AP Chemistry · Handsworth Secondary 2026–27

Unit 3 · Properties of Substances & Mixtures

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 3, the heaviest-weighted unit on the AP exam (18–22%), taught here in 8 blocks. From BC Chemistry 11 you already know solutions, molarity, and the basics of gases; we move quickly there and spend our depth on the AP story: how intermolecular forces set physical properties, the ideal gas law & KMT, real-gas deviations, separations, and the Beer–Lambert law (our Brass/Cu lab). We work in ~15-minute instruction bursts, then you practice while I circulate.

Blocks:  1 → A  ·  2 → B  ·  3 → C  ·  4 → D  ·  5 → E  ·  6 → F  ·  7 → G  ·  8 → H + review

Name:Block:Date:

A · Intermolecular & Interparticle Forces CED 3.1 · LO 3.1.A

Forces between molecules (intermolecular) decide melting points, boiling points, and vapour pressure, not the covalent bonds inside a molecule (intramolecular). Three IMFs to rank, weakest to strongest for similar-size molecules.

The three forces: London dispersion (temporary fluctuating dipoles; in every substance; grows with more electrons / bigger, more polarizable electron cloud / more contact area) < dipole–dipole (permanent dipoles in polar molecules) < hydrogen bonding (a special, strong dipole–dipole: H bonded to N, O, or F attracted to N, O, or F on another molecule).
Don't forget dispersion: for large molecules, London dispersion can be the strongest net force even though it's "weakest per contact." A bigger electron cloud beats a small permanent dipole.
Worked example
Which has the higher boiling point: CH₄ or H₂O? Name the dominant IMF in each.
CH₄ is nonpolar (symmetric tetrahedral) → only London dispersion, and it's a small molecule (10 electrons).
H₂O is bent & polar with O–H bonds → hydrogen bonding (strong) on top of dipole–dipole and dispersion.
H₂O boils far higher (100 °C vs −161 °C): stronger IMFs (H-bonding) need more energy to separate the molecules.

Watch the trap: "stronger" / "weaker" alone is not an answer on the AP exam, name the actual force (e.g. "hydrogen bonding") and say why it's stronger than the force in the other substance.

Practice A

  1. Rank the IMFs and predict which boils higher: F₂, HCl, or HF. Explain each choice.
  2. The noble gases He → Xe have rising boiling points down the group. Which IMF is responsible, and why does it increase?
  3. Why does CH₃OH (methanol) boil higher than CH₃OCH₃ (dimethyl ether), even though both are C₂H₆O?

B · Solids, Liquids & Gases: Properties from Particles CED 3.2–3.3 · LO 3.2.A / 3.3.A

A substance's macroscopic properties (hardness, conductivity, melting point) follow from its particle-level structure and the forces between particles. Match the solid type to its behaviour.

Solid types: ionic (high mp, brittle, conducts only when molten/dissolved) · covalent network (diamond, SiO₂, very high mp, hard, non-conducting) · molecular (low mp, soft, non-conducting, held by IMFs) · metallic (malleable, conducts via a sea of mobile electrons).
States: solid = particles fixed, vibrate in place (crystalline = ordered; amorphous = not). Liquid = close-packed but moving/colliding. Gas = far apart, constant random motion, no fixed shape or volume. Solid & liquid have similar molar volumes (both close-packed); gas is far larger.
Worked example
A solid melts at 801 °C and conducts electricity only after melting. What type is it, and why those properties?
Conducts only when molten → ions become mobile → ionic solid (this is NaCl).
Strong electrostatic attraction between oppositely charged ions → high melting point; in the solid the ions are locked, so no conduction until melted (or dissolved) frees them to move.

Not tested: understanding or interpreting phase diagrams will not be assessed on the AP Exam (CED 3.3 Exclusion). Learn the particle pictures, skip the P–T diagram.

Practice B

  1. Graphite conducts electricity and is soft/slippery, yet diamond is a hard insulator, both are pure carbon. Explain using bonding/structure.
  2. Classify and justify: (a) Cu wire, (b) dry ice CO₂(s), (c) SiO₂ (quartz).
  3. Draw 8 particles each for the solid, liquid, and gas phases of H₂O, showing spacing and motion.

C · The Ideal Gas Law & Partial Pressures CED 3.4 · LO 3.4.A

For an ideal gas, four macroscopic variables are locked together. Know the constant and its units, and Dalton's law for mixtures.

Core: PV = nRT with R = 0.08206 L·atm/(mol·K). Always use kelvin (K = °C + 273.15). Dalton: P_total = P_A + P_B + … and P_A = X_A · P_total, where X_A = n_A / n_total.
Worked example
What volume does 0.500 mol of an ideal gas occupy at 25.0 °C and 1.20 atm?
T = 25.0 + 273.15 = 298.15 K; rearrange PV = nRT → V = nRT / P.
V = (0.500 mol)(0.08206 L·atm·mol⁻¹·K⁻¹)(298.15 K) / (1.20 atm)
numerator = 0.500 × 0.08206 × 298.15 = 12.23 L·atm
V = 12.23 / 1.20 = 10.2 L
Worked example, partial pressure
A flask holds 2.0 mol N₂ and 3.0 mol O₂ at a total pressure of 5.0 atm. Find the partial pressure of O₂.
X(O₂) = 3.0 / (2.0 + 3.0) = 0.60
P(O₂) = 0.60 × 5.0 atm = 3.0 atm

Practice C

  1. How many moles of gas are in a 2.50 L container at 350 K and 2.00 atm?
  2. A gas at 1.00 atm and 300 K is heated to 600 K in a sealed rigid container. Find the new pressure.
  3. A mixture is 0.25 mol He, 0.25 mol Ne, 0.50 mol Ar at 4.0 atm total. Find each partial pressure.

D · Kinetic Molecular Theory & Real-Gas Deviations CED 3.5–3.6 · LO 3.5.A / 3.6.A

KMT is the particle-level "why" behind the gas law. Real gases break the ideal model under two specific conditions, know which and why.

KMT assumes: gas particles are in constant random motion, have negligible volume, exert no attractions, and collide elastically. Temperature link: Kelvin T is proportional to average kinetic energy; KE = ½mv². At the same T, lighter particles move faster on average (Maxwell–Boltzmann distribution).
Real gases deviate most at high pressure & low temperature: (1) at high P, particle volume is no longer negligible; (2) at low T / near condensation, intermolecular attractions matter. Gases behave most ideally at low P and high T.
Worked example
At the same temperature, which moves faster on average: He or O₂? Why?
Same T → same average KE = ½mv². He has the smaller molar mass (4.00 vs 32.00 g/mol).
He moves faster: with equal KE, smaller m means larger v².

Practice D

  1. Two gases share a container at the same temperature. State what is equal for both and what differs, using KMT.
  2. Under which conditions does CO₂ behave most ideally, (a) 200 atm & 200 K or (b) 1 atm & 500 K? Explain.
  3. Explain, in terms of attractions, why a real gas often exerts less pressure than the ideal gas law predicts near its condensation point.

E · Solutions, Molarity & Dilution CED 3.7–3.8 · LO 3.7.A / 3.8.A

A solution is a homogeneous mixture, uniform throughout. Molarity is the lab's standard concentration. Dilution adds solvent: moles of solute stay the same.

Core: M = n_solute / L_solution (mol/L). Dilution: M₁V₁ = M₂V₂ (the moles of solute before = after). Particulate pictures show relative concentration (how crowded) and interactions, not exact counts.
Worked example, molarity
Dissolve 5.85 g of NaCl in enough water to make 250.0 mL of solution. Find the molarity.
M(NaCl) = 22.99 + 35.45 = 58.44 g/mol → n = 5.85 / 58.44 = 0.1001 mol
V = 250.0 mL = 0.2500 L
M = 0.1001 / 0.2500 = 0.400 mol/L
Worked example, dilution
What volume of 12.0 M HCl is needed to make 500.0 mL of 0.100 M HCl?
M₁V₁ = M₂V₂ → V₁ = M₂V₂ / M₁ = (0.100)(500.0 mL) / (12.0)
V₁ = 4.17 mL of the stock acid (then dilute to 500.0 mL).

Not tested: molality, percent-by-mass, and percent-by-volume calculations are not assessed (CED 3.8 Exclusion). Colligative properties are also excluded. Stick to molarity.

Practice E

  1. Find the molarity of a solution made from 4.00 g of NaOH (M = 40.00 g/mol) in 500.0 mL.
  2. How many grams of glucose (M = 180.16 g/mol) are in 2.00 L of 0.150 M solution?
  3. You dilute 25.0 mL of 6.00 M HNO₃ to 250.0 mL. Find the new molarity.

F · Separations & Solubility CED 3.9–3.10 · LO 3.9.A / 3.10.A

You can't separate a true solution by filtration, the particles are too small and uniformly mixed. Separations exploit differences in intermolecular interactions.

Methods: filtration separates an undissolved solid from a liquid (not a solution). Distillation separates liquids by differences in vapour pressure / boiling point (IMF strength). Chromatography (paper, TLC, column) separates by how strongly each component sticks to the stationary phase vs. travels with the mobile phase, reveals relative polarity.
Solubility, "like dissolves like": substances with similar intermolecular interactions are miscible/soluble. Polar & ionic dissolve in polar solvents (water); nonpolar dissolves in nonpolar solvents.
Worked example
On a paper chromatogram run with a polar solvent, dye X travels farther up than dye Y. Which dye is more strongly attracted to the (polar) paper, and which is likely more polar?
The paper (stationary phase) is polar; the more polar dye sticks to it more and moves less.
Y moves less → Y is held more strongly by the polar paper → Y is the more polar dye; X travels farther because it prefers the moving solvent.

Practice F

  1. Why can't you separate salt water into salt and water by filtration? What method works, and on what property?
  2. Predict: is I₂ (nonpolar) more soluble in water or in hexane (nonpolar)? Explain with "like dissolves like."
  3. You must separate two liquids that boil at 78 °C and 100 °C. Name the technique and the property it exploits.

G · Spectroscopy, the EM Spectrum & Photons CED 3.11–3.12 · LO 3.11.A / 3.12.A

Different regions of light probe different kinds of molecular change. The photon's energy is set by its frequency, this is the bridge to the Beer–Lambert lab.

Region → transition: microwave → molecular rotation; infrared (IR) → molecular vibration; ultraviolet/visible (UV-Vis) → electronic transitions. UV-Vis is what a spectrophotometer uses for coloured solutions.
Photon equations: c = λν with c = 3.00×10⁸ m/s; E = hν with h = 6.626×10⁻³⁴ J·s. Higher frequency (shorter wavelength) = higher energy per photon. Absorbing a photon raises the species' energy by exactly that amount.
Worked example
A photon of green light has wavelength λ = 500. nm. Find its frequency and energy.
λ = 500. nm = 5.00×10⁻⁷ m; ν = c/λ = (3.00×10⁸) / (5.00×10⁻⁷) = 6.00×10¹⁴ Hz
E = hν = (6.626×10⁻³⁴ J·s)(6.00×10¹⁴ s⁻¹)
E = 3.98×10⁻¹⁹ J per photon

Practice G

  1. Which has more energy per photon: UV light (λ = 200 nm) or IR light (λ = 2000 nm)? Justify with c = λν and E = hν.
  2. A Cu²⁺ solution looks blue. What region of the EM spectrum and what kind of transition is the spectrophotometer using?
  3. Find the energy of a photon with frequency 5.00×10¹⁴ Hz.

H · The Beer–Lambert Law (Brass / Cu Lab) CED 3.13 · LO 3.13.A

This is the math behind our spectrophotometry lab: dissolve a brass sample, measure how much light the blue Cu²⁺ solution absorbs, and back out its concentration from a calibration line.

Core: A = εbcabsorbance A (no units) = molar absorptivity ε (L·mol⁻¹·cm⁻¹) × path length b (cm) × concentration c (mol/L). With b and the wavelength (set to λ_max for best sensitivity) held constant, A is directly proportional to c: a straight calibration line through the origin.
Worked example
A 1.00 cm cell holds a Cu²⁺ solution with A = 0.450 at λ_max. The molar absorptivity is ε = 90.0 L·mol⁻¹·cm⁻¹. Find c.
A = εbc → c = A / (εb)
c = 0.450 / (90.0 L·mol⁻¹·cm⁻¹ × 1.00 cm)
c = 5.00×10⁻³ mol/L
Worked example, calibration line
Standards give a best-fit line A = 90.0·c (b = 1.00 cm). An unknown brass solution reads A = 0.315. Find c.
c = A / 90.0 = 0.315 / 90.0
c = 3.50×10⁻³ mol/L Cu²⁺

Lab tip (CED skill 2.E, sources of error): a fingerprinted or bubbled cuvette, wrong wavelength, or a concentration so high that A > ~1 (off the linear range) all distort A. Blank the instrument first.

Practice H

  1. A solution has c = 2.00×10⁻³ M, b = 1.00 cm, ε = 95.0 L·mol⁻¹·cm⁻¹. Find A.
  2. Two solutions of the same species in the same 1.00 cm cell read A = 0.20 and A = 0.60. Compare their concentrations.
  3. Why does the spectrophotometer get set to λ_max for the analysis instead of any random wavelength?
AP Chemistry · Unit 3, Properties of Substances & Mixtures · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 3.1–3.13) and BC Chemistry 11 (solutions & gases). Atomic masses from the IUPAC periodic table; R = 0.08206 L·atm·mol⁻¹·K⁻¹, c = 3.00×10⁸ m/s, h = 6.626×10⁻³⁴ J·s.