AP Unit 3 Solutions

1. All samples exhibit dispersion; the polar molecules also provide permanent-dipole interactions.
2. Methanol self hydrogen bonds through its O–H donor and O acceptor. Ether lacks a qualifying donor, but accepts a hydrogen bond donated by water.
3. Around Cl⁻, water’s Hδ+ regions point inward, giving ion–dipole attraction.
4. The dotted intermolecular hydrogen-bond or hydration contact does not replace an intramolecular O–H covalent bond.

1. The polar molecule has dipole interactions in addition to dispersion, but category names do not fix the total attraction. A larger, more easily distorted electron cloud can give stronger summed London dispersion. Greater contact area also allows more simultaneous attractive contacts; a compact isomer may offer less contact than an extended one.
2. Accessible delocalized π density enhances polarizability, allowing instantaneous charge fluctuations and induced dipoles to produce stronger dispersion contacts, other factors comparable. It does not alone determine an exact boiling point: overall size, geometry and other interactions must also be considered.
3. London dispersion denotes interactions involving temporary, fluctuating dipoles and the dipoles they induce. Van der Waals is a broader grouping in common conventions, including other neutral-particle interactions; it must not be treated as a synonym for London dispersion.

1. The polar site induces a dipole in the nonpolar aromatic electron cloud. Oppositely charged distant sites can also attract. These noncovalent contacts stabilize folds while the covalent backbone remains connected.
2. For part (a), the dipole-induced-dipole strength increases P₁–N₁ < P₁–N₂ < P₂–N₂. First the nonpolar partner becomes more polarizable at unchanged permanent dipole; then the permanent dipole increases at unchanged polarizability. The induced dipole responds to the polar partner, giving attraction rather than requiring a permanent dipole on N.
3. For part (b), at the stated same separation and favorable orientation, larger permanent dipoles give a stronger dipole–dipole contribution for the P₂ pair. An attractive end-to-end sketch places the δ− end of one molecule beside the δ+ end of the other. An unfavorable close approach places like ends together; their repulsion changes the interaction. A dipole magnitude alone therefore cannot specify the interaction without orientation.
4. For part (c), both samples have London dispersion attractions. Under the stated comparable-size/polarizability and contact conditions, polar molecules also have attractive dipole–dipole contributions, so their total attractions are typically greater. Polarity does not replace dispersion. This comparison does not rank a very large, highly polarizable nonpolar molecule against a much smaller polar molecule.
5. For part (d), the full ionic charge produces ion–dipole attraction to the opposite partial charge; it tends to exceed a comparable dipole–dipole attraction involving partial charges. The δ− end faces the cation. Charge magnitude, molecular dipole, distance and orientation affect strength, so an unfavorable orientation or unlike sizes/distances cannot be ignored.

1. For similar size/shape, stronger net attractions in X make escape into vapor less favorable, so its vapor pressure is generally lower.
2. A higher temperature is needed for X’s vapor pressure to reach the shared external pressure, so it generally boils higher.
3. Melting rearranges contacts and depends on packing, symmetry and the solid structures, so force categories alone do not fix exact melting points.

1. A: ionic, NaCl. B: molecular, solid CO₂. C: diamond is the supplied identity; connected carbon atoms form a covalent network rather than separate molecules. D: metallic, copper. Several interacting particles reveal an extended lattice, intermolecular contacts, network connectivity or mobile carriers; an isolated species cannot show those features.
2. Opposite ions attract strongly throughout A. Disrupting these contacts requires substantial energy, so ionic solids tend to have high melting temperatures. Separating particles from the condensed material into a gas also requires overcoming strong ionic attractions, giving low vapor pressures and typically high boiling temperatures at a specified pressure; some ionic compounds decompose before boiling. A shifted layer can bring like charges together, whose repulsion causes fracture. In solid NaCl the ions are not free to translate; in the melt they are mobile charge carriers. Its current is carried by ions, not an electron sea.
3. In B, melting rearranges relatively weak intermolecular contacts between intact CO₂ molecules. It does not require breaking their C–O bonds. Molecular solids generally melt at lower temperatures than ionic or network solids; the molecules have neither mobile ions nor freely mobile electrons, so solid CO₂ ordinarily does not conduct. Exact melting rankings for all molecular solids require packing and structure information; large molecules and polymers need additional qualifications.
4. The covalent network in C requires many strong bonds to be disrupted, explaining a high melting or decomposition temperature; fixed bonding geometry makes a three-dimensional network rigid and hard. AP network examples contain nonmetals/metalloids: elemental diamond or graphite, and binary SiO₂ or SiC. Graphite has strong covalent networks within layers but weaker interlayer contacts, allowing layers to slide. Delocalized electrons within graphite layers permit conduction; diamond lacks freely mobile carriers in its ordinary network.
5. In D, delocalized electrons carry electrical charge and transfer thermal energy; lattice vibrations also contribute to heat transfer. Metal cores can rearrange while remaining bound by the mobile electron distribution, so a pure metal is often malleable and ductile. Interstitial guests hinder that rearrangement and generally reduce malleability/ductility while mobile electrons preserve conduction.

1. Heating can disrupt or rearrange interchain or intrachain noncovalent contacts, increasing segment mobility while connectivity remains.
2. Backbone cleavage would change molecular connectivity or chain-size distribution; structural or molecular-size measurements could test that.
3. Mechanical softness alone does not establish which bonds changed.

1. A crystalline-solid sketch has close particles in a repeating regular arrangement, with small arrows for local vibration and no overall neighbor exchange. An amorphous-solid sketch has close irregular positions, also local vibration and restricted overall translation. Both keep their sample shape and volume on the modeled timescale; disorder alone does not make a liquid.
2. A liquid sketch has closely spaced intact molecules and arrows indicating motion, collision and changing neighbors. It has a definite volume but adapts to the vessel’s shape. A gas sketch preserves molecule identity/icon size but has much larger separations and random container-wide motion; it adopts the container’s volume and shape.
3. For an ordinary substance at fixed amount, solid and liquid molar volumes are generally similar because particles are close in both phases. Equal-amount whole-sample panels conserve particle counts. Equal-volume windows can have fewer gas particles, but must be labeled as number-density samples, not particle loss.
4. The warmer liquid snapshot should show a spread of motion arrows with greater typical speed, not every arrow the same length. Higher average kinetic energy promotes continuing collisions and contact rearrangement against intermolecular attractions; particles remain close and change neighbors at both temperatures. This does not require covalent-bond cleavage. Without density/thermal-expansion data, exact spacing cannot be inferred; the sketches need not invent an exact expansion ratio.

1. Particle number and intrinsic size remain unchanged; number per volume doubles and average spacing decreases.
2. Constant temperature preserves the speed distribution for the same species.
3. Wall collisions occur more frequently per area; ideal P ∝ 1/V, so pressure doubles.

1. Convert temperatures: 300.15 K and 400.15 K.
2. Use P₂ = P₁(V₁/V₂)(T₂/T₁) = 0.950(1.50/3.00)(400.15/300.15).
3. For fixed n,V, P = (nR/V)T: plot P on the vertical axis against Kelvin T on the horizontal with slope nR/V.

1. Initial total P = 0.300RT/V = 1.4672 atm, and P_He = 0.100RT/V = 0.4891 atm.
2. After addition, total amount is 0.400 mol, so X_He = 0.250 and P_total = 1.9563 atm.
3. P_He remains 0.4891 atm because its amount, T and V remain unchanged; equivalently 0.250 × 1.9563.

1. Plot P against 1/V. The values are (1.00 L⁻¹,4.00 atm), (0.500,2.00), (0.250,1.00).
2. Slope is 4.00 atm L, representing nRT in the ideal law.
3. Dry-gas partial pressure = 0.980 − 0.031 = 0.949 atm.

1. The 675 K curve has its peak at a higher speed, a lower peak height and a broader spread. Area across a speed interval represents the fraction of particles with speeds in that interval; a density value at one exact speed is not a particle fraction. Both complete distributions have unit total area. Lower peak height reflects a broader distribution, not particle loss. The finite plotting window omits the remote tails, so distinguish its visible area from the complete normalized area.
2. Mean translational kinetic energy scales with Kelvin temperature: 675/300 = 2.25. For the same molecular mass, 〈KE〉 = ½m〈v²〉, so the rms-speed ratio is √2.25 = 1.50. Do not equate a 2.25 energy ratio with a 2.25 speed ratio.
3. At fixed amount and volume, P ∝ T, giving P₆₇₅/P₃₀₀ = 2.25. Faster particles on average strike walls more often and transfer more momentum per collision; the combined time-averaged transfer gives the pressure increase.

1. Equal temperature gives equal mean translational kinetic energy, not equal speed. Since m〈v²〉 is equal, v_rms,H₂/v_rms,O₂ = √(32.00/2.00) = 4.00. H₂’s speed distribution lies at higher typical speeds. The normalized ideal translational kinetic-energy distributions at the same T match regardless of molecular mass.
2. A valid constructed example has eight intact H₂ pairs in one box and eight intact O₂ pairs in an identically sized box. Arrow directions are varied, and lengths vary within both samples. At the same arrow scale, H₂ has larger typical speed, consistent with the rms ratio; this does not require every H₂ particle to be faster than every O₂ particle. The range of arrow lengths represents the distribution. Positions and icon dimensions are schematic, not measured trajectories or molecular sizes.
3. The pressures are equal because n, V and T are equal. A lighter particle’s greater speed gives a different collision frequency, but its smaller mass gives a smaller momentum transfer at a given speed. The time-averaged wall transfer depends on m〈v²〉, which is equal at common T. The model therefore preserves equal pressure.

For a text layout, specify eight paired H labels and eight paired O labels, then assign at least several arrow lengths and directions in each box. Identify larger typical H₂ speed on the same scale and preserve equal n, V and T.

1. The measured pressure is 0.900 of ideal, consistent with attractive interactions reducing net wall momentum transfer.
2. The absence of liquid supports using the stated amount as gas n.
3. Lower density increases typical separations, reducing attractions and particle-volume effects, so the sample generally approaches the ideal model.

1. The ratio below 1 is consistent with attraction-dominated reduction of pressure.
2. The ratio above 1 is consistent with excluded particle volume becoming important under strong compression.
3. The ideal model neglects both effects, which can compete; attractions do not impose a universal upper bound on real pressure.

1. V = 0.400 L; analytical concentration = 0.0200/0.400 = 0.0500 M.
2. Ca²⁺ amount is 0.0200 mol; Cl⁻ amount is 0.0400 mol, giving 0.0500 M and 0.100 M.
3. Total ion amount = 0.0600 mol, so count = 3.6132 × 10²².

1. Conserved solute amount = 0.200 mol L⁻¹ × 0.0150 L = 0.00300 mol.
2. Final M = 0.00300/0.1000 = 0.0300 M.
3. Above the mark means a final volume larger than intended at fixed solute amount, so the actual concentration is below 0.0300 M.

1. Molarity uses final solution volume, which may differ from starting water volume. Make the total solution volume 1.00 L for a known dissolved 0.100 mol solute, provided no reaction changes its amount. An undissolved second phase makes the overall sample heterogeneous even if its liquid portion is homogeneous.
2. (a) A solid solution: the specified one-phase alloy has uniform macroscopic composition. The atoms need not be identical for the mixture to be homogeneous.
3. (b) A liquid solution: completely mixed components form a single liquid phase of uniform macroscopic composition at the stated conditions.
4. (c) A gas solution: the gases are uniformly intermixed without a condensed phase.
5. (d) A heterogeneous mixture: location changes the amount/proportion of sand, so the whole sample is not one uniform solution.

1. The second-to-first salt-concentration ratio is (6/2.0)/(4/1.0) = 0.75. Both ion ratios are 1:2 and both windows have total charge zero.
2. Water Oδ− ends favor Mg²⁺; Hδ+ regions favor chloride.
3. More total icons in the larger window does not mean a larger concentration.
4. The created sketches must label both represented volumes and the same icon-amount scale, retain the specified 1:2 ion ratios, and show intact bent water. Use solid O–H bonds and dashed ion–dipole contacts; water oxygen faces Mg²⁺ while hydrogen regions face Cl⁻.

1. Glucose remains intact neutral molecules dispersed among water, with hydrogen-bond/dipole contacts where appropriate.
2. KCl gives equal numbers of separated K⁺ and Cl⁻ icons, hydrated with opposite water ends facing each charge.
3. KCl provides mobile ions and therefore electrolyte conduction. Neutral glucose does not provide comparable ionic carriers.
4. For example, with the same amount per icon and equal window volumes, four glucose molecule icons correspond to four K⁺ plus four Cl⁻ icons for equal analytical solute amounts. A different common count works if the 1:1 ionic ratio and equal analytical amounts are retained. Label covalent structure versus noncovalent contacts and point O ends toward K⁺, H ends toward Cl⁻.

1. R_f,A = 0.25; R_f,B = 0.625 ≈ 0.63 at the supplied precision.
2. A travels less and is more strongly retained relative to the mobile phase, consistent with stronger polar interactions with silica in this system.
3. Ordinary filtration does not retain these dissolved molecules separately.

1. Use filtration to retain the suspended solid while the liquid solution passes.
2. Use distillation of the filtrate to enrich the component with greater vapor tendency under the mixture conditions.
3. The two liquids both may enter the vapor, particularly if their volatilities are similar; a single collected fraction need not be pure.

1. R_f depends on competition involving both stationary and mobile phases, so changing the solvent can change migration for the same substance.
2. Run unknown and reference under the same stationary material, mobile solvent and conditions, preferably in the same run.
3. Matching migration supports identity but is not unique proof; additional evidence can distinguish coincident migration.

1. Both can form hydrogen bonds at O–H, but the longer hydrocarbon region contributes a larger nonpolar area.
2. In water, this increased mismatch is not necessarily compensated by one unchanged O–H site, so the long-chain alcohol generally has lower solubility.
3. The larger nonpolar region has stronger compatible dispersion contacts with a hydrocarbon solvent, favoring relative compatibility there.

1. Separating Na⁺ and Cl⁻ disrupts their ionic lattice. Making space also rearranges water–water hydrogen-bond contacts or n-hexane–n-hexane dispersion contacts. Water forms stabilizing ion–dipole hydration: oxygen δ− regions face Na⁺, while Hδ+ regions face Cl⁻. For this salt, hydration and the complete energy/arrangement balance permit appreciable dissolution. n-Hexane provides dispersion and possible ion-induced polarization, but lacks permanent polar regions for comparable ion–dipole stabilization; these contacts do not sufficiently compensate the lattice/solvent changes in this case.
2. Polarity is not a guarantee for all salts. A strong lattice and other unfavorable changes may limit dissolution even though hydration is favorable.
3. Cooling surroundings indicates heat absorption during the stated dissolution. Changes in available arrangements also contribute to favorability, so an endothermic dissolution can occur. “Like dissolves like” is a compatibility heuristic, not a full thermodynamic law. This specified n-hexane result must not be generalized to every nonaqueous solvent: polar nonaqueous solvents can stabilize ions differently.

1. A typically probes rotational-level changes; B probes vibrational-level changes; C probes electronic-level changes.
2. Absorption adds photon energy to the relevant available level; emission removes the corresponding energy.
3. These are state changes and need not change atomic connectivity or eject an electron.

1. An ordinary molecular IR band corresponds to a vibrational transition, not the AP electronic-transition association.
2. UV/visible absorption is associated with an electronic-level change, not nuclear mass or isotope identity.
3. An elemental mass spectrum with the stated singly charged monatomic-ion model gives isotope masses and relative abundances.

1. Frequency ν = E/h = 6.79143 × 10¹⁴ s⁻¹.
2. Wavelength λ = hc/E = 4.41733 × 10⁻⁷ m = 441.733 nm.
3. Emission lowers the atom’s energy by 4.50 × 10⁻¹⁹ J.

1. Direct-to-ground photon energy is 5.0 × 10⁻¹⁹ J; highest-to-middle is 2.0 × 10⁻¹⁹ J.
2. Because λ ∝ 1/E, the larger direct gap gives shorter wavelength.
3. λ_{to ground}/λ_{to middle} = 2.0/5.0 = 0.40.

1. Photon energy is hc/λ = 3.313 × 10⁻¹⁹ J for this wavelength.
2. Molar photon energy = 3.313 × 10⁻¹⁹ × 6.022 × 10²³ = 199.509 kJ mol⁻¹.
3. Unchanged wavelength means unchanged ν and photon energy. Greater intensity normally supplies more photons per time.

1. A = (250.)(1.00)(0.00200) = 0.500, dimensionless.
2. I/I₀ = 10^−A = 10^−0.500 = 0.316.
3. Doubling b and halving c keeps bc unchanged, so A remains 0.500.

1. Diluted c = (0.255 − 0.015)/400. = 0.000600 mol L⁻¹.
2. Original c is larger by 50.00/10.00 = 5.000, giving 0.00300 M.
3. Diluted Δc ≈ 0.008/400 = 0.000020 M; multiply by 5 for original Δc = 0.00010 M.
4. Use the intercept and state that standard/slope/volume uncertainties were omitted by the task.

1. Choose 525 nm among the measured wavelengths: its absorbance is largest, 0.900, at the same b and c, implying the largest measured ε. The calibration slope dA/dc = εb is therefore greatest there, so a small concentration change gives the largest absorbance change under the stated comparable-noise/no-interference conditions. This identifies a measured maximum, not a finer exact peak. A spectrum at one concentration cannot establish linearity over other concentrations; standards must validate the usable range.
2. (i) Doubling b increases A at the same true c; the original 1.00 cm slope gives a high inferred c. (ii) Dilution reduces c and A; reporting it without recovering the dilution gives a low original concentration. (iii) Lower ε gives lower A at unchanged c; applying the larger original εb slope gives a low inferred c. Retain wavelength, path length and blank conditions across standards and unknown, and measure inside a validated calibration range.

1. The high standard indicates possible deviation from the simple response or another error; investigate rather than relabel the measurement as linear.
2. Dilute the unknown by a measured factor so its absorbance falls inside the validated range, measure under matched wavelength/path/blank conditions, then recover original c using the dilution factor.
3. Replicate measurements and assess blank/standard preparation; retain justified calibration limits and precision.