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AP Chemistry › Unit 3

AP Chemistry Unit 3: Properties of Substances and Mixtures

18 to 22% of the exam · 180 practice questions · 5 free-response questions

Unit 3 explains physical properties through the forces between particles. You will compare intermolecular forces, use the ideal gas law and kinetic molecular theory, work with solutions and molarity, and connect light to molecules through spectroscopy and the Beer-Lambert law. Questions often ask you to rank boiling points or vapor pressures and justify the ranking, or to draw particle diagrams. A common mistake is confusing bonds within molecules with forces between them. Another is calling a force strong or weak without naming it and comparing it to the others present.

Practice this unit

White belt · 60 questionsRegular · 60 questionsBlack belt · 60 questionsFree response

White belt covers core terms and foundational ideas, Regular is set at exam level, and Black belt has harder multi-step questions.

Topics in this unit

Free-response questions

Sample questions

These are taken from the Regular set, with the answer and explanation shown.

Topic 3.2 Properties of Solids

Properties of four solids
SolidMelting point (°C)Conducts as a solid?Conducts as a liquid?Other observations
W801NoYesBrittle; dissolves in water
X1,085YesYesCan be hammered into thin sheets
Y1,713NoNoVery hard; insoluble in water
Z114NoNoSoft; dissolves in hexane

Hypothetical data for four unknown solids. The values match those of common substances.

1. Solid Z melts at only 114 °C. Which statement best explains this low melting point?

  1. AMelting Z overcomes only weak forces between molecules.
  2. BMelting Z breaks weak covalent bonds within its molecules.
  3. CMelting Z separates metal cations from a weak electron sea.
  4. DMelting Z overcomes weak attractions between its ions.

Answer: A. Z is a molecular solid. Its molecules are held together only by weak intermolecular forces, here dispersion forces. Little energy is needed to overcome them, and the bonds inside each molecule stay intact.

Common trap: Thinking the bonds inside the molecule break. Intramolecular covalent bonds are much stronger than intermolecular forces.

Topic 3.6 Deviation from Ideal Gas Law

Real-gas behavior of He and CH₄ at 300 K
01002003004005000.60.81.01.21.41.6PV/nRTPressure (atm)ideal gasAB

Curves calculated from a simple real-gas model. For an ideal gas, PV/nRT = 1 at every pressure.

2. One curve on the graph is for He and the other is for CH₄. Which curve is for CH₄, and why?

  1. ACurve A, because the weaker attractions in CH₄ push PV/nRT above 1.
  2. BCurve A, because the larger size of CH₄ molecules pushes PV/nRT above 1.
  3. CCurve B, because the smaller size of CH₄ molecules pulls PV/nRT below 1.
  4. DCurve B, because the stronger attractions in CH₄ pull PV/nRT below 1.

Answer: D. CH₄ has 10 electrons spread over a larger molecule, so its dispersion forces are much stronger than helium's. Attractions make the real pressure lower than ideal, so PV/nRT drops below 1. Only Curve B dips below 1.

Common trap: Choosing Curve A because CH₄ molecules are bigger. Size matters mostly at very high pressure; at moderate pressure, the stronger attractions of CH₄ show first.

Topic 3.11 Spectroscopy and the Electromagnetic Spectrum

3. A chemist measures how much 600 nm light is absorbed by a blue solution of copper(II) ions. What type of transition does the absorbed light cause?

  1. AAn electronic transition
  2. BA rotational transition
  3. CA vibrational transition
  4. DA nuclear transition

Answer: A. Light of 600 nm is in the visible region. Visible photons move electrons to higher energy levels, an electronic transition. The solution looks blue because it absorbs orange-red light.

Common trap: Choosing vibrational. Vibrations are caused by lower-energy infrared radiation, not visible light.

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