AP Chemistry Unit 3 practice questions
Try 10 multiple-choice questions from Unit 3 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.
Question 1
The Henry's law constant for oxygen gas () in water at is mol/(L·atm). Calculate the concentration of dissolved in a solution at equilibrium with air where the partial pressure of is 0.21 atm.
M
M
M
M
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Answer: (A)
Applying Henry's Law () requires multiplying the constant () by the pressure (0.21). The result is M, limited to two significant figures by the given values.
Question 2
Potassium nitrate () has a solubility of 85.0 g per 100.0 g of water at and 32.0 g per 100.0 g of water at . A student prepares a saturated solution using 50.0 g of water at and then cools it to . What mass of will precipitate?
26.5 g
16.0 g
53.0 g
58.5 g
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Answer: (A)
Calculating the precipitate requires determining the mass dissolved at (42.5 g) and (16.0 g) scaled to 50.0 g of water, then finding the difference (42.5 g - 16.0 g = 26.5 g).
Question 3
At , the solubility of sucrose (, molar mass 342.3 g/mol) is 204 g per 100. mL of solution. What is the molarity of a saturated sucrose solution at this temperature?
5.96 M
0.596 M
1.68 M
16.8 M
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Answer: (A)
Calculating molarity requires converting mass to moles (204 g / 342.3 g/mol = 0.596 mol) and dividing by volume in liters (0.596 mol / 0.100 L = 5.96 M).
Question 4
The solubility of ammonium chloride () at is 37.0 g per 100.0 g of water. If a student dissolves 12.0 g of in 50.0 g of water at , how much additional is required to saturate the solution?
6.5 g
18.5 g
25.0 g
30.5 g
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Answer: (A)
Calculating the required addition involves finding the saturation limit for 50.0 g of water (18.5 g) and subtracting the amount already dissolved (18.5 g - 12.0 g = 6.5 g).
Question 5
A structural model of liquid water (H₂O) and liquid hydrogen sulfide (H₂S) shows both molecules with bent geometries and similar dipole moments. However, the model does not depict lone pair electrons. How well does this model explain why H₂O has a significantly higher boiling point than H₂S?
Fully, because the bent geometry creates the net dipole responsible for the attraction.
Fully, because the similar dipole moments imply similar strengths of intermolecular forces.
Poorly, because it omits the lone pairs needed to explain strong hydrogen bonding in water.
Poorly, because it fails to show the stronger London dispersion forces in hydrogen sulfide.
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Answer: (C)
Identifies the model's key deficiency: hydrogen bonding requires interactions between H atoms and lone pairs on highly electronegative atoms (O), which the model omits.
Question 6
A student uses ball-and-stick models to represent n-pentane (a straight chain) and neopentane (a compact sphere). Both models contain the same number of carbon and hydrogen atoms. The student claims the models predict identical boiling points. Which evaluation of this claim is most accurate?
Correct, because isomers with identical molecular formulas have identical intermolecular forces.
Correct, because both molecules are nonpolar and rely solely on London dispersion forces.
Incorrect, because the straight chain allows more surface contact for stronger dispersion forces.
Incorrect, because the compact sphere creates a stronger permanent dipole moment in the isomer.
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Answer: (C)
Evaluates the connection between shape and property; the linear model allows greater surface area contact, leading to stronger LDF and a higher boiling point.
Question 7
A representation of oxygen gas (O₂) dissolving in water depicts water molecules as permanent dipoles and O₂ molecules as neutral spheres with fixed electron distributions. To what extent does this representation explain the solubility of O₂ in water?
Completely, because nonpolar molecules dissolve via simple mixing with polar solvents.
Completely, because the fixed electron distribution prevents repulsion from water dipoles.
Incompletely, because it fails to show the induction of a temporary dipole in the O₂ molecule.
Incompletely, because it fails to show the formation of hydrogen bonds between O₂ and water.
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Answer: (C)
Identifies the missing mechanism; for a nonpolar species to dissolve in a polar solvent, the solvent must induce a dipole in the solute, which requires a distortable electron cloud.
Question 8
A kinetic model of a liquid in a closed container shows particles as spheres with a single uniform velocity vector at 298 K. The model is used to explain why vapor pressure increases at 350 K. Which limitation of the model most hinders its ability to explain this property?
It fails to show the distribution of kinetic energies needed to overcome attractive forces.
It fails to show that individual particles have different speeds at the same temperature.
It fails to show that some particles have sufficient kinetic energy to overcome intermolecular attractions.
It fails to show that the average kinetic energy of particles increases with temperature change.
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Answer: (A)
Evaluates the model's inability to explain evaporation; vapor pressure depends on the fraction of particles with enough energy to escape IMFs, which requires a Boltzmann distribution, not uniform velocity.
Question 9
A student presents a model comparing hydrogen fluoride (HF) and hydrogen chloride (HCl). The model accurately depicts bond lengths but shows the intermolecular attraction between molecules as identical dashed lines. How well does this model account for the difference in boiling points between HF (293 K) and HCl (188 K)?
Well, because both molecules are linear polar acids with similar dipole strengths.
Well, because the bond length difference compensates for the difference in forces.
Poorly, because it does not distinguish hydrogen bonding in HF from dipole forces in HCl.
Poorly, because it does not show the stronger London dispersion forces present in HF.
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Answer: (C)
Critiques the model's lack of differentiation; HF exhibits hydrogen bonding while HCl only exhibits dipole-dipole forces, a distinction crucial for explaining the boiling point difference.
Question 10
According to Coulomb's law (), if the charge of a dissolved metal cation doubles while its distance from a water molecule remains constant, how does the attractive force change?
It increases by a factor of 2; force is directly proportional to charge magnitude
It increases by a factor of 4; force is directly proportional to charge squared
It decreases by a factor of 2; force is inversely proportional to charge magnitude
It decreases by a factor of 4; force is inversely proportional to charge squared
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Answer: (A)
Analyzing the equation shows a linear relationship between force and charge. Doubling one charge () while holding other variables constant results in a doubled force ().
What's on the AP Chemistry Unit 3 practice test
The standard version takes about 60 minutes and follows the timing and directions of the AP exam, sized to one unit:
- 20 Multiple Choice Questions (30 minutes)
- 1 FRQ – FRQ 1 – Long Answer (15 minutes)
- 1 FRQ – FRQ 4 – Short Answer (15 minutes)
Questions are drawn from 2330 multiple-choice questions and 119 FRQs written for Unit 3, Properties of Substances and Mixtures.
Practice AP Chemistry Unit 3 FRQs
Answer the unit's free-response questions one at a time and get a score on every part.
- FRQ 1 – Long Answer (10 questions)
- FRQ 2 – Long Answer (8 questions)
- FRQ 3 – Long Answer (7 questions)
- FRQ 4 – Short Answer (33 questions)
- FRQ 5 – Short Answer (30 questions)
- FRQ 6 – Short Answer (17 questions)
- FRQ 7 – Short Answer (14 questions)
Review Unit 3 before you start
- 3.1 Intermolecular and Interparticle Forces
- 3.2 Properties of Solids
- 3.3 Solids, Liquids, and Gases
- 3.4 Ideal Gas Law
- 3.5 Kinetic Molecular Theory
- 3.6 Deviation from Ideal Gas Law
- 3.7 Solutions and Mixtures
- 3.8 Representations of Solutions
- 3.9 Separation of Solutions and Mixtures
- 3.10 Solubility
- 3.11 Spectroscopy and the Electromagnetic Spectrum
- 3.12 Properties of Photons
- 3.13 Beer-Lambert Law