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Colligative Properties Quiz

12 questions chemistry Grades 9-12

The question sheet

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  1. Molality is defined as the ratio of:

    • Grams of solute to grams of solvent
    • Moles of solute to kilograms of solvent
    • Moles of solute to total moles
    • Moles of solute to liters of solution
    Reveal answer

    Answer: Moles of solute to kilograms of solvent

    Source evidence

    PDF page 629: mol solute m = kg solvent Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.

  2. Why are mole fraction and molality preferred for colligative property calculations?

    • They depend on solution volume
    • They do not vary with temperature
    • They vary with temperature
    • They are easier to measure
    Reveal answer

    Answer: They do not vary with temperature

    Source evidence

    PDF page 628: a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species: mol solute M = L solution Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality. The mole fraction, X, of a component is the ratio of its molar amount to the total number of moles of all solution components: mol A X = A total mol of all components By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one. Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:

    PDF page 629: mol solute m = kg solvent Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.

  3. Dissolving a nonvolatile solute in a volatile liquid causes the liquid's vapor pressure to:

    • Fluctuate randomly
    • Increase
    • Decrease
    • Remain the same
    Reveal answer

    Answer: Decrease

    Source evidence

    PDF page 631: As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates: liquid ⇌ gas Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporizationcondensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure 11.18). While this interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the more dispersed nature of matter in a solution, compared to separate solvent and solute phases, serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.

  4. For a solution containing only nonvolatile solutes, the vapor pressure above it is due to:

    • Both solute and solvent equally
    • Neither component
    • The solute only
    • The solvent only
    Reveal answer

    Answer: The solvent only

    Source evidence

    PDF page 632: i solution i i i i A nonvolatile substance is one whose vapor pressure is negligible (P* ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:

    PDF page 633: P = X P* solution solvent solvent

  5. Distillation separates solution components based on differences in their:

    • Molar masses
    • Vapor pressures
    • Densities
    • Colors
    Reveal answer

    Answer: Vapor pressures

    Source evidence

    PDF page 633: Solutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult’s law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected—effectively separating it from component B. Distillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in Figure 11.19.

  6. In distillation of two volatile liquids, the vapor is enriched in:

    • Neither liquid
    • The more volatile liquid
    • Both equally
    • The less volatile liquid
    Reveal answer

    Answer: The more volatile liquid

    Source evidence

    PDF page 633: Solutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult’s law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected—effectively separating it from component B. Distillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in Figure 11.19.

  7. Boiling point elevation constants (Kb) depend on:

    • The container shape
    • The identity of the solvent
    • The color of the solution
    • The mass of the solute
    Reveal answer

    Answer: The identity of the solvent

    Source evidence

    PDF page 635: As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Vapor pressure increases with temperature, and so a solution will require a higher temperature than will pure solvent to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, ΔTb, is called boiling point elevation and is directly proportional to the molal concentration of solute species: ΔT = K m b b where Kb is the boiling point elevation constant, or the ebullioscopic constant and m is the molal concentration (molality) of all solute species. Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table 11.2.

  8. Freezing point depression, ΔTf, is directly proportional to:

    • The molal concentration of the solute
    • The container volume
    • The molar mass of the solute
    • The mass of the solute
    Reveal answer

    Answer: The molal concentration of the solute

    Source evidence

    PDF page 638: The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the freezing point depression and is directly proportional to the molal concentration of the solute ΔT = K m

  9. The osmotic pressure of a dilute solution is given by which equation?

    • Π = X P*
    • Π = Kf m
    • Π = MRT
    • Π = Kb m
    Reveal answer

    Answer: Π = MRT

    Source evidence

    PDF page 641: When osmosis is carried out in an apparatus like that shown in Figure 11.24, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure (Π) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation Π = MRT where R is the universal gas constant.

  10. A solution injected into the body should be isotonic, meaning it has:

    • Higher osmotic pressure than blood serum
    • No osmotic pressure
    • Lower osmotic pressure than blood serum
    • The same osmotic pressure as blood serum
    Reveal answer

    Answer: The same osmotic pressure as blood serum

    Source evidence

    PDF page 643: Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure 11.27.

  11. According to Raoult's law, the partial pressure of a solution component equals the pure vapor pressure multiplied by its:

    • Mole fraction
    • Mass percentage
    • Molar mass
    • Molality
    Reveal answer

    Answer: Mole fraction

    Source evidence

    PDF page 632: The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. P = X P* A A A where P is the partial pressure exerted by component A in the solution, P* is the vapor pressure of pure A, and X

  12. In the solution phase diagram, the solid-liquid curve is displaced left of the pure solvent, representing:

    • Vapor pressure raising
    • Boiling point elevation
    • Osmotic pressure
    • Freezing point depression
    Reveal answer

    Answer: Freezing point depression

    Source evidence

    PDF page 640: the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed of solvent only, and so transitions between these phases are not subject to colligative effects.

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