AP Chemistry Unit 3 FRQ Practice

• Part A (1 point): London dispersion forces. Butane is a nonpolar hydrocarbon. See London dispersion forces.
• Part B(i) (1 point): Butane −0.5°C, acetone 56°C, 1-propanol 97°C.
• Part B(ii) (1 point): Molar masses are similar, so dispersion forces are comparable. Butane has only dispersion forces. Acetone is polar, so it adds dipole-dipole forces. 1-Propanol has an O–H bond, so its molecules also form hydrogen bonds, the strongest of these attractions. Stronger forces take more energy to overcome. Review intermolecular forces.
• Part C (1 point): Molecules in the liquid are in close contact, while those in the vapor are much farther apart. Attractions keep liquid molecules together in a fixed volume; vapor molecules move mostly independently and spread out to fill the container. See solids, liquids, and gases.
• Part D(i) (1 point): 0.580 g ÷ 58.12 g/mol = 9.98 × 10⁻³ mol.
• Part D(ii) (1 point): P = nRT/V = (9.98 × 10⁻³ mol)(0.08206 L·atm/(mol·K))(298.15 K) ÷ 1.00 L = 0.244 atm. The temperature must be in kelvins. Check it with the gas law calculator.
• Part E(i) (1 point): The assumption that gas particles do not attract one another.
• Part E(ii) (1 point): When molecules are crowded together, dispersion forces between them become significant. They pull a molecule back as it approaches the wall, so it strikes with less force, and the pressure is lower. See real gases.
• Part F (1 point): Neon adds (0.0100 mol)(0.08206)(298.15) ÷ 1.00 L = 0.245 atm. The total is 0.244 + 0.245 = 0.489 atm, because partial pressures add. See partial pressure.
• Part G (1 point): Neon atoms move faster on average. At the same temperature, both gases have the same average kinetic energy, and neon (20.18 g/mol) is less massive than butane. See kinetic molecular theory.

The most common lost point is Part B(ii): “1-propanol has stronger bonds” earns nothing. Boiling does not break covalent bonds, so the answer must name the forces between molecules, the distinction made in intermolecular vs intramolecular forces.

• Part A (1 point): (0.200 mol/L)(0.250 L) = 0.0500 mol; (0.0500 mol)(101.11 g/mol) = 5.06 g. The molarity calculator shows the same steps.
• Part B (1 point): Dissolve the solid in less than 250 mL of water, transfer it to a 250 mL volumetric flask, and add water to the calibration mark, then mix. Adding the solid to 250 mL of water does not earn the point, because the total volume must be 250 mL.
• Part C (1 point): The oxygen ends of the water molecules face the K⁺ ion, because oxygen carries a partial negative charge that is attracted to the positive ion. See particulate diagrams.
• Part D (1 point): Hexane is nonpolar and has only dispersion forces, so it cannot form ion-dipole attractions strong enough to replace the strong ionic attractions in the solid. See AP Chemistry solubility.

• Part A (1 point): c = A ÷ slope = 0.420 ÷ 2.80 M⁻¹ = 0.150 M. See the Beer-Lambert law.
• Part B (1 point): 630 nm. A blue solution transmits blue light and absorbs orange-red light, so absorbance is greater at 630 nm and changes more with concentration. See AP Chemistry spectroscopy.
• Part C (1 point): E = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (6.30 × 10⁻⁷ m) = 3.15 × 10⁻¹⁹ J. The wavelength calculator shows each step.
• Part D (1 point): Too low. The water dilutes the sample in the cuvette, so its absorbance is lower than it should be, and absorbance is directly proportional to concentration. The same reasoning runs through the percent copper in brass lab.

For Part B, “630 nm because it is red” is not enough: the justification must connect the solution’s color to the light it absorbs.