Unit 3: Substances and Mixtures

Intermolecular forces

Intermolecular forces (IMFs) are attractions between molecules or ions. They are much weaker than intramolecular (within a molecule) covalent or ionic bonds, but they control many everyday properties: boiling point, melting point, vapor pressure, surface tension, and solubility in a given solvent.

London dispersion forces

London dispersion forces arise from temporary electron fluctuations that create instantaneous dipoles, which in turn induce dipoles in neighbors. They are present in every substance (atoms and molecules). In nonpolar species they are often the only IMF and dominate cohesive energy. Strength grows with polarizability: larger electron clouds, higher molar mass in a homologous series, and greater surface area (e.g. linear versus branched alkanes) increase the amount of london disperson forces.

Example. Pentane and neopentane have the same molecular formula and electron count. Why can the less compact molecule have a higher boiling point?

Electron count alone does not determine dispersion attractions. The less compact pentane shape permits greater contact between neighboring molecules, strengthening their combined dispersion interactions. More energy is then needed to separate them into the gas phase. Branching changes molecular shape without changing molar mass, so this comparison isolates a limitation of the mass-only shortcut.

Dipole–dipole interactions

Polar molecules possess a permanent dipole moment, since the more electronegative element always tries to pull in electrons, creating a negatively-charged area around that element and a positive-charged area on the other element. Dipole–dipole attractions align partially positive ends toward partially negative ends. They are typically stronger than dispersion between comparable-sized molecules, though dispersion still contributes a little bit of force.

Example. A student predicts that every polar molecule must boil above every nonpolar molecule. Explain why this rule can fail without denying dipole-dipole attractions.

Both classes experience dispersion forces. A large, highly polarizable nonpolar molecule can have stronger total intermolecular attractions than a small polar molecule, even though only the latter has permanent dipole-dipole interactions. Comparing named force types alone ignores their magnitudes, molecular size, and shape; polarity is most useful when the other factors are comparable.

Hydrogen bonding

Hydrogen bonding is an especially strong dipole–dipole interaction (often treated as its own category) when hydrogen is bonded to F, O, or N: small, very electronegative atoms that leave a highly exposed proton. It explains the anomalously high boiling points of $\text{H}_2\text{O}$ , alcohols ( $\text{-OH}$ ), and amines in appropriate contexts, since they all include hydrogen bonds.

Example. Dimethyl ether and ethanol both contain oxygen and have formula $\mathrm{C_2H_6O}$ . Which can hydrogen-bond to itself, and can both hydrogen-bond with water?

Ethanol has an O-H bond and oxygen lone pairs, so one ethanol molecule can donate a hydrogen bond to another. Dimethyl ether has lone pairs but no O-H, N-H, or F-H bond, so it cannot form the same self-associated hydrogen-bond network. Both can accept hydrogen bonds from water. The relevant distinction is donor versus acceptor ability, not simply whether oxygen is present.

Ion–dipole interactions

Ion–dipole forces act between an ion and the partial charges of a polar solvent. They are central to dissolving ionic compounds in water and are typically stronger than neutral-molecule IMFs, and therefore these are usually the strongest type of IMF.

Example. Predict how water molecules orient around $\mathrm{Mg^{2+}}$ versus $\mathrm{Cl^-}$ . Would the larger charge of magnesium by itself prove that magnesium chloride is more soluble than every singly charged salt?

The partially negative oxygen end points toward magnesium; partially positive hydrogen ends point toward chloride. A higher ionic charge can strengthen hydration, but it also strengthens attractions within an ionic lattice. Dissolution depends on the competing energetic changes and entropy, so hydration strength alone does not establish a universal solubility ranking.

Molecular polarity and physical properties

Bond polarity compares electronegativity across a bond; molecular polarity is the vector sum of bond dipoles in three dimensions. Symmetric molecules (e.g. $\text{CO}_2$ , $\text{CCl}_4$ ) can be nonpolar even with polar bonds because dipoles cancel. Lone pairs change geometry (VSEPR), which changes whether dipoles cancel, so lone pairs matter indirectly even though polarity is a property of the whole molecule, not “of” a lone pair in isolation. A good way to tell if a molecule is polar if its dipoles are not symmetrical, or if one dipole is significantly stronger than another.

Trends: among similar molecules, greater polarity and stronger IMFs tend to raise boiling point and lower vapor pressure; the opposite trend holds for weaker IMFs.

Example. A student says that $\mathrm{CO_2}$ has dipole-dipole attractions because every C=O bond is polar. Evaluate the argument.

The two equal bond dipoles oppose each other in a linear molecule, giving zero permanent molecular dipole. Pure carbon dioxide still has dispersion forces, but polar bonds alone do not establish permanent dipole-dipole attraction. The vector arrangement of those bonds matters.

Different Solids and Properties

Selected bulk properties

Electrical conductivity is the ability for a substance to conduct electricity, basically meaning they have an active flow of electrons going through
Malleability and ductility describe the ability to deform without brittle fracture
Viscosity is a liquid’s resistance to flow; it rises with stronger IMFs and often with molecular size or hydrogen bonding
Crystalline solids have long-range order and often show sharp melting points. Amorphous solids (many glasses) lack that order and soften over a temperature range rather than melting at a single temperature.

Example. Two liquids have similar molar masses. One has lower vapor pressure at a fixed temperature and a higher normal boiling point. Explain how these two observations support the same interpretation.

Lower vapor pressure means fewer molecules escape into the equilibrium vapor at that temperature, consistent with stronger liquid-phase attractions. Boiling requires vapor pressure to reach external pressure, so that liquid must be heated more to boil at one atmosphere. Neither observation requires breaking covalent bonds within the molecules; both concern separating molecules from one another.

Types of crystalline solids

Chemists often classify solids by the particles at lattice sites and the forces holding them together.

Ionic solids combine cations and anions in a regular crystal lattice. They tend to have high melting points, are brittle, conduct when molten or dissolved (mobile ions), and often dissolve in polar solvents. Ionic solids can be hydrated (water in the crystal, often written with $\cdot\, x\,\text{H}_2\text{O}$ in the formula) or anhydrous (no water in the lattice). Ionic compounds are not described as “polar molecules” because they are not discrete polar molecules in the solid; they dissociate into ions in water.
Molecular solids are composed of discrete molecules held by IMFs. They are usually softer, lower-melting, and do not conduct as pure solids. Molecular solids are either polar or nonpolar, based on the distribution of dipoles.
Covalent network solids (e.g. diamond, quartz) extend covalent bonding through a three-dimensional network. They are typically very hard, high-melting, and insoluble (Exception: graphite is a network solid that conducts along planes).
Metallic solids feature cations in a sea of delocalized electrons, giving variable melting points, conductivity, and malleability. They look and feel metallic, so if a solid looks like a traditional metal, it is probably a metallic solid.

Example. Diamond does not conduct electricity, yet it has a very high melting temperature. Why does lack of conductivity not establish that a solid is molecular?

Diamond is a covalent-network solid: strong covalent bonds extend throughout its structure, but its electrons are not free to carry charge through the solid. A molecular solid can also lack mobile charges, although its particles are held together mainly by weaker intermolecular forces. Conductivity probes charge mobility; melting behavior probes the interactions that must be disrupted. Both observations are needed for a useful classification.

Identifying the type of solid

Simple tests help distinguish the main types of crystalline solids, as the flowchart shows. Use several tests to check the classification, since a real sample may not fit every expected property.

Heating in a test tube: Look first for condensation near the cool top of the tube. Water driven off from a crystal can indicate a hydrated ionic solid (often you are done after also checking conductivity in water). If there is no such hint and the sample melts at modest temperature, it is likely a molecular or metallic solid; test electrical conductivity on the solid (and again if you have a melt): metallic solids conduct; molecular solids do not. Many ionic salts do not melt cleanly over a burner, they may sit unchanged or decompose, so “does not melt” is not enough to prove molecular.
Solid conductivity: If the solid conducts electricity, it is metallic (or graphite, a special network solid). If it does not conduct and you already know it melts easily, treat it as molecular and go to tests 4–5.
Aqueous Conductivity: If the solid does not dissolve, skip this step for now. If the solution conducts well, the solid is almost certainly ionic (anhydrous or hydrated—combine with step 1). If the solid dissolves but the solution does not conduct (or only very weakly), it is a molecular nonelectrolyte, usually polar (e.g. sugar). This step does not distinguish polar from nonpolar if the solid is insoluble in water.
Water solubility (for molecular solids): Use only when you have already ruled out metallic and ionic behavior. Dissolves in water → polar molecular (for typical neutral molecules). Does not dissolve in water → likely nonpolar molecular, but covalent network solids such as diamond or quartz are also insoluble and non-conducting; they usually do not melt in a test tube and are extremely hard—unlike most molecular crystals.
Hexane solubility (confirmation for molecular solids): Hexane is nonpolar. If the solid dissolves in hexane, it behaves like a nonpolar species (nonpolar molecular). If it does not dissolve in hexane but did dissolve in water, that matches polar molecular. If it dissolves in neither, reconsider ionic (if it never dissolved), network covalent, or a very high–molar-mass molecular solid.

Caveat: Covalent network solids are easy to confuse with nonpolar molecular solids in water/hexane tests alone: hardness, melting behavior, and structure from other evidence matter.

Example. A solid is brittle, does not conduct as a solid, and conducts when dissolved in water. Explain why these observations favor an ionic solid over a molecular nonelectrolyte.

Fixed ions explain the lack of solid conductivity. Dissolving releases mobile ions, allowing current; a dissolved molecular nonelectrolyte would remain mostly neutral molecules. Brittleness also fits a lattice in which shifting layers can place like charges next to each other. The observations support a model together rather than identifying it from one property.

Separating mixtures

Chromatography separates components by differential affinity for a mobile phase and a stationary phase based on their differing polarity. It is typically done on cellulose or silica based paper (both polar) and involves dipping the bottom of the paper in a solvent (that is both polar and nonpolar) which allows the components to separate.

The retention factor is

R_f = \frac{\text{distance traveled by the spot}}{\text{distance traveled by the solvent front}},

with values between $0$ and $1$ for typical thin-layer work. The retention factor is a measure of how far a molecular moved compared to the solvent, and can determine the polarity of the molecule. Usually (if using polar paper and a polar stationary phase) nonpolar molecules will go higher than polar molecules, since they have a weaker attachment to the paper and thus will move up with the solvent more.

Other methods of separation include distillation (uses differences in boiling point (hence vapor pressure)) and evaporation/crystallization (removes or concentrates solvent to isolate solute).

Example. A mixture contains sand, salt, and water. A student filters it and calls the filtrate pure water. Design a sequence that recovers all three components.

Filtration removes sand but dissolved ions pass through the filter. Distill the filtrate to collect condensed water and leave salt behind, then dry the residue. Simply evaporating the water would recover salt but would not collect the water.

Temperature, kinetic energy, and the Maxwell–Boltzmann distribution

Temperature (in Kelvin) is proportional to the average translational kinetic energy of ideal-gas molecules, but varies per molecule. However, the KE can be mapped as a distribution (called the Maxwell-Boltzmann distribution). A Maxwell–Boltzmann curve plots fraction versus speed or energy. Lighter gases at the same $T$ have higher average speed, and raising $T$ broadens the curve and increases the fraction with energy above a given activation energy. Note that macroscopic kinetic energy $\frac{1}{2}mv^2$ applies to bulk motion; do not confuse it with thermal motion of molecules inside a sample. An example of a Maxwell-Boltzmann distribution can be seen below:

Example. Helium and xenon are at the same temperature. Which has the greater average translational kinetic energy, and which has the greater typical speed?

Their average translational kinetic energies are equal because each is $3k_BT/2$ . Since kinetic energy is $mv^2/2$ , the lighter helium atoms have higher typical speeds. Equal temperature does not imply equal speed when particle masses differ.

Kinetic molecular theory

Kinetic molecular theory sets down some assumptions for an ideal gas:

The gas molecules have negligible volume compared to the container
The gas molecules have no IMFs
The gas molecules perform elastic collisions with each other and the walls of the container
The gas molecules move around in constant, random motion
For one mole of a monatomic ideal gas, average translational kinetic energy (per mole) is

KE_{avg} = \frac{3}{2}RT.

Per molecule, the equation becomes

KE_{avg} = \frac{3}{2} k_B T

with Boltzmann’s constant $k_B = 1.38 \times 10^{-23} \text{ J/K}$ . Temperature in this model is proportional to the mean kinetic energy of random motion. In addition, if you are solving for velocity, the average RMS (root mean square, one way to taking an average) speed of the molcules is:

v_{RMS} = \sqrt{\frac{3RT}{M}}.

Always remember that kinetic energy is distributed along a Maxwell-Boltzmann distribution, so we always talk about the average kinetic energy and RMS velocity.

Example. An ideal gas is compressed to half its volume at constant temperature. Explain the pressure change without claiming that individual molecules acquire more kinetic energy.

Pressure doubles from $P=nRT/V$ . Average kinetic energy stays fixed because temperature stays fixed. Molecules strike the walls more frequently per unit area in the smaller volume, producing greater pressure without a sustained increase in average kinetic energy.

Pressure, STP, and the Ideal Gas Law

Pressure measures force per unit area. Useful conversions include

1 \text{ atm} = 760 \text{ mmHg} = 760 \text{ torr} = 101.325 \text{ kPa} = 14.7 \text{ psi} = 1.013 \text{bar}

STP (standard temperature and pressure) is commonly taken as $0^\circ\text{C}$ ( $273.15 \text{ K}$ ) and $1 \text{ atm}$ , with one mole of an ideal gas under those conditions occupies about $22.4 \text{ L}$ (Modern IUPAC defines a slightly different standard pressure, but for AP chemistry purposes this is the standard pressure). For the remainder of the unit and AP Chemistry, we will use the following notations, along with their standard units:

The Gas Laws and the Ideal Gas Law

There are many gas laws that are useful for the AP exam (Remember to ALWAYS use Kelvin!):

Boyle’s law: $P_1 V_1 = P_2 V_2$ (or equivalently $P \propto \frac{1}{V}$ ) at fixed $n$ and $T$ .
Charles’s law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (or equivalently $V \propto T$ ) at fixed $n, P$ (use Kelvin for $T$ ).
Gay-Lussac’s law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (or equivalently $P \propto T$ ) at fixed $n$ and $V$ .
Avogadro’s law: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$ (or equivalently $V \propto n$ ) at fixed $P$ and $T$ .

The combined gas laws can merge to form the Ideal Gas Law (IGL):

PV = nRT.

Use a value of $R$ whose pressure and volume units match the problem:

$R = 0.08206\ \text{L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$
$R = 8.314\ \text{J}/(\text{mol}\cdot\text{K}) = 8.314\ \text{kPa}\cdot\text{L}/(\text{mol}\cdot\text{K})$
$R = 62.36\ \text{L}\cdot\text{torr}/(\text{mol}\cdot\text{K})$
$R = 0.08314\ \text{L}\cdot\text{bar}/(\text{mol}\cdot\text{K})$

Molarity in the gas phase is often written $C = n/V$ , giving

P = CRT

for an ideal gas at temperature $T$ .

Two useful rearrangements combine the ideal gas law with molar mass $M$ :

d = \frac{PM}{RT}

for gas density $d$ , and

M = \frac{dRT}{P}

when density is measured directly. These are just $PV=nRT$ with $n=m/M$ and $d=m/V$ .

Dalton’s law of partial pressures: for a mixture of ideal gases,

P_{\text{total}} = \sum P_i, \qquad P_i = x_i P_{\text{total}},

where $x_i$ is the mole fraction of gas $i$ . When collecting a gas over water, include the vapor pressure of water:

P_{\text{total}} = P_{\text{gas}} + P_{\text{H}_2\text{O}}.

Example. A rigid vessel contains equal moles of He and Ne. Half the He is removed at constant temperature. Find the new total pressure as a fraction of the old pressure and the new Ne mole fraction.

Initially there are $2n$ total moles; afterward there are $1.5n$ . Thus $P_f/P_i=3/4$ . Neon now represents $n/(1.5n)=2/3$ of the mixture, but its partial pressure stays unchanged because its own mole count, temperature, and volume did not change.

Diffusion and effusion

Diffusion is mixing driven by random molecular motion, while effusion is escape through a small opening. Graham’s law compares rates for gases at the same temperature:

\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}},

where $M$ is molar mass. Graham’s Law basically states that lighter molecules move faster on average and effuse faster.

Example. Gas X takes twice as long as helium to effuse the same amount through the same opening under matching conditions. Find its molar mass and explain why the time ratio is inverted in the rate ratio.

Equal amounts give $r_X/r_{He}=t_{He}/t_X=1/2$ . Graham’s law gives $1/2=\sqrt{4.00/M_X}$ , so $M_X=16.0\ \mathrm{g/mol}$ . Longer time means lower rate; substituting the time ratio directly as a rate ratio would predict the wrong mass.

Real gases

Real gases deviate when volume is finite and attractions matter. The van der Waals equation is a textbook correction:

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT.

The compressibility factor is a measure of the effects of IMFs on a gas:

Z = \frac{PV}{nRT}

which equals $1$ for an ideal gas (by Ideal Gas Law). $Z < 1$ often reflects attractive effects dominating at moderate pressure, and $Z > 1$ can appear when repulsive effects dominate at high pressure. Note that this is not very important for an AP context.

Example. At moderate pressure, a gas has measured pressure below $nRT/V$ . Which ideal-gas assumption is most likely failing, and why can very high pressure produce the opposite deviation?

Attractions reduce momentum transferred to the walls, lowering measured pressure. At very high density, finite particle volume reduces the space available for motion, which can raise pressure above the ideal prediction. These competing effects explain why a gas need not deviate in the same direction at every pressure.

Phase behavior and phase diagrams

Vapor pressure is the pressure of vapor in equilibrium with a condensed phase at a given temperature; it rises with $T$ and reflects IMF strength (volatile liquids have high vapor pressure at a given $T$ ). A phase diagram plots pressure versus temperature; the triple point is where solid, liquid, and gas coexist. The critical point ends the liquid–vapor boundary; above the critical temperature there is no distinct liquid phase at any pressure.

Example. A solid is heated at constant pressure below its triple-point pressure. Can a liquid plateau appear on the heating curve? Explain from phase stability.

Below the triple-point pressure the liquid is not a stable equilibrium phase. The solid sublimes to gas rather than melting to liquid. A phase-change plateau can still occur, but it represents sublimation rather than melting.

Colligative properties of solutions

Colligative properties depend on the concentration of solute particles, not on their chemical identity. When you dissolve a solid into a liquid, it increases the IMFs/interactions between the molecules. Resulting in the following formulas (Use molality $m$ (moles solute per kilogram solvent) in the standard formulas):

Boiling point elevation:

\Delta T_b = i K_b m.

Freezing point depression:

\Delta T_f = i K_f m.

$K_b$ and $K_f$ are solvent constants. The van’t Hoff factor $i$ is the moles of dissolved particles produced per mole of formula units added (for molecular solids $i = 1$ ). Although this is not always the case, for ionic solids, you can assume that $i$ is the number of ions that result after one molecule of the solid dissolves.

Example. An ideal $0.10\ m$ solution of a salt that produces three ions and an ideal $0.30\ m$ glucose solution use the same solvent. Compare their freezing points.

The relevant product is $im$ . It is $3(0.10)=0.30\ m$ for the fully dissociated salt and $1(0.30)=0.30\ m$ for glucose. They have equal predicted freezing-point depressions, despite different formula-unit concentrations. Ion association would weaken that equality in a real salt solution.

Vapor pressure of solutions and volatile mixtures

All liquids tend to evaporate, since their energy can be mapped to a Maxwell-Boltzmann distribution. The pressure generated by the evaporation is called vapor pressure. Raoult’s law for a nonvolatile solute states that:

P_{\text{solution}} = x_{\text{solvent}}\, P^\circ_{\text{solvent}},

lowering vapor pressure relative to the pure solvent. For two volatile components,

P_{\text{total}} = x_A P^\circ_A + x_B P^\circ_B

when the mixture is ideal. Positive deviations from Raoult’s law mean weaker attractions between unlike molecules than the average of like–like interactions; negative deviations mean stronger attractions between unlike partners.

Henry’s law relates gas solubility in a liquid to partial pressure:

C = k_H P,

with $k_H$ a constant for a given solute–solvent pair at fixed $T$ .

Example. An ideal liquid mixture has mole fraction $x_A=0.50$ , with pure vapor pressures $P_A^*=100$ and $P_B^*=40\ \mathrm{torr}$ . Determine whether its vapor is also half A.

Raoult’s law gives partial pressures $50$ and $20\ \mathrm{torr}$ , so total pressure is $70\ \mathrm{torr}$ . The vapor mole fraction of A is $50/70=0.714$ . The more volatile component is enriched in the vapor, which is the basis for separation by distillation.

Osmotic pressure

For dilute solutions, osmotic pressure $\Pi$ obeys

\Pi = i M R T,

with $M$ in $\text{mol/L}$ and $R$ matched to the units of $\Pi$ (commonly $0.0821 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$ when $\Pi$ is in atm). Osmosis is net flow of solvent through a semipermeable membrane toward higher solute concentration. Note that this will likely not appear on the AP Chemistry exam.

Example. A membrane passes water but no solute. On one side is ideal $0.10\ M$ glucose; on the other is fully dissociated $0.060\ M$ NaCl. Predict the initial net water movement at equal temperature and pressure.

Effective particle concentrations are $0.10\ M$ and $2(0.060)=0.120\ M$ . Since $\Pi=iMRT$ , water initially moves toward the NaCl side. Comparing formula-unit molarities alone would give the opposite prediction.

Light and solutions: Beer–Lambert law

Spectrophotometry/colorimetry uses the Beer–Lambert Law (Or alternatively Beer’s Law): absorbance is proportional to concentration for a fixed path length:

A = \varepsilon l c,

where $\varepsilon$ is the molar absorptivity, $l$ the path length, and $c$ the concentration. Transmittance $T$ (fraction of light passing) relates to absorbance by

A = -\log_{10}(T)

Transmittance rarely shows up on the AP exam. Beer’s law is very applicable for measuring equilibrium/kinetics, since absorbance is directly proportional to concentration. When doing colorimetry, always calibrate beforehand and set the wavelength to the wavelength that is closest to the OPPOSITE of the color of the solution to get maximum absorbance.

Example. An unknown gives absorbance $0.60$ in a $2.0\ \mathrm{cm}$ cuvette. A $0.020\ M$ standard gives $0.40$ in a $1.0\ \mathrm{cm}$ cuvette. Determine the unknown concentration.

Taking a ratio of $A=\varepsilon lc$ gives $c_u=0.020(0.60/0.40)(1.0/2.0)=0.015\ M$ . The unknown absorbs more overall because its path is longer, even though it is less concentrated. Absorbance alone is not a concentration comparison unless path lengths match.

Practice

MCQ

A $10.0\ \mathrm{mL}$ colored sample is diluted to $50.0\ \mathrm{mL}$ . Its absorbance is then 0.240 in a 2.00 cm cuvette. What would the original sample’s absorbance be in a 1.00 cm cuvette, assuming linear Beer-Lambert behavior?

(A) $0.120$
(B) $0.600$
(C) $1.20$
(D) $2.40$

Equal moles of He and Ne share a rigid container at the same temperature. Which comparison is correct?

(A) He has greater partial pressure and greater mean speed
(B) Ne has greater partial pressure and smaller mean speed
(C) Partial pressures and mean speeds are both equal
(D) Partial pressures are equal, but He has greater mean speed

An ideal liquid mixture contains equal moles of A and B, whose pure vapor pressures are 80 and 20 torr. What fraction of the vapor is A?

(A) $0.20$
(B) $0.50$
(C) $0.80$
(D) $1.00$

A gas mixture is collected over water at total pressure 760 torr. Water vapor pressure is 24 torr. If the collected volume is used with 760 torr to calculate dry gas moles, what is the error?

(A) Dry gas moles are overestimated
(B) Dry gas moles are underestimated
(C) No error because water is liquid
(D) No error because partial pressures cannot be subtracted

Two gases X and Y effuse under matching conditions. X effuses three times as fast as Y. Which molar-mass relation follows?

(A) $M_X=3M_Y$
(B) $M_X=M_Y/3$
(C) $M_X=9M_Y$
(D) $M_X=M_Y/9$

An ideal gas is compressed isothermally from V to V/3. Which explanation correctly accounts for the pressure increase?

(A) Average kinetic energy triples
(B) Wall collisions become more frequent while average kinetic energy stays fixed
(C) Molecular mass triples
(D) Intermolecular attractions triple the pressure

FRQ

A student measures the absorbance of several solutions of $\text{Cu}^{2+}$ at the same wavelength and path length.

$(A)$ Explain why absorbance can be used to determine concentration.

$(B)$ A solution has absorbance $0.420$ . A calibration line has equation $A=15.0c$ , where $c$ is in $\text{mol/L}$ . Calculate the concentration.

$(C)$ Explain why the wavelength should be chosen near the color most strongly absorbed by the solution.

$(D)$ Original extension. The measured solution was prepared by diluting $10.0\ \text{mL}$ of a stock to $100.0\ \text{mL}$ . Calculate the stock concentration. Then explain the direction of error if fingerprints on the cuvette reduced transmitted light.

The 2026 AP Chemistry exam included a spectrophotometry particle-diagram question about absorbance and ion concentration. (Adapted from College Board, 2026 AP Chemistry FRQ 6.)

$(A)$ Explain why a solution with greater concentration of colored ions has greater absorbance at a fixed wavelength.

$(B)$ If a calibration curve has equation $A=4.00c$ and an unknown solution has $A=0.120$ , calculate $c$ .

$(C)$ In a particle diagram of equal volume, how should the number of colored ions compare between a $0.020\ M$ solution and a $0.040\ M$ solution?

$(D)$ Original extension. Predict the absorbance of the $0.0300\ M$ solution if the cuvette path length is halved. Would halving the concentration in the original cuvette have the same optical effect? Explain using particle encounters along the light path.