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All 9 units · Practice test

AP Chemistry Full-Length Practice Test

Sixty exam-style multiple-choice questions covering all nine units in the same proportions as the real exam, with instant scoring and an explanation for every answer.

This is Section I of a complete practice exam: 60 original four-choice questions, each unit weighted to match the course description’s exam weightings as closely as 60 whole questions allow. On exam day you have 90 minutes for these, with a calculator, periodic table, and equation sheet available, so set a timer for 90 minutes and answer every question before pressing Check My Answers. There is no penalty for guessing.

For Section II, work the 2026 AP Chemistry free-response questions in 105 minutes and grade them with our worked answers. Then put both results into the AP Chem Score Calculator for a predicted 1–5 score. To practice a single unit instead, start with the Unit 1 practice test.

1 How many moles of oxygen atoms are in 0.50 mol of Al2(SO4)3?
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Answer: C. Each formula unit has 3 sulfate groups with 4 O atoms each, 12 O atoms in all. 0.50 mol × 12 = 6.0 mol O. Choice A counts the O in only one sulfate group.

2 The mass spectrum of an element has two peaks: mass 107 at 51.8% and mass 109 at 48.2%. Which element is it?
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Answer: B. 107(0.518) + 109(0.482) = 108.0 with whole-number masses, closest to silver (107.87). The average must sit between 107 and 109, which rules out the others.

3 In the photoelectron spectrum of silicon, which peak is tallest?
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Answer: D. Silicon is 1s2 2s2 2p6 3s2 3p2. Peak height is proportional to the number of electrons in the subshell, and 2p holds 6.

4 The second ionization energy of sodium is about nine times its first. Which statement best explains this?
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Answer: A. Na loses its single 3s electron first. The next electron comes from the filled n = 2 shell, much closer to the nucleus, so the Coulombic attraction on it is far greater. Na and Na+ have the same 11 protons.

5 Which species has the smallest radius?
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Answer: D. All four have 10 electrons in the same configuration. Mg2+ has the most protons (12), so it pulls those electrons in most tightly.

6 Which bond has the greatest ionic character?
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Answer: B. Ionic character increases with the electronegativity difference. Potassium has one of the lowest electronegativities and fluorine the highest, so K–F has by far the largest difference.

7 Which ionic compound has the greatest lattice energy?
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Answer: C. By Coulomb’s law, attraction grows with the product of the charges and shrinks with distance. Mg2+ and O2− have charges of 2 (product 4, versus 1 for the others) and small radii.

8 In steel, small carbon atoms occupy the spaces between iron atoms. Compared with pure iron, how does this change the metal?
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Answer: B. Atoms that fit in the gaps between larger atoms make an interstitial alloy. They hinder the layers of metal atoms from sliding past each other, so the alloy is harder and less malleable, while it still conducts.

9 In a Lewis diagram of the nitrate ion, NO3−, with one N=O double bond and two N–O single bonds, what is the formal charge on nitrogen?
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Answer: D. Formal charge = valence electrons − nonbonding electrons − bonds = 5 − 0 − 4 = +1. The two singly bonded O atoms are −1 each, giving the ion’s overall −1.

10 What is the molecular geometry of SF4?
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Answer: A. Sulfur has 4 bonding pairs and 1 lone pair, 5 electron domains in a trigonal bipyramidal arrangement. With the lone pair in an equatorial position, the atoms form a seesaw.

11 What is the hybridization of the carbon atom in formaldehyde, CH2O?
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Answer: C. Carbon has three electron domains (two C–H bonds and one C=O double bond, which counts once), so it is sp2 with bond angles near 120°.

12 Which substance has the highest boiling point?
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Answer: D. Methanol has an O–H bond, so its molecules hydrogen bond to one another. The others have only dispersion forces or weaker dipole-dipole attractions.

13 I2 is a solid at room temperature but Br2 is a liquid. Which explanation is best?
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Answer: B. Both are nonpolar, so only London dispersion forces hold the molecules together. More electrons mean a more polarizable cloud and stronger attractions. Melting does not break the covalent bond, so choice C is irrelevant.

14 Which solid has the highest melting point?
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Answer: C. SiO2 is a covalent network solid: melting it means breaking strong Si–O covalent bonds throughout the crystal. It melts above 1,700 °C, compared with 801 °C for NaCl.

15 What volume does 0.500 mol of an ideal gas occupy at 273 K and 1.00 atm? (R = 0.08206 L·atm/(mol·K))
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Answer: B. V = nRT/P = (0.500)(0.08206)(273) ÷ 1.00 = 11.2 L. Choice C is the volume of a full mole.

16 A container holds 2.0 mol of N2 and 1.0 mol of O2 at a total pressure of 6.0 atm. What is the partial pressure of O2?
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Answer: A. The mole fraction of O2 is 1.0 ÷ 3.0 = 1/3, and its partial pressure is (1/3)(6.0 atm) = 2.0 atm.

17 Samples of He, Ne, Ar, and Kr are at the same temperature. Which statement is true?
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Answer: C. Average kinetic energy depends only on temperature, so it is the same for all four. Since KE = ½mv2, the lightest atoms, helium, must move fastest.

18 Under which conditions does a real gas deviate most from ideal behavior?
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Answer: A. At high pressure the particles’ own volume is a significant fraction of the container, and at low temperature they move slowly enough for attractions to matter. Both effects grow under these conditions.

19 What mass of NaOH (40.00 g/mol) is needed to make 250. mL of 0.200 M NaOH solution?
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Answer: D. Moles = (0.250 L)(0.200 mol/L) = 0.0500 mol, and 0.0500 mol × 40.00 g/mol = 2.00 g. Choice A uses 1.00 L instead of 0.250 L.

20 A student dilutes 25.0 mL of 6.0 M HCl with water to a total volume of 150. mL. What is the new concentration?
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Answer: A. M1V1 = M2V2: (6.0 M)(25.0 mL) = M2(150. mL), so M2 = 1.0 M. Choice B uses only the volume ratio and leaves out the original 6.0 M.

21 Which substance is most soluble in hexane, C6H14?
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Answer: C. Hexane is nonpolar, so it dissolves substances held together by similar dispersion forces. Ionic solids such as the other three cannot be stabilized by a nonpolar solvent.

22 Absorption of which type of radiation causes transitions between molecular vibrational energy levels?
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Answer: B. Infrared radiation matches vibrational energy changes. Microwaves cause rotational transitions, and ultraviolet and visible light cause electronic transitions.

23 What is the energy of one photon of light with a wavelength of 500. nm? (h = 6.626 × 10−34 J·s, c = 2.998 × 108 m/s)
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Answer: C. E = hc/λ = (6.626 × 10−34)(2.998 × 108) ÷ (500. × 10−9 m) = 3.97 × 10−19 J. Choice B leaves the wavelength in nanometers without the 10−9.

24 A 0.020 M solution has an absorbance of 0.40. Measured in the same cuvette at the same wavelength, what concentration has an absorbance of 0.30?
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Answer: D. Absorbance is directly proportional to concentration (A = εbc), so c = (0.30/0.40)(0.020 M) = 0.015 M.

25 Which process is a chemical change?
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Answer: A. Combustion breaks and forms covalent bonds to make new substances, CO2 and H2O. The others change only the arrangement of the same molecules.

26 Solutions of AgNO3 and NaCl are mixed and a white precipitate forms. Which is the net ionic equation?
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Answer: C. Sodium and nitrate salts are always soluble, so Na+ and NO3− are spectator ions. Only the ions that form the precipitate remain. Choice A is the molecular equation.

27 What mass of CO2 forms when 8.02 g of CH4 burns completely? CH4 + 2 O2 → CO2 + 2 H2O
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Answer: B. 8.02 g ÷ 16.04 g/mol = 0.500 mol CH4, which makes 0.500 mol CO2. 0.500 mol × 44.01 g/mol = 22.0 g. Mass is not conserved molecule for molecule, so choice A is wrong.

28 For 2 H2 + O2 → 2 H2O, a mixture of 4.0 mol H2 and 3.0 mol O2 reacts completely. How much water forms?
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Answer: D. 4.0 mol H2 needs only 2.0 mol O2, so H2 runs out first and 1.0 mol O2 is left over. 4.0 mol H2 makes 4.0 mol H2O.

29 It takes 20.0 mL of 0.150 M NaOH to neutralize 25.0 mL of HCl. What is the concentration of the HCl?
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Answer: A. Moles NaOH = (0.0200 L)(0.150 M) = 0.00300 mol, which equals moles HCl in a 1:1 reaction. 0.00300 mol ÷ 0.0250 L = 0.120 M. Choice B inverts the volume ratio.

30 In the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), which species is oxidized?
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Answer: B. Zinc’s oxidation number rises from 0 to +2, so zinc loses electrons and is oxidized. Cu2+ gains those electrons and is reduced.

31 For A + B → products, doubling [A] makes the initial rate four times faster, and doubling [B] does not change it. What is the rate law?
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Answer: C. A factor of 4 for a factor of 2 means second order in A (22 = 4). No change when [B] doubles means zero order in B, so B does not appear.

32 A first-order reaction has a rate constant of 0.0231 min−1. What is its half-life?
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Answer: B. For a first-order reaction t1/2 = 0.693/k = 0.693 ÷ 0.0231 min−1 = 30.0 min, independent of concentration.

33 A reactant in a first-order reaction starts at 0.800 M. What is its concentration after three half-lives?
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Answer: D. Each half-life halves the concentration: 0.800 → 0.400 → 0.200 → 0.100 M. Choice A divides by 3 instead.

34 A proposed mechanism is: Step 1 (slow) NO2 + NO2 → NO3 + NO; Step 2 (fast) NO3 + CO → NO2 + CO2. Which rate law is consistent with it?
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Answer: A. The slow step limits the rate, and the rate law of an elementary step follows its molecularity: two NO2 molecules collide. CO enters only after the slow step, and NO3 is an intermediate, so neither appears.

35 How does a catalyst increase the rate of a reaction?
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Answer: C. A catalyst offers an alternative mechanism with a lower energy barrier, so more collisions are successful. It does not change ΔH, K, or the temperature.

36 When solid NH4NO3 dissolves in water, the solution becomes colder. Which statement is correct?
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Answer: B. The temperature of the water (the surroundings) drops because energy flows from it into the dissolving system, so the process is endothermic.

37 How much heat is absorbed when 50.0 g of water warms from 22.0 °C to 30.0 °C? (c = 4.18 J/(g·°C))
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Answer: C. q = mcΔT = (50.0 g)(4.18 J/(g·°C))(8.0 °C) = 1,670 J = 1.7 kJ. Choices A and B use a single temperature instead of the change.

38 The enthalpy of fusion of water is 6.01 kJ/mol. How much energy is needed to melt 36.0 g of ice at 0 °C?
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Answer: A. 36.0 g ÷ 18.02 g/mol = 2.00 mol, and 2.00 mol × 6.01 kJ/mol = 12.0 kJ. The temperature stays at 0 °C while the ice melts.

39 Using bond enthalpies H–H 436, Cl–Cl 242, and H–Cl 431 kJ/mol, estimate ΔH for H2(g) + Cl2(g) → 2 HCl(g).
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Answer: D. Bonds broken minus bonds formed: (436 + 242) − 2(431) = 678 − 862 = −184 kJ/mol. Forgetting that 2 H–Cl bonds form gives +247 (choice C).

40 Given ΔHf° values CH4(g) −75, CO2(g) −394, and H2O(l) −286 kJ/mol, what is ΔH° for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)?
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Answer: B. Products minus reactants: [−394 + 2(−286)] − [−75 + 0] = −966 + 75 = −891 kJ/mol. O2 is an element in its standard state, so its value is 0. Choice A forgets the coefficient 2 on water.

41 For A → B, ΔH = +50 kJ. What is ΔH for 2 B → 2 A?
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Answer: C. Reversing a reaction changes the sign of ΔH (−50 kJ), and doubling it doubles the value: −100 kJ.

42 What is the equilibrium-constant expression Kc for N2(g) + 3 H2(g) ⇌ 2 NH3(g)?
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Answer: A. Products over reactants, each concentration raised to its coefficient. Choice C is the expression for the reverse reaction, 1/K.

43 For a reaction mixture, Q = 0.15 and K = 4.0 at the same temperature. What happens as the system moves toward equilibrium?
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Answer: D. Q < K means the ratio of products to reactants is too small, so the forward reaction runs until Q rises to K. K is fixed at a given temperature.

44 For A(g) ⇌ 2 B(g), the equilibrium concentrations are [A] = 0.50 M and [B] = 0.20 M. What is Kc?
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Answer: B. Kc = [B]2/[A] = (0.20)2 ÷ 0.50 = 0.040 ÷ 0.50 = 0.080. Choice A forgets to square [B]; choice D is the inverse.

45 For N2(g) + 3 H2(g) ⇌ 2 NH3(g), ΔH < 0, which change increases the amount of NH3 at equilibrium?
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Answer: C. Decreasing the volume raises the pressure, and the system shifts toward fewer gas molecules: 2 on the product side versus 4 on the reactant side. Raising the temperature favors the endothermic reverse reaction, and a catalyst does not shift equilibrium.

46 The Ksp of AgCl is 1.8 × 10−10. What is its molar solubility in pure water?
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Answer: A. AgCl ⇌ Ag+ + Cl−, so Ksp = s2 and s = √(1.8 × 10−10) = 1.3 × 10−5 M.

47 How does the molar solubility of PbI2 in 0.10 M NaI compare with its molar solubility in pure water?
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Answer: C. The I− already in solution is a common ion. With Ksp = [Pb2+][I−]2 fixed, less PbI2 can dissolve. Ksp stays the same; the solubility does not.

48 What is the pH of 0.0010 M HCl?
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Answer: C. HCl is a strong acid, so [H3O+] = 0.0010 M = 1.0 × 10−3 M and pH = −log(1.0 × 10−3) = 3.00.

49 What is the pH of 0.010 M NaOH at 25 °C?
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Answer: B. [OH−] = 0.010 M, so pOH = 2.00 and pH = 14.00 − 2.00 = 12.00. Choice A is the pOH.

50 What is the pH of 0.10 M acetic acid? (Ka = 1.8 × 10−5)
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Answer: D. [H3O+] ≈ √(KaC) = √((1.8 × 10−5)(0.10)) = 1.3 × 10−3 M, so pH = 2.87. Choice A treats acetic acid as strong; choice B is its pKa.

51 What is the conjugate base of H2PO4−?
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Answer: B. A conjugate base has one fewer H+ than its acid: H2PO4− loses H+ to become HPO42−. H3PO4 is its conjugate acid.

52 During the titration of a weak acid with NaOH, at which point does pH equal the pKa of the acid?
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Answer: A. Halfway to equivalence, half the acid has become its conjugate base, so [HA] = [A−] and pH = pKa + log(1) = pKa.

53 Which mixture of equal moles in water forms a buffer?
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Answer: D. A buffer needs a weak acid and its conjugate base in comparable amounts. Acetic acid and acetate qualify; the other pairs contain a strong acid or base and an ion that does not react appreciably with water.

54 A buffer has [A−]/[HA] = 10 for an acid with pKa = 4.74. What is the pH?
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Answer: C. pH = pKa + log([A−]/[HA]) = 4.74 + log(10) = 4.74 + 1 = 5.74.

55 A weak acid HA has a pKa of 4.2. In a solution buffered at pH 6.2, which form predominates?
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Answer: B. When pH > pKa, the deprotonated form predominates. The pH is 2 units above the pKa, so [A−]/[HA] = 102 = 100.

56 Which process has a negative entropy change, ΔS < 0?
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Answer: C. Condensing a gas to a liquid sharply reduces the number of ways the particles can be arranged and move. The other three increase dispersal of matter, two of them by forming a gas.

57 A reaction has ΔH° = −100 kJ/mol and ΔS° = −200 J/(mol·K). Under standard conditions, when is it thermodynamically favorable?
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Answer: D. ΔG° = ΔH° − TΔS° is negative while |ΔH°| exceeds T|ΔS°|, that is, while T < 100,000 J/mol ÷ 200 J/(mol·K) = 500 K. Note the unit conversion from kJ to J.

58 If ΔG° for a reaction is negative, what must be true about its equilibrium constant K?
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Answer: A. From ΔG° = −RT ln K, a negative ΔG° requires ln K > 0, so K > 1 and products are favored at equilibrium.

59 Given E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = −0.76 V, what is the standard cell potential of a galvanic cell made from these half-cells?
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Answer: B. Cu2+ has the higher reduction potential, so it is reduced and Zn is oxidized: E°cell = 0.34 − (−0.76) = +1.10 V. Potentials are not multiplied by coefficients, so choice D is wrong.

60 A current of 2.00 A passes through a Cu2+ solution for 965 s. What mass of copper is deposited? (F = 96,485 C/mol e−)
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Answer: C. q = (2.00 A)(965 s) = 1,930 C = 0.0200 mol e−. Each Cu2+ needs 2 electrons, so 0.0100 mol Cu × 63.55 g/mol = 0.636 g. Choice A forgets to divide by 2.

How the Full-Length Practice Test Is Weighted by Unit

UnitExam weightingQuestionsReview
Unit 1: Atomic Structure and Properties7–9%1–5Unit review
Unit 2: Compound Structure and Properties7–9%6–11Unit review
Unit 3: Properties of Substances and Mixtures18–22%12–24Unit review
Unit 4: Chemical Reactions7–9%25–30Unit review
Unit 5: Kinetics7–9%31–35Unit review
Unit 6: Thermochemistry7–9%36–41Unit review
Unit 7: Equilibrium7–9%42–47Unit review
Unit 8: Acids and Bases11–15%48–55Unit review
Unit 9: Thermodynamics and Electrochemistry7–9%56–60Unit review

Units 3 and 8 carry the most questions here, as they do on the real exam. After you check your answers, the result box lists every topic you missed with a link to its guide; group those misses by unit and start with the unit where you lost the most. Your number correct out of 60 goes straight into the MCQ calculator, and the FRQ calculator handles Section II. For other ways to practice, see AP Chemistry practice tests, which lists the official College Board practice sources.

Frequently Asked Questions

How long is the AP Chemistry multiple-choice section?

Ninety minutes for 60 questions, about 90 seconds per question. It is worth 50 percent of the exam score, and a calculator is allowed.

How is this practice test weighted?

Each unit gets roughly its share of the real exam: 13 questions on Unit 3 and 8 on Unit 8, the two heaviest units, and 5 or 6 on each of the other seven units.

What score on this practice test is good?

There is no official conversion for an unofficial test, but enter your number correct in the MCQ calculator alongside a free-response estimate to see what composite score and 1 to 5 result it points to.

Are these real AP Chemistry questions?

No. They are original questions written in the AP style for this site and checked against the course description. For released questions, College Board provides practice exams through AP Classroom.

These are original practice questions written for this site, not released College Board questions. AP® is a trademark registered by the College Board, which is not affiliated with and does not endorse this site.