AP Chemistry — Practice Questions flashcards

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Multiple-choice practice questions, one best answer each, on atomic structure, bonding, IMFs and gases, stoichiometry, kinetics, thermochemistry, equilibrium, acids and bases, and electrochemistry.

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Practice deck · part of AP Chemistry · written against: College Board AP Chemistry Course and Exam Description (nine units). Updated 2026-10-10.

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The mass spectrum of a hypothetical element shows two peaks: one at 50 amu with a relative abundance of 75% and one at 54 amu with a relative abundance of 25%. What is the average atomic mass of the element?

A) 53.0 amu
B) 51.5 amu
C) 51.0 amu
D) 52.0 amu

C) 51.0 amu

The average is weighted by abundance: (0.75)(50) + (0.25)(54) = 37.5 + 13.5 = 51.0 amu, close to the more common isotope. The simple average of the two masses, 52.0 amu, ignores the abundances.
Atomic Structure and Properties › Moles and Mass Spectra
Br₂ (160 g/mol) and ICl (162 g/mol) have the same number of electrons (70 each). Which statement correctly predicts and explains which has the higher boiling point?

A) ICl, because it has dipole–dipole forces as well as dispersion forces
B) Br₂, because its nonpolar bond produces stronger dispersion forces
C) ICl, because its molecules can form hydrogen bonds with one another
D) Neither, because boiling point depends only on the number of electrons

A) ICl, because it has dipole–dipole forces as well as dispersion forces

With equal electron counts, the dispersion forces are comparable, and the polar I–Cl bond adds dipole–dipole attractions. Hydrogen bonding is impossible here because neither molecule contains hydrogen.
Intermolecular Forces and Properties › Intermolecular Forces
Aqueous barium nitrate and aqueous potassium sulfate are mixed, and a white precipitate forms. Which is the net ionic equation?

A) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
B) Ba²⁺(aq) + 2 NO₃⁻(aq) → Ba(NO₃)₂(s)
C) K⁺(aq) + NO₃⁻(aq) → KNO₃(s)
D) 2 K⁺(aq) + SO₄²⁻(aq) → K₂SO₄(s)

A) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Potassium and nitrate salts are always soluble, so K⁺ and NO₃⁻ are spectators and the precipitate must be barium sulfate. An equation forming KNO₃ as a solid contradicts that solubility rule.
Chemical Reactions › Equations and Reaction Types
A 20.0 g metal sample at 75.0 °C is placed in 50.0 g of water at 23.00 °C in an insulated cup. The final temperature is 25.00 °C. Using 4.184 J/(g·°C) for water, what is the specific heat of the metal?

A) 1.05 J/(g·°C)
B) 10.5 J/(g·°C)
C) 0.167 J/(g·°C)
D) 0.418 J/(g·°C)

D) 0.418 J/(g·°C)

The water gains (50.0)(4.184)(2.00) = 418 J, which the metal loses over a 50.0 °C drop: 418 ÷ (20.0 × 50.0) = 0.418 J/(g·°C). Using the water's 2.00 °C change for the metal gives the impossible 10.5.
Thermochemistry
For N₂O₄(g) ⇌ 2 NO₂(g), $K_c$ = 4.0 at a certain temperature. A flask at that temperature contains 0.50 M N₂O₄ and 1.0 M NO₂. Which statement is correct?

A) Q = 0.50, so the reaction proceeds to the right, forming more NO₂
B) Q = 4.0, so the mixture is already at equilibrium and does not shift
C) Q = 2.0, so the reaction proceeds to the right, forming more NO₂
D) Q = 2.0, so the reaction proceeds to the left, forming more N₂O₄

C) Q = 2.0, so the reaction proceeds to the right, forming more NO₂

Q = (1.0)² ÷ 0.50 = 2.0, which is less than K, so the forward reaction runs until Q rises to K. The value 0.50 comes from inverting the expression and putting the reactant on top.
Equilibrium › Equilibrium Constants
Which process has a positive entropy change, ΔS > 0?

A) 2 H₂(g) + O₂(g) → 2 H₂O(l)
B) N₂(g) + 3 H₂(g) → 2 NH₃(g)
C) CaCO₃(s) → CaO(s) + CO₂(g)
D) H₂O(g) → H₂O(l)

C) CaCO₃(s) → CaO(s) + CO₂(g)

Decomposing a solid into another solid plus a gas creates gas particles with many more available arrangements. Making ammonia lowers entropy because four moles of gas become two.
Thermodynamics and Electrochemistry › Entropy and Free Energy

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Topic outline follows the College Board AP Chemistry Course and Exam Description (effective Fall 2024), Units 1–9 (outline only; no course or exam text reproduced).
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