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Quiz: Phase Changes, Solutions, and Gas Laws

Test your understanding of phase changes, solution chemistry, Beer-Lambert Law, and colligative properties with these questions.

1. During which segment of a heating curve does the temperature of a substance remain constant even though heat is continuously being added?

  1. When the solid is warming below its melting point
  2. When the liquid is warming between the melting and boiling points
  3. At the melting and boiling point plateaus during phase changes
  4. When the gas is warming above its boiling point
Show Answer

The correct answer is C. During a phase change — at the melting point plateau and the boiling point plateau — all the energy added goes into breaking intermolecular forces rather than increasing molecular kinetic energy, so temperature stays constant. In the sloped segments (solid warming, liquid warming, gas warming), temperature rises as heat is added and the specific heat equation \(q = mc\Delta T\) applies. Flat plateaus are the signature feature that distinguishes phase changes from single-phase heating.

Concept Tested: Heating Curves / Phase Changes

2. For the same substance, which relationship between enthalpy values is ALWAYS true?

  1. \(\Delta H_{vap} > \Delta H_{fus}\) because vaporization must fully overcome all intermolecular forces
  2. \(\Delta H_{fus} > \Delta H_{vap}\) because melting requires breaking more bonds
  3. \(\Delta H_{fus} = \Delta H_{vap}\) because both processes are endothermic
  4. \(\Delta H_{vap} < \Delta H_{fus}\) at high temperatures only
Show Answer

The correct answer is A. The enthalpy of vaporization is always greater than the enthalpy of fusion for the same substance. Melting only disrupts the ordered lattice into a mobile liquid — molecules remain in close contact. Vaporization must completely overcome all IMFs, separating molecules to distances where attractions are negligible. For water, \(\Delta H_{vap} = 40.7\) kJ/mol is more than six times larger than \(\Delta H_{fus} = 6.01\) kJ/mol. Option B states exactly the opposite, a common misconception.

Concept Tested: Heat of Fusion / Heat of Vaporization

3. On the phase diagram of water, the solid-liquid boundary curve has an unusual negative slope (leans to the left). What does this mean physically?

  1. Water cannot exist as a liquid above 0°C at any pressure.
  2. Increasing pressure raises the melting point of ice, unlike most substances.
  3. Increasing pressure lowers the melting point of ice, so ice can melt at temperatures below 0°C under high pressure.
  4. The triple point of water is located at a higher pressure than that of most other substances.
Show Answer

The correct answer is C. For water, higher pressure lowers the melting point — the solid-liquid boundary leans to the left, opposite to most substances. This occurs because ice is less dense than liquid water; pressure favors the denser liquid phase, promoting melting at lower temperatures. Option B states the opposite of the correct relationship. Option D is actually true for water (triple point ≈ 0.006 atm) but does not explain the slope of the solid-liquid boundary.

Concept Tested: Phase Diagrams / Water's Unusual Behavior

4. Dry ice (\(\ce{CO2}\)) sublimes at atmospheric pressure rather than melting into a liquid. Which feature of the \(\ce{CO2}\) phase diagram best explains this observation?

  1. The critical point of \(\ce{CO2}\) is below room temperature.
  2. The triple point of \(\ce{CO2}\) is at 5.11 atm, which is above atmospheric pressure.
  3. \(\ce{CO2}\) has no vapor pressure at room temperature.
  4. The boiling point of \(\ce{CO2}\) is above its melting point at 1 atm.
Show Answer

The correct answer is B. The triple point of \(\ce{CO2}\) is at 5.11 atm and −57°C. Because atmospheric pressure (1 atm) is far below the triple point pressure, the liquid phase of \(\ce{CO2}\) is never stable at 1 atm — the solid converts directly to gas (sublimation) without passing through the liquid phase. At pressures above 5.11 atm, liquid \(\ce{CO2}\) can exist. This is a direct reading of the \(\ce{CO2}\) phase diagram.

Concept Tested: Triple Point / Sublimation / Phase Diagrams

5. A liquid sealed in a container reaches dynamic equilibrium with its vapor. Which statement correctly describes the equilibrium vapor pressure?

  1. The vapor pressure increases indefinitely as more liquid evaporates over time.
  2. The vapor pressure at equilibrium is higher for liquids with stronger intermolecular forces.
  3. At equilibrium, the rate of evaporation equals the rate of condensation, and vapor pressure is constant.
  4. Vapor pressure is independent of temperature and depends only on the identity of the liquid.
Show Answer

The correct answer is C. Dynamic equilibrium means that the rate of evaporation equals the rate of condensation — molecules leave and return to the liquid at the same rate. At this point the vapor pressure reaches a constant value for a given temperature. Option B is backwards: stronger IMFs produce lower vapor pressure (molecules are held more tightly in the liquid). Option D is incorrect because vapor pressure increases significantly with temperature, as described by the Clausius-Clapeyron equation.

Concept Tested: Vapor Pressure / Dynamic Equilibrium

6. The "like dissolves like" rule predicts that iodine (\(\ce{I2}\)), a nonpolar solid, will dissolve readily in hexane (\(\ce{C6H14}\)), a nonpolar solvent, but not in water. What is the best molecular-level explanation?

  1. Iodine molecules are too large to fit between water molecules in the liquid structure.
  2. Hexane provides London dispersion interactions with \(\ce{I2}\) that are comparable in strength to those within pure iodine, while water's hydrogen bonds are far stronger than any interaction it could form with \(\ce{I2}\).
  3. Water is a polar molecule, so it repels nonpolar iodine through dipole-dipole repulsion.
  4. Iodine is insoluble in hexane because both are nonpolar and nonpolar substances never interact with each other.
Show Answer

The correct answer is B. Dissolving is favorable when the solute-solvent interactions are energetically competitive with the pure-component interactions being disrupted. Hexane-\(\ce{I2}\) London dispersion forces are similar to \(\ce{I2}\)-\(\ce{I2}\) and hexane-hexane dispersion forces, making dissolution thermodynamically favorable. Water's strong hydrogen bonds are far superior to any interaction water could form with nonpolar \(\ce{I2}\), so water molecules strongly prefer each other over the iodine solute. Option C misuses the word "repulsion" — the issue is insufficient attraction, not literal repulsion.

Concept Tested: Like Dissolves Like / Solubility

7. A student dissolves 11.7 g of \(\ce{NaCl}\) (molar mass = 58.5 g/mol) in enough water to make 500.0 mL of solution. What is the molarity of the solution?

  1. 0.400 M
  2. 0.200 M
  3. 0.800 M
  4. 2.00 M
Show Answer

The correct answer is A. First calculate moles of \(\ce{NaCl}\): \(n = \frac{11.7 \text{ g}}{58.5 \text{ g/mol}} = 0.200 \text{ mol}\). Then apply the molarity definition: \(M = \frac{n}{V} = \frac{0.200 \text{ mol}}{0.5000 \text{ L}} = 0.400 \text{ M}\). Option B uses moles as the molarity without dividing by volume. Option C doubles the correct answer, a common error from failing to convert mL to L correctly.

Concept Tested: Molarity / Concentration Units

8. A solution of a colored dye is analyzed with a spectrophotometer. Using Beer-Lambert Law (\(A = \varepsilon l c\)) with \(\varepsilon = 5000\) L mol\(^{-1}\) cm\(^{-1}\) and a 1.00 cm path length, the measured absorbance is 0.750. What is the concentration of the dye?

  1. \(3.75 \times 10^{3}\) mol/L
  2. \(6.67 \times 10^{-3}\) mol/L
  3. \(7.50 \times 10^{3}\) mol/L
  4. \(1.50 \times 10^{-4}\) mol/L
Show Answer

The correct answer is D. Rearranging Beer-Lambert Law: \(c = \frac{A}{\varepsilon l} = \frac{0.750}{(5000 \text{ L mol}^{-1} \text{cm}^{-1})(1.00 \text{ cm})} = 1.50 \times 10^{-4} \text{ mol/L}\). Option A results from multiplying \(A \times \varepsilon \times l\) rather than dividing. Option B results from a unit-conversion error in the molar absorptivity. Concentrations measured by spectrophotometry are typically in the \(10^{-4}\) to \(10^{-6}\) mol/L range for strongly absorbing dyes.

Concept Tested: Beer-Lambert Law / Spectrophotometry

9. Which of the following solutions will have the LOWEST freezing point, assuming complete dissociation of all electrolytes?

  1. 1.0 m glucose (\(\ce{C6H12O6}\)), \(i = 1\)
  2. 1.0 m \(\ce{NaCl}\), \(i = 2\)
  3. 1.0 m \(\ce{MgCl2}\), \(i = 3\)
  4. 1.0 m \(\ce{AlCl3}\), \(i = 4\)
Show Answer

The correct answer is D. Freezing point depression is \(\Delta T_f = K_f m i\). With all solutions at the same molality (1.0 m), the solution with the largest van 't Hoff factor \(i\) produces the greatest depression. \(\ce{AlCl3}\) dissociates into 4 ions (\(\ce{Al^{3+}}\) + 3 \(\ce{Cl^-}\)), giving \(i = 4\) and the largest \(\Delta T_f\). Glucose is a nonelectrolyte (\(i = 1\)) and produces the smallest depression. This question tests understanding that colligative properties depend on the number of particles, not their identity.

Concept Tested: Freezing Point Depression / van 't Hoff Factor / Colligative Properties

10. A student measures the freezing point depression of an unknown nonelectrolyte dissolved in water. The solution contains 6.00 g of solute dissolved in 200. g of water, and the freezing point is \(-0.930°C\). Given \(K_f = 1.86\) °C·kg/mol for water, what is the approximate molar mass of the unknown?

  1. 30.0 g/mol
  2. 60.0 g/mol
  3. 120 g/mol
  4. 186 g/mol
Show Answer

The correct answer is B. First find molality: \(m = \frac{\Delta T_f}{K_f \cdot i} = \frac{0.930}{1.86 \times 1} = 0.500\) mol/kg. Then find moles of solute: \(n = 0.500 \text{ mol/kg} \times 0.200 \text{ kg} = 0.100 \text{ mol}\). Finally, molar mass = \(\frac{6.00 \text{ g}}{0.100 \text{ mol}} = 60.0\) g/mol. This is a classic molar mass determination by freezing point depression — a standard AP lab technique. Option A results from using 100 g instead of 200 g of solvent.

Concept Tested: Freezing Point Depression / Molar Mass Determination / Colligative Properties