Topic 3.7: Solutions and Mixtures Quiz

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Master AP Chemistry Topic 3.7 with this free practice quiz. Test your skills on molarity, solute vs solvent, dilutions, and volumetric flasks to prepare for your AP exam.

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Solutions and Mixtures Quiz

Master solution concentrations with these 25 MCQs on AP Chemistry Topic 3.7: Solutions and Mixtures. This quiz evaluates your ability to calculate molarity and analyze the macroscopic properties and composition of various liquid mixtures.

Covered Subtopics: Molarity calculations, Macroscopic properties of solutions, Particulate models of mixtures.

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1. How many total moles of ions are present in 200 mL of a 0.10 M solution of MgCl₂?

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2. Which macroscopic observation best indicates that a liquid mixture is a true homogeneous solution rather than a suspension?

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3. Brass is a uniform alloy composed of copper and zinc. In precise chemical terms, brass is best classified as a:

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4. What volume of a 12.0 M stock HCl solution is required to prepare exactly 2.0 L of a 0.60 M HCl solution?

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5. A 0.20 M aqueous solution of Al(NO₃)₃ is prepared. What is the concentration of nitrate (NO₃⁻) ions in the solution?

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6. If a student accidentally adds water past the calibration mark of a volumetric flask while preparing a standard solution, what is the result?

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7. When solid NaCl dissolves in liquid water, how do the water molecules orient themselves around the Na⁺ ions?

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8. Which of the following 1.0 M aqueous solutions will have the highest electrical conductivity?

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9. To convert the mass percent of a solute in an aqueous solution to molarity, which additional physical property must be known?

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10. Which of the following is a key defining characteristic of all true chemical mixtures?

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11. What is the molarity of a solution containing 0.50 moles of NaCl dissolved in enough water to make 250 mL of solution?

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12. A particulate drawing of a mixture shows solute particles evenly distributed and completely surrounded by solvent particles. This represents a:

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13. Why is the molarity of a solution temperature-dependent, while its mass percent is not?

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14. What is the proper laboratory technique for preparing 1.0 L of a 1.0 M NaCl solution from solid NaCl?

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15. If 100 mL of 0.20 M KCl is mixed with 100 mL of 0.40 M KCl, what is the final concentration of the resulting mixture?

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16. How many grams of solid NaOH (molar mass 40.0 g/mol) are required to prepare 500 mL of a 0.10 M solution?

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17. An aqueous solution of Cu(NO₃)₂ appears distinctly blue. If the solution is diluted with pure water, what happens to the macroscopic color?

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18. Solution A is 1.0 M NaCl. Solution B is 2.0 M NaCl. If 50 mL of A and 50 mL of B are mixed, what is the final concentration?

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19. A student adds pure water to an aqueous solution in a beaker. At the particulate level, what happens to the solute particles?

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20. If 50.0 mL of a 6.0 M HCl stock solution is diluted to a final volume of 300.0 mL, what is the new concentration?

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21. A mixture contains solid AgCl and aqueous NaNO₃. Which physical separation technique would best isolate the solid AgCl?

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22. A student extracts multiple 10 mL samples from different depths of a thoroughly mixed 1.0 L aqueous solution. What should they find regarding the molarity of these samples?

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23. In a solution prepared by vigorously mixing 20 mL of liquid ethanol and 80 mL of liquid water, which substance is defined as the solvent?

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24. In a 0.50 M aqueous solution of sucrose (C₁₂H₂₂O₁₁), what are the primary solute species present distributed throughout the water?

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25. Which of the following aqueous solutions contains the highest concentration of total dissolved particulate species?

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Introduction to Solutions and Mixtures

A mixture contains two or more substances physically combined without necessarily undergoing a chemical reaction. The substances retain their chemical identities, and their proportions can vary. Mixtures can be classified as homogeneous or heterogeneous, depending on whether their composition is uniform throughout.

A homogeneous mixture has the same composition and properties throughout a sample at the macroscopic level. A solution is a homogeneous mixture consisting of one or more solutes dissolved in a solvent.

For example, when sodium chloride dissolves in water, the resulting solution contains dissolved Na⁺ and Cl⁻ ions distributed throughout the liquid.

A heterogeneous mixture has a non-uniform composition, with distinguishable regions or phases. Sand mixed with water is an example because the sand remains a separate solid phase.

The solvent is the component that dissolves the solute and is usually present in the greater amount. The solute is the substance dissolved in the solvent. Solutions can involve different physical states, including gases dissolved in liquids, liquids mixed with liquids, and solids dissolved in liquids.

Molarity: Measuring Solution Concentration

Concentration describes the amount of solute present in a given amount of solution. AP Chemistry commonly uses molarity to express solution concentration.

Molarity (M) is the number of moles of solute dissolved per liter of total solution.

M = nsolute / Vsolution

where:

M = molarity in mol L⁻¹, or mol/L nsolute = amount of solute in moles Vsolution = total volume of solution in liters

The volume refers to the entire solution, not just the solvent. For example, a solution prepared by dissolving salt in water has a volume determined by the final mixture, which may differ from the initial volume of water.

A 1.0 M NaCl solution contains 1.0 mol of NaCl per liter of solution. A 0.50 M solution contains 0.50 mol of NaCl per liter. Therefore, at the same volume, the more concentrated solution contains more solute particles.

Molarity is temperature-dependent because the volume of a solution can change with temperature, even when the amount of solute remains constant. The number of moles, however, does not change simply because the solution is heated or cooled.

Calculating Moles, Volume, and Molarity

The molarity equation can be rearranged to calculate any of its three quantities.

nsolute = M × Vsolution

Vsolution = nsolute / M

Always convert the solution volume into liters before using these relationships. Since 1 L = 1000 mL, a volume of 250 mL equals 0.250 L.

Worked example: Calculating molarity

A student dissolves 0.200 mol of glucose in enough water to make 500.0 mL of solution. Calculate the molarity.

Step 1: Convert the volume to liters.

500.0 mL ÷ 1000 = 0.5000 L

Step 2: Apply the molarity equation.

M = nsolute / Vsolution

M = 0.200 mol / 0.5000 L

M = 0.400 M

The solution has a glucose concentration of 0.400 mol/L.

Worked example: Calculating moles of solute

A student needs 250.0 mL of a 0.150 M KNO₃ solution. Calculate the moles of KNO₃ required.

Step 1: Convert the volume.

250.0 mL = 0.2500 L

Step 2: Rearrange the equation.

nsolute = M × Vsolution

nsolute = 0.150 mol/L × 0.2500 L

nsolute = 0.0375 mol KNO₃

The student needs 0.0375 mol of KNO₃ to prepare the required solution.

If the question asks for the mass instead, use the molar mass of KNO₃, approximately 101.1 g/mol.

Mass = moles × molar mass

Mass = 0.0375 mol × 101.1 g/mol

Mass = 3.79 g KNO₃

The student would dissolve this mass in water and then add enough water to bring the total solution volume to 250.0 mL.

Calculating the Number of Solute Particles

Molarity can also be used to calculate the number of solute particles in a solution. This requires connecting molar quantities to Avogadro’s constant.

Avogadro’s constant = 6.022 × 10²³ particles/mol

The calculation has two stages:

  1. Calculate the moles of solute using nsolute = M × Vsolution.
  2. Multiply the number of moles by Avogadro’s constant.

Number of particles = moles × 6.022 × 10²³ particles/mol

The meaning of “particles” depends on the substance. For a molecular compound such as glucose, the particles are molecules. For an ionic compound such as NaCl, the dissolved solute is represented by ions rather than intact NaCl molecules.

Worked example: Calculating dissolved particles

Calculate the number of glucose molecules in 0.200 L of a 0.300 M glucose solution.

Step 1: Calculate moles.

n = M × V

n = 0.300 mol/L × 0.200 L

n = 0.0600 mol

Step 2: Convert moles to molecules.

Number of molecules = 0.0600 × 6.022 × 10²³

Number of molecules = 3.61 × 10²² glucose molecules

Therefore, the solution contains approximately 3.61 × 10²² glucose molecules.

Counting ions in ionic solutions

When an ionic compound dissolves, it may dissociate into its constituent ions. For soluble sodium chloride, the process is represented by:

NaCl(s) → Na⁺(aq) + Cl⁻(aq)

Each formula unit of NaCl produces one Na⁺ ion and one Cl⁻ ion. If a solution contains 0.100 mol of dissolved NaCl, it contains 0.100 mol of Na⁺ ions and 0.100 mol of Cl⁻ ions, assuming complete dissociation.

The total amount of dissolved ions is therefore 0.200 mol of ions.

For calcium chloride, the balanced dissolution equation is:

CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq)

Each formula unit produces one calcium ion and two chloride ions. Consequently, dissolving 0.100 mol of CaCl₂ produces 0.100 mol of Ca²⁺ ions and 0.200 mol of Cl⁻ ions, assuming complete dissociation.

The total amount is 0.300 mol of ions. This illustrates why students must distinguish between the number of solute formula units and the total number of dissolved particles.

Dilution: Changing Concentration Without Changing Solute Amount

A solution can be diluted by adding more solvent. Its volume increases, and its molarity decreases, but the amount of solute remains constant if none is lost or removed.

Because the initial and final amounts of solute are equal:

M₁V₁ = M₂V₂

Here, M₁ and V₁ represent the initial molarity and volume, while M₂ and V₂ represent the final molarity and volume. Both volumes must use the same units.

For example, 100.0 mL of a 2.00 M solution is diluted to a final volume of 500.0 mL.

M₂ = M₁V₁ / V₂

M₂ = (2.00 × 100.0) / 500.0

M₂ = 0.400 M

The final solution has a molarity of 0.400 M. The moles of solute remain unchanged; only the volume and concentration change.

Applying Solutions and Mixtures in AP Chemistry

Solution problems frequently connect with stoichiometry, chemical reactions, and particle diagrams. Molarity allows students to convert the volume of a solution into moles, which can then be used in balanced chemical equations to determine reactant amounts or product yields.

For example, if 25.0 mL of 0.200 M AgNO₃ reacts with excess NaCl, the amount of AgNO₃ is:

n = 0.200 mol/L × 0.0250 L

n = 0.00500 mol

The precipitation reaction is:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

The mole ratio between AgNO₃ and AgCl is 1:1. Therefore, 0.00500 mol of AgCl can form if NaCl is present in excess and the reaction proceeds completely.

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