DAT Quiz: Solutions And Colligative Properties
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Solutions And Colligative PropertiesQuestion 1 of 20

A solution is prepared by dissolving 15.0 g of sucrose in 85.0 g of water. What is the mass percent of sucrose in the solution?

1.50%
15.0%
17.6%
85.0%
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DAT Quiz: Solutions And Colligative Properties

Practice Solutions And Colligative Properties in DAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solutions And Colligative Properties, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A solution is prepared by dissolving 15.0 g of sucrose in 85.0 g of water. What is the mass percent of sucrose in the solution?

  1. 1.50%
  2. 15.0% (correct answer)
  3. 17.6%
  4. 85.0%
Explanation: Mass percent problems test your understanding of concentration calculations, which are fundamental in chemistry. When you see "mass percent," you're looking for the mass of solute divided by the total mass of solution, multiplied by 100%. To solve this problem, you need to identify what makes up the solution. You have 15.0 g of sucrose (the solute) dissolved in 85.0 g of water (the solvent). The total mass of the solution is therefore 15.0 g + 85.0 g = 100.0 g. The mass percent formula is: Mass percent=mass of solutetotal mass of solution×100%\text{Mass percent} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100\% Substituting your values: Mass percent=15.0 g100.0 g×100%=15.0%\text{Mass percent} = \frac{15.0 \text{ g}}{100.0 \text{ g}} \times 100\% = 15.0\% This confirms answer choice B is correct. Looking at the wrong answers: Choice A (1.50%) appears to result from incorrectly dividing 15.0 by 1000 instead of 100, or perhaps confusing decimal placement. Choice C (17.6%) comes from the common error of dividing the solute mass by only the solvent mass (15.0/85.0 × 100% ≈ 17.6%). Choice D (85.0%) represents the mass percent of water, not sucrose—essentially solving for the wrong component. Remember this key point: in mass percent calculations, always use the total solution mass in the denominator, not just the solvent mass. The solution includes both solute and solvent combined.

Question 2

Each of the following is a colligative property of a solution EXCEPT one. Which one is the EXCEPTION?

  1. Boiling point elevation
  2. Freezing point depression
  3. Density of the solution (correct answer)
  4. Osmotic pressure
Explanation: Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. These properties change predictably when you dissolve any substance in a solvent, regardless of what that substance is. The correct answer is C because density is not a colligative property. Density depends on both the mass and volume of the solution, which are directly related to the chemical identity and properties of the solute. Different solutes of the same molarity will produce solutions with different densities because they have different molecular masses and may interact differently with the solvent. Choice A is incorrect because boiling point elevation is a classic colligative property. Adding any solute raises the boiling point proportionally to the number of particles dissolved. Choice B is wrong because freezing point depression is another fundamental colligative property—the freezing point decreases based on particle concentration, which is why we use salt to melt ice. Choice D is incorrect because osmotic pressure is the third major colligative property, where pressure develops across a semipermeable membrane proportional to the concentration of solute particles. Remember the "big three" colligative properties: boiling point elevation, freezing point depression, and osmotic pressure. If you see a solution property question, ask yourself: "Does this depend only on how many particles are dissolved, or does it matter what those particles actually are?" Colligative properties care only about quantity, not identity.

Question 3

At 30 °C, the vapor pressure of pure benzene is 120 torr. If 1.0 mole of a non-volatile, non-electrolyte solute is dissolved in 4.0 moles of benzene, what is the new vapor pressure of the solution?

  1. 24 torr
  2. 96 torr (correct answer)
  3. 115 torr
  4. 150 torr
Explanation: This question tests your understanding of Raoult's law, which describes how adding a non-volatile solute decreases the vapor pressure of a solution. When you encounter vapor pressure problems involving solutions, think about how the solute particles interfere with the solvent's ability to evaporate. Raoult's law states that the vapor pressure of a solution equals the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent: Psolution=Ppure×XsolventP_{solution} = P_{pure} \times X_{solvent} First, calculate the mole fraction of benzene (the solvent). With 4.0 moles of benzene and 1.0 mole of solute, the total moles = 5.0. The mole fraction of benzene is Xbenzene=4.05.0=0.8X_{benzene} = \frac{4.0}{5.0} = 0.8 Now apply Raoult's law: Psolution=120 torr×0.8=96 torrP_{solution} = 120 \text{ torr} \times 0.8 = 96 \text{ torr} Answer B (96 torr) is correct. Answer A (24 torr) incorrectly uses the mole fraction of the solute instead of the solvent, giving 120×0.2=24120 \times 0.2 = 24 torr. Answer C (115 torr) appears to subtract a small arbitrary amount from the original vapor pressure without proper calculation. Answer D (150 torr) mistakenly suggests that adding solute increases vapor pressure, which violates Raoult's law entirely. Remember: non-volatile solutes always decrease vapor pressure. Calculate the solvent's mole fraction (not the solute's), then multiply by the pure solvent's vapor pressure. Watch for answer choices that use the wrong mole fraction—this is a common trap.

Question 4

A solution is made by mixing 2.0 moles of ethanol (C₂H₅OH) with 8.0 moles of water. What is the mole fraction of ethanol in the solution?

  1. 0.20 (correct answer)
  2. 0.25
  3. 0.80
  4. 4.00
Explanation: Mole fraction problems test your understanding of solution composition using moles rather than mass or volume. When you see "mole fraction," you're looking for the ratio of moles of one component to the total moles in the solution. To find the mole fraction of ethanol, you need to divide the moles of ethanol by the total moles in solution. First, calculate the total moles: 2.0 moles ethanol + 8.0 moles water = 10.0 total moles. Then apply the mole fraction formula: χethanol=moles of ethanoltotal moles=2.010.0=0.20\chi_{ethanol} = \frac{\text{moles of ethanol}}{\text{total moles}} = \frac{2.0}{10.0} = 0.20 Looking at the wrong answers: Choice B (0.25) likely comes from incorrectly using just the water as the denominator: 2.0/8.0 = 0.25. This ignores that mole fraction requires total moles, not just one component. Choice C (0.80) represents the mole fraction of water, not ethanol—this is the result of flipping the components (8.0/10.0 = 0.80). Choice D (4.00) comes from incorrectly dividing water by ethanol (8.0/2.0 = 4.00), which isn't even a valid mole fraction since these values must be between 0 and 1. Remember that all mole fractions in a solution must add up to 1.0. As a quick check, verify that your ethanol mole fraction (0.20) plus water's mole fraction (0.80) equals 1.0. This confirms answer A is correct and helps catch calculation errors on similar problems.

Question 5

A chemist needs to prepare a solution with a concentration that will not change if the ambient temperature fluctuates. Which of the following concentration units is the most suitable choice?

  1. Molality (mol/kg solvent) (correct answer)
  2. Molarity (mol/L solution)
  3. Normality (equivalents/L solution)
  4. Volume percent (% v/v)
Explanation: When you encounter questions about concentration units and temperature effects, think about how each unit is defined and whether volume changes will affect the concentration calculation. Molality (A) is defined as moles of solute per kilogram of solvent. Since this unit is based on mass rather than volume, and mass doesn't change with temperature fluctuations, molality remains constant regardless of thermal expansion or contraction of the solution. This makes it the ideal choice for temperature-sensitive applications. Molarity (B) is moles of solute per liter of solution. Since liquids expand when heated and contract when cooled, the volume of solution changes with temperature. As the denominator in the molarity calculation changes, the concentration value will fluctuate even though the actual amount of solute remains constant. Normality (C) suffers from the same volume-dependence problem as molarity, since it's defined as equivalents per liter of solution. Temperature changes will alter the solution volume and thus the normality value. Volume percent (D) is also problematic because it's based on the ratio of solute volume to solution volume. Both volumes can change with temperature, making this unit unsuitable for temperature-stable applications. Study tip: Remember that mass-based concentration units (molality, mass percent) are temperature-independent, while volume-based units (molarity, normality, volume percent) will vary with temperature due to thermal expansion and contraction. When you see temperature stability mentioned, immediately think "mass-based units."

Question 6

An ideal solution is best defined as a solution that:

  1. has a total volume not equal to component volumes.
  2. contains a solute that completely dissociates into ions.
  3. exhibits a significant negative enthalpy of mixing.
  4. obeys Raoult's Law across all concentrations. (correct answer)
Explanation: When you encounter questions about ideal solutions, focus on the fundamental thermodynamic properties that define ideality. An ideal solution represents a theoretical standard where molecular interactions between different components are essentially identical to those within pure components. The correct answer is D because Raoult's Law is the defining characteristic of an ideal solution. This law states that the partial vapor pressure of each component equals its mole fraction multiplied by its pure component vapor pressure: Pi=xiPi∗P_i = x_i P_i^*. An ideal solution obeys this relationship at all concentrations, from pure solvent to pure solute. Let's examine why the other options are incorrect. Choice A describes non-ideal behavior - ideal solutions actually exhibit zero volume change upon mixing because molecular sizes and interactions remain unchanged. Choice B describes strong electrolytes that dissociate completely, but this has nothing to do with solution ideality; in fact, ionic solutions typically deviate significantly from ideal behavior due to strong electrostatic interactions. Choice C refers to exothermic mixing processes, but ideal solutions have zero enthalpy of mixing (ΔHmix=0\Delta H_{mix} = 0) because intermolecular forces don't change when components mix. Remember that ideal solutions are defined by three key properties: they obey Raoult's Law, have zero enthalpy of mixing, and show zero volume change upon mixing. On the DAT, when you see "ideal solution," immediately think Raoult's Law - it's the most fundamental and testable concept that distinguishes ideal from real solution behavior.

Question 7

When preparing pasta, a cook adds salt to the boiling water. From a chemical perspective, the primary effect of adding salt is:

  1. a significant decrease in the cooking time required.
  2. an increase in the vapor pressure of the water.
  3. a decrease in the specific heat of the water.
  4. an increase in the boiling point of the water. (correct answer)
Explanation: When you encounter questions about adding substances to water, think about colligative properties—characteristics that depend on the number of dissolved particles, not their identity. Salt dissolving in water creates a classic example of these properties in action. Adding salt (sodium chloride) to boiling water demonstrates boiling point elevation. When salt dissolves, it dissociates into Na⁺ and Cl⁻ ions, which disrupt the water molecules' ability to escape into the vapor phase. More energy is now required to overcome these intermolecular interactions, so the water must reach a higher temperature before it can boil. This is why option D is correct—the boiling point increases. Let's examine why the other choices miss the mark. Option A suggests significantly decreased cooking time, but the boiling point elevation from typical cooking amounts of salt (1-2 tablespoons per quart) only raises the temperature by 1-2°F—barely noticeable in cooking time. Option B incorrectly states that vapor pressure increases; actually, dissolved particles decrease vapor pressure at any given temperature, which is precisely why the boiling point rises. Option C claims the specific heat decreases, but adding salt actually slightly increases the solution's specific heat capacity compared to pure water. Remember this pattern for the DAT: when you see questions about dissolving substances in water, immediately think colligative properties. The four key effects are freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure. Salt in cooking water is a textbook case of boiling point elevation.

Question 8

Osmosis is the net movement of solvent molecules across a semipermeable membrane. This movement occurs from a region of:

  1. higher solute concentration to a region of lower solute concentration.
  2. lower temperature to a region of higher temperature.
  3. lower osmotic pressure to a region of higher osmotic pressure.
  4. higher solvent concentration to a region of lower solvent concentration. (correct answer)
Explanation: When you encounter osmosis questions, focus on the fundamental driving force: water moves to equalize concentrations across a membrane that blocks solute movement. Osmosis occurs because water molecules naturally move from areas where they are more concentrated (fewer dissolved particles) to areas where they are less concentrated (more dissolved particles). Think of it this way: if one side of a membrane has pure water and the other has sugar water, the pure water side has a higher concentration of water molecules. Water will flow toward the sugar water side to dilute it. This makes choice D correct—solvent (water) moves from higher solvent concentration to lower solvent concentration. Choice A represents a common misconception. While it's true that osmosis equalizes solute concentrations, the actual movement is of water molecules, not solute particles. The semipermeable membrane prevents solute movement. Choice B incorrectly brings temperature into the equation. While temperature affects the rate of molecular movement, osmosis is driven by concentration differences, not temperature gradients. Choice C confuses cause and effect. Water moves from lower osmotic pressure to higher osmotic pressure, but this describes the pressure relationship, not the concentration gradient that drives the movement. Osmotic pressure is the result of solute concentration differences. Remember this key relationship: higher solute concentration means lower solvent concentration in that region. Water always moves toward where there's less water (more solute). This inverse relationship between solute and solvent concentrations is crucial for mastering osmosis problems on the DAT.

Question 9

The process of reverse osmosis, used for water desalination, involves applying an external pressure to a saltwater solution. For fresh water to be produced, this applied pressure must be:

  1. greater than the osmotic pressure of saltwater. (correct answer)
  2. less than the osmotic pressure of saltwater.
  3. exactly equal to the osmotic pressure of saltwater.
  4. equal to the atmospheric pressure on the solution.
Explanation: When you encounter reverse osmosis questions, remember that this process works by forcing water molecules to move against their natural tendency. In normal osmosis, water moves from areas of low solute concentration to high solute concentration, creating osmotic pressure that resists further water movement. To understand why the applied pressure must be greater than the osmotic pressure of saltwater (A), think about what you're trying to accomplish. You want to push pure water molecules through a semipermeable membrane, leaving the salt behind. This means forcing water to move in the opposite direction of natural osmosis - from the salty side to the pure water side. To overcome the natural osmotic pressure and actually drive this reverse process, you need to apply external pressure that exceeds the osmotic pressure. Option B is incorrect because applying less pressure than the osmotic pressure would result in normal osmosis continuing - water would actually flow toward the salt solution, not away from it. Option C is wrong because equal pressures would create equilibrium with no net water movement in either direction, producing no fresh water. Option D misses the point entirely - atmospheric pressure has nothing to do with overcoming the specific osmotic pressure created by dissolved salts. For DAT questions on colligative properties and membrane processes, always ask yourself: "What natural tendency am I trying to overcome?" If you're working against a natural process like osmosis, you need to apply a force greater than the opposing force to drive the reaction in your desired direction.

Question 10

Under which condition is the molarity of an aqueous solution most likely to be approximately equal to its molality?

  1. When the solution is very dilute. (correct answer)
  2. When the solution is highly concentrated.
  3. When the solute has a very high molar mass.
  4. When the solution is at a high temperature.
Explanation: When you encounter questions comparing molarity and molality, focus on how the volume of water changes with concentration and temperature. Molarity (M) measures moles of solute per liter of solution, while molality (m) measures moles of solute per kilogram of solvent. These values converge when 1 liter of solution contains approximately 1 kilogram of water, which occurs when the solution's density is close to 1 g/mL (pure water's density). In very dilute aqueous solutions (A), the small amount of solute has minimal effect on the solution's total volume and density. Since the solution is mostly water, 1 liter of solution contains very close to 1 kilogram of water, making molarity ≈ molality. This is the correct answer. Highly concentrated solutions (B) create the opposite effect. Large amounts of solute significantly alter the solution's volume and density, causing molarity and molality to diverge substantially. High molar mass solutes (C) actually worsen the approximation. Even small molar quantities of high-mass solutes can dramatically change solution properties, making the density deviate further from 1 g/mL. High temperatures (D) decrease water's density below 1 g/mL and can cause solution expansion, making 1 liter of solution contain even less than 1 kilogram of water. Remember this pattern: molarity equals molality when solution density equals water's density (1 g/mL). This happens most reliably in dilute aqueous solutions where solute effects are minimized. On the DAT, "dilute" conditions often create ideal approximations.

Question 11

The addition of a non-volatile solute to a pure solvent results in a solution with a lower vapor pressure. This is primarily because the solute particles:

  1. occupy surface area, reducing the rate of solvent evaporation. (correct answer)
  2. form strong chemical bonds with the solvent, removing the solvent from the liquid phase.
  3. increase the kinetic energy of the solvent molecules, causing them to escape more slowly.
  4. increase the overall density of the solution, preventing solvent molecules from escaping.
Explanation: This question tests your understanding of vapor pressure depression, one of the key colligative properties. When you encounter colligative property questions, focus on how the physical presence of solute particles affects the behavior of solvent molecules. Vapor pressure depression occurs because non-volatile solute particles physically occupy space at the liquid's surface where solvent molecules would normally evaporate. Think of it like a crowded dance floor - the more people (solute particles) taking up space, the harder it is for dancers (solvent molecules) to reach the exit (evaporate). The solute particles don't prevent evaporation entirely, but they reduce the surface area available for solvent molecules to escape, thereby lowering the rate of evaporation and reducing vapor pressure. Choice A correctly identifies this mechanism - solute particles occupy surface area and reduce the rate of solvent evaporation. Choice B is wrong because non-volatile solutes typically don't form strong chemical bonds with the solvent; if they did, this would be a chemical change rather than the physical process that defines colligative properties. Choice C incorrectly suggests that solute particles increase kinetic energy - they actually don't affect the kinetic energy of solvent molecules. Choice D misunderstands the mechanism entirely; while density may change, vapor pressure depression isn't about density preventing molecular escape. Remember that colligative properties depend only on the number of particles present, not their identity. When you see vapor pressure depression questions, visualize solute particles physically blocking evaporation sites at the surface - this mental model will guide you to the right answer.

Question 12

The solubility of a gas such as carbon dioxide in a liquid like water is greatest under which conditions?

  1. High pressure and high temperature
  2. High pressure and low temperature (correct answer)
  3. Low pressure and high temperature
  4. Low pressure and low temperature
Explanation: When you encounter gas solubility questions, think about Henry's Law and Le Chatelier's principle - two fundamental concepts that govern how gases dissolve in liquids. Gas solubility increases with higher pressure because more gas molecules are forced into contact with the liquid surface, driving more dissolution. This is why carbonated beverages are bottled under pressure. Conversely, lower temperatures favor gas solubility because dissolution is typically an exothermic process - removing heat shifts the equilibrium toward the dissolved state. Answer B correctly combines high pressure (forcing more gas into solution) with low temperature (favoring the dissolution process). Think of cold soda staying fizzy longer than warm soda. Answer A is wrong because high temperature works against solubility - heat provides energy for gas molecules to escape the liquid phase, which is why hot soda goes flat quickly. Answer C combines the worst conditions: low pressure reduces the driving force for dissolution while high temperature promotes gas escape. Answer D gets temperature right but fails on pressure - low pressure means fewer gas molecules available to dissolve, like opening a soda bottle where CO₂ immediately escapes due to reduced pressure. For DAT success, remember the simple rule: "Cold and squeezed" gases dissolve best. High pressure pushes gas in, low temperature keeps it there. This principle applies to everything from respiratory gas exchange (where cool conditions in lungs favor oxygen uptake) to industrial processes. When you see gas solubility questions, immediately consider both pressure and temperature effects independently, then combine them.

Question 13

Ethylene glycol (C₂H₆O₂) is commonly used as an automotive antifreeze. Its effectiveness is primarily due to its ability to:

  1. react chemically with water to form new compounds.
  2. significantly lower the freezing point of water. (correct answer)
  3. form strong hydrogen bonds releasing thermal energy.
  4. increase the specific heat capacity of the liquid.
Explanation: When you encounter questions about antifreeze, think about colligative properties—properties that depend on the number of dissolved particles rather than their chemical identity. Ethylene glycol works as antifreeze because it demonstrates freezing point depression, a key colligative property. Ethylene glycol significantly lowers water's freezing point by disrupting the orderly crystal structure that ice requires. When dissolved in water, ethylene glycol molecules interfere with water molecules' ability to form the regular hydrogen-bonded network needed for ice formation. This means the solution must be cooled to a much lower temperature before it can freeze—exactly what you want in automotive coolant. Looking at the incorrect options: Choice A is wrong because ethylene glycol doesn't chemically react with water to form new compounds; it simply dissolves and forms a solution through intermolecular forces. Choice C misses the mark because while ethylene glycol does form hydrogen bonds with water, this isn't the primary mechanism for antifreeze effectiveness—it's the disruption of ice crystal formation that matters. Choice D is incorrect because changing specific heat capacity isn't the main antifreeze mechanism, and ethylene glycol doesn't dramatically increase water's specific heat anyway. The correct answer is B—ethylene glycol significantly lowers the freezing point of water through the colligative property of freezing point depression. Study tip: Remember that antifreeze questions on the DAT typically test colligative properties. When you see "antifreeze effectiveness," immediately think "freezing point depression" rather than chemical reactions or thermal properties.

Question 14

Of the following aqueous solutions, which will have the highest vapor pressure at a given temperature?

  1. 0.2 m C₁₂H₂₂O₁₁ (sucrose)
  2. 0.1 m NaCl
  3. 0.1 m CaCl₂
  4. 0.1 m C₁₂H₂₂O₁₁ (sucrose) (correct answer)
Explanation: When you encounter vapor pressure questions involving different solutions, you're dealing with Raoult's Law and colligative properties. The key insight is that vapor pressure decreases when solute particles are added to a solvent, and this decrease depends on the total number of particles in solution. To find the highest vapor pressure, you need to identify which solution has the fewest total particles. Let's count the particles each solution produces: Choice D (0.1 m sucrose) produces the fewest particles. Sucrose is a molecular compound that doesn't ionize, so 0.1 m sucrose creates only 0.1 m total particles. This means the smallest decrease in vapor pressure, resulting in the highest vapor pressure. Choice A (0.2 m sucrose) also doesn't ionize, but creates 0.2 m total particles—twice as many as choice D, causing greater vapor pressure depression. Choice B (0.1 m NaCl) ionizes completely into Na⁺ and Cl⁻ ions, producing 0.1×2=0.20.1 \times 2 = 0.2 m total particles, the same as choice A. Choice C (0.1 m CaCl₂) ionizes into Ca²⁺ and two Cl⁻ ions, producing 0.1×3=0.30.1 \times 3 = 0.3 m total particles—the most of all options, causing the greatest vapor pressure depression. Remember: more dissolved particles mean lower vapor pressure. When comparing solutions, always calculate the total molality of particles by multiplying the initial molality by the number of ions the compound produces. Molecular compounds like sucrose don't ionize, while ionic compounds do.

Question 15

A 0.1 m aqueous solution of NaCl and a 0.1 m aqueous solution of glucose (C₆H₁₂O₆) are prepared. Compared to the glucose solution, the NaCl solution will have:

  1. a higher freezing point and a lower boiling point.
  2. a lower freezing point and a higher boiling point. (correct answer)
  3. a lower freezing point and a lower boiling point.
  4. the same freezing point and the same boiling point.
Explanation: When you encounter a question comparing solutions with different solutes, think about colligative properties—physical properties that depend on the number of particles in solution, not their identity. Both solutions have the same molality (0.1 m), but they behave very differently. Glucose is a molecular compound that dissolves without breaking apart, so 0.1 m glucose produces 0.1 m total particles. NaCl, however, is an ionic compound that dissociates completely: NaCl→Na++Cl−\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-. This means 0.1 m NaCl produces 0.2 m total particles—twice as many as glucose. More particles create stronger colligative effects. For freezing point depression, ΔTf=Kf×m×i\Delta T_f = K_f \times m \times i, where i is the number of particles formed. The NaCl solution has twice the effective particle concentration, so it will freeze at a lower temperature than the glucose solution. Similarly, for boiling point elevation, ΔTb=Kb×m×i\Delta T_b = K_b \times m \times i, the NaCl solution will boil at a higher temperature due to its greater particle count. Answer A incorrectly suggests NaCl has a higher freezing point—this ignores the dissociation effect. Answer C wrongly claims NaCl has a lower boiling point, when more particles actually elevate the boiling point. Answer D assumes both solutions behave identically, failing to account for NaCl's ionic dissociation versus glucose's molecular nature. Remember: ionic compounds typically dissociate in water, multiplying their colligative effects compared to molecular compounds. Always count the total particles, not just the initial compound concentration.

Question 16

A solution of 0.1 M MgCl₂ is compared to a solution of 0.1 M NaCl. The MgCl₂ solution will exhibit:

  1. a higher vapor pressure than the NaCl solution.
  2. a greater boiling point elevation than the NaCl solution. (correct answer)
  3. a higher freezing point than the NaCl solution.
  4. the same osmotic pressure as the NaCl solution.
Explanation: When you encounter questions comparing solutions with different solutes, focus on colligative properties—properties that depend only on the number of dissolved particles, not their identity. The key insight is understanding how many particles each compound produces when it dissolves. MgCl₂ dissociates into three ions when it dissolves: one Mg²⁺ and two Cl⁻ ions. So a 0.1 M MgCl₂ solution actually contains 0.3 M total particles. NaCl dissociates into only two ions: one Na⁺ and one Cl⁻, giving a 0.1 M NaCl solution 0.2 M total particles. Since MgCl₂ produces more dissolved particles, it will have a greater effect on all colligative properties. Boiling point elevation is directly proportional to the number of dissolved particles. With 50% more particles than the NaCl solution, the MgCl₂ solution will exhibit greater boiling point elevation, making choice B correct. Choice A is wrong because vapor pressure decreases as the number of dissolved particles increases—MgCl₂ will have lower vapor pressure than NaCl. Choice C is incorrect because more dissolved particles lower the freezing point more dramatically—MgCl₂ will have a lower freezing point than NaCl. Choice D is wrong because osmotic pressure is also proportional to particle concentration, so MgCl₂ will have higher osmotic pressure than NaCl. Remember this pattern: count the total ions produced when compounds dissociate. More ions means greater effects on boiling point elevation, freezing point depression, and osmotic pressure, but lower vapor pressure.

Question 17

When 34.2 g of an unknown non-electrolyte is dissolved in 500.0 g of water, the freezing point of the solution is -0.372 °C. What is the molar mass of the unknown substance? (K₟ for water = 1.86 °C/m)

  1. 171 g/mol
  2. 180 g/mol
  3. 342 g/mol (correct answer)
  4. 684 g/mol
Explanation: This question tests freezing point depression, a colligative property where dissolved solutes lower the freezing point of solvents. When you see freezing point changes with unknown substances, you're solving for molar mass using the relationship between molality and freezing point depression. Start with the freezing point depression equation: ΔTf=Kf×m\Delta T_f = K_f \times m, where ΔTf=0.372°C\Delta T_f = 0.372°C, Kf=1.86°C/mK_f = 1.86°C/m, and mm is molality. Solving for molality: m=0.3721.86=0.200 mol/kgm = \frac{0.372}{1.86} = 0.200 \text{ mol/kg} Since molality equals moles of solute per kilogram of solvent, and you have 0.500 kg of water: moles of unknown = 0.200×0.500=0.100 mol0.200 \times 0.500 = 0.100 \text{ mol} Finally, molar mass = massmoles=34.2 g0.100 mol=342 g/mol\frac{\text{mass}}{\text{moles}} = \frac{34.2 \text{ g}}{0.100 \text{ mol}} = 342 \text{ g/mol} Answer A (171 g/mol) results from incorrectly doubling the molality calculation or halving the final result. Answer B (180 g/mol) comes from using the wrong freezing point constant or making calculation errors in the molality step. Answer D (684 g/mol) occurs when you forget to account for the 0.500 kg of water instead of 1.00 kg, effectively doubling your final answer. For colligative property problems, always work systematically: calculate the property change, find molality, determine moles of solute, then calculate molar mass. Double-check that you're using kilograms for the solvent mass, not grams.

Question 18

What is the theoretical van 't Hoff factor (i) for a solution of ammonium phosphate, (NH₄)₃PO₄, assuming complete dissociation?

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 8
Explanation: The van 't Hoff factor (i) represents how many particles a compound produces when it dissolves in solution. To find this value, you need to identify all the ions formed during complete dissociation. When ammonium phosphate (NH4)3PO4(NH_4)_3PO_4 dissolves completely in water, it breaks apart according to this equation: (NH4)3PO4→3NH4++PO43−(NH_4)_3PO_4 \rightarrow 3NH_4^+ + PO_4^{3-} From one formula unit of ammonium phosphate, you get 3 ammonium ions and 1 phosphate ion, for a total of 4 particles. Therefore, the theoretical van 't Hoff factor is 4, making C correct. Let's examine why the other options are wrong. A) 2 would only account for two particles - perhaps if you incorrectly thought the compound split into just two large pieces rather than individual ions. B) 3 represents a common error where students count only the ammonium ions (3) and forget to include the phosphate ion in their total. D) 8 might result from incorrectly counting atoms instead of ions, or from some other miscalculation involving the subscripts in the formula. Remember this key strategy: for van 't Hoff factor problems, write out the complete dissociation equation and count every ion produced. Don't just look at subscripts in the formula - actually break apart the compound and tally all the separate particles. This systematic approach prevents the common mistake of undercounting or overcounting ions.

Question 19

If the addition of a small seed crystal to a clear solution causes rapid and extensive crystallization of the solute, the original solution must have been:

  1. supersaturated. (correct answer)
  2. unsaturated.
  3. saturated.
  4. an ideal solution.
Explanation: When you encounter questions about crystal formation and solubility, focus on understanding the three states of solution saturation and what triggers crystallization. A supersaturated solution contains more dissolved solute than it can theoretically hold at equilibrium under normal conditions. These solutions are inherently unstable and exist in a delicate balance. The key insight here is that supersaturated solutions are "waiting" to crystallize but need a nucleation site to begin the process. When you add a seed crystal, it provides the perfect surface for solute molecules to organize and rapidly form crystals, causing extensive crystallization as the excess solute precipitates out. Let's examine why the other options fail: Option B (unsaturated) is incorrect because an unsaturated solution has room for more solute to dissolve, so adding a seed crystal would simply cause it to dissolve rather than trigger crystallization. Option C (saturated) is wrong because a saturated solution is at equilibrium - adding a seed crystal might cause minor crystal growth, but not the rapid, extensive crystallization described. Option D (ideal solution) refers to a theoretical solution following Raoult's law perfectly, which has nothing to do with crystallization behavior. The correct answer is A) supersaturated, because only supersaturated solutions exhibit this dramatic crystallization response to seed crystals. Study tip: Remember the crystallization trigger test - if a small disturbance causes massive crystal formation, the solution was supersaturated. This is a classic sign of an unstable, supersaturated system returning to equilibrium.

Question 20

Four solutions are placed in separate containers, each separated from a reservoir of pure water by a semipermeable membrane. Which solution will generate the greatest osmotic pressure?

  1. 0.20 M AlCl₃ (correct answer)
  2. 0.25 M MgBr₂
  3. 0.30 M NaCl
  4. 0.50 M glucose
Explanation: Osmotic pressure questions test your understanding of colligative properties - properties that depend on the number of particles in solution, not their identity. When you see osmotic pressure problems, immediately think about how many particles each solute produces when dissolved. Osmotic pressure follows the equation π=iMRT\pi = iMRT, where i is the van't Hoff factor (number of particles produced per formula unit), M is molarity, R is the gas constant, and T is temperature. Since R and T are constant here, you need to find which solution has the highest i×Mi \times M value. Let's calculate for each option: A) AlCl₃ dissociates into 4 ions (1 Al³⁺ + 3 Cl⁻), so i×M=4×0.20=0.80i \times M = 4 \times 0.20 = 0.80. B) MgBr₂ produces 3 ions (1 Mg²⁺ + 2 Br⁻), giving 3×0.25=0.753 \times 0.25 = 0.75. C) NaCl creates 2 ions (1 Na⁺ + 1 Cl⁻), yielding 2×0.30=0.602 \times 0.30 = 0.60. D) Glucose doesn't dissociate (it's molecular), so 1×0.50=0.501 \times 0.50 = 0.50. Choice A generates the highest osmotic pressure with an effective particle concentration of 0.80 M. Choice B is close but slightly lower at 0.75 M. Choice C seems tempting with higher molarity, but produces fewer total particles. Choice D tricks students who forget that molecular compounds don't dissociate. Remember: for osmotic pressure problems, multiply molarity by the number of ions produced. Don't be fooled by the highest molarity alone - ionic compounds amplify their effect through dissociation.