and 1.00 atm. If the temperature is increased to 450 K, what is the new pressure in the container?

  • A. 0.67 atm
  • B. 1.00 atm
  • C. 1.50 atm
  • D. 2.25 atm
Show answer and explanation

Answer: C. Correct. Since the container is rigid (constant volume) and sealed (constant moles), we use Gay-Lussac's Law: P₁/T₁ = P₂/T₂. P₂ = P₁ × (T₂/T₁) = 1.00 atm × (450 K / 300 K) = 1.00 × 1.50 = 1.50 atm. Alternatively, from PV = nRT with constant V and n: P is directly proportional to T. T increases by a factor of 1.5, so P does too. ✓

Study tip

Never say 'hydrogen bonds break inside the molecule' — on ranking questions, name the specific IMF between molecules and compare its strength; breaking covalent bonds is a different (and wrong) claim.