# Solutions and Mixtures

> AP Chemistry · Unit 3: Intermolecular Forces and Properties
> Source: https://www.owlsprep.com/study/ap-chemistry-u3-solutions-and-mixtures/

This aligned study guide covers core concepts of mixtures and solutions for AP Chemistry, including concentration units, interconversions, energetics of dissolution, factors affecting solubility, and Henry's Law.

**Prerequisites:** [Types of intermolecular forces](https://www.owlsprep.com/study/ap-chemistry-u3-intermolecular-forces/); [Basic mole and molar mass calculations](https://www.owlsprep.com/study/ap-chemistry-u1-mole-concept/); [Enthalpy change sign conventions](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-change/)

## Learning objectives

- Distinguish between homogeneous and heterogeneous mixtures
- Calculate and interconvert between common concentration units
- Explain the three-step energetics of solution formation
- Predict how pressure and temperature affect solute solubility
- Apply Henry's Law to calculate solubility of gaseous solutes

## Classification of Mixtures and Solutions

A mixture is a physical combination of two or more pure substances, where no covalent or ionic chemical bonds form between components, though new intermolecular interactions do develop. Mixtures are split into two broad categories for AP Chemistry: heterogeneous mixtures (non-uniform composition with distinct visible phases) and homogeneous mixtures (uniform composition at the molecular level), which are called solutions.

Unlike pure substances, solutions can have variable composition, making concentration calculations a core recurring skill for this topic. This subtopic contributes 3-5% of total AP Chemistry exam score, as part of Unit 3 which makes up 18-22% of the overall exam, appearing in both multiple-choice and free-response sections.

**Solution** — A homogeneous mixture of two or more substances, with uniform composition throughout at the molecular level

*Example:* Salt water, clean air, sucrose dissolved in water

## Concentration Units and Interconversions

Concentration describes the relative amount of solute (the minor component of a solution) dissolved in solvent (the major component). AP Chemistry requires mastery of five common concentration units:

1. **Mass percent**: Ratio of solute mass to total solution mass, multiplied by 100
2. **Mole fraction**: Ratio of moles of a component to total moles of all solution components
3. **Molarity ($M$)**: Moles of solute per liter of *total solution*
4. **Molality ($m$)**: Moles of solute per kilogram of *pure solvent*
5. **ppm/ppb**: Used for very dilute solutions of trace solutes

$$\% \text{ mass} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100\% = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 100\%$$

$$\chi_A = \frac{n_A}{n_{\text{total}}}$$

The sum of all mole fractions in any solution equals 1, and mole fraction is temperature independent.

$$M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$$

Molarity is temperature dependent because liquid volume expands and contracts with temperature changes.

$$m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$$

Molality is temperature independent, so it is the preferred unit for colligative property calculations where temperature changes occur.

$$\text{ppm} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^6 \\ \text{ppb} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^9$$

**Worked example:** A solution is prepared by dissolving 12.0 g of urea ($CH_4N_2O$, molar mass = 60.0 g/mol) in 180. g of pure water. The final volume of the solution is 188 mL. Calculate (a) molality and (b) molarity of urea.

1. First calculate moles of urea:

   $$n_{\text{urea}} = \frac{12.0\ \text{g}}{60.0\ \text{g/mol}} = 0.200\ \text{mol}$$
2. For (a) molality: Convert solvent mass to kg and apply the formula:

   $$180. \text{g} = 0.180\ \text{kg} \\ m = \frac{0.200\ \text{mol}}{0.180\ \text{kg}} = 1.11\ m$$
3. For (b) molarity: Convert total solution volume to liters and apply the formula:

   $$188\ \text{mL} = 0.188\ \text{L} \\ M = \frac{0.200\ \text{mol}}{0.188\ \text{L}} = 1.06\ M$$

**Check your understanding**

Test your understanding of concentration interconversion:

1. A solution of methanol ($CH_3OH$, molar mass 32 g/mol) in water has a mole fraction of methanol equal to 0.10. What is the molality of methanol in this solution?

   - 0.10 m
   - 6.2 m
   - 1.0 m
   - 5.6 m

   *Answer:* 6.2 m

   *Why:* If $\chi_{methanol} = 0.10$, $\chi_{water} = 0.90$. Assuming 1 total mole of solution, you have 0.10 mol methanol and 0.90 mol water (16.2 g = 0.0162 kg solvent). Molality = 0.10 mol / 0.0162 kg ≈ 6.2 m.

> **Exam tip:** Always explicitly label whether you are using solvent mass or total solution mass when starting calculations to avoid common errors

*Calculator:* allowed

## Energetics of Solution Formation

Solution formation occurs in three distinct steps, each with a characteristic enthalpy change, based on breaking and forming intermolecular attractions:

1. Step 1: Separate solute particles from each other, overcoming solute-solute intermolecular attractions: $\Delta H_1 > 0$ (always endothermic, energy is required to separate attracted particles)
2. Step 2: Separate solvent particles from each other to make space for solute, overcoming solvent-solvent intermolecular attractions: $\Delta H_2 > 0$ (also always endothermic)
3. Step 3: Mix solute and solvent particles, forming new solute-solvent intermolecular attractions: $\Delta H_3 < 0$ (always exothermic, energy is released when new attractions form)

$$\Delta H_{\text{soln}} = \Delta H_1 + \Delta H_2 + \Delta H_3$$

If the magnitude of the exothermic $\Delta H_3$ is larger than the sum of $\Delta H_1 + \Delta H_2$, $\Delta H_{\text{soln}}$ is negative (exothermic, releases heat). If $\Delta H_1 + \Delta H_2 > |\Delta H_3|$, $\Delta H_{\text{soln}}$ is positive (endothermic, absorbs heat). The 'like dissolves like' rule comes directly from this relationship.

**Worked example:** Predict the sign of $\Delta H_{\text{soln}}$ for octane (nonpolar $C_8H_{18}$) dissolved in water, and justify your answer.

1. Step 1 (separate octane molecules): $\Delta H_1$ is small positive, because only weak London dispersion forces between octane molecules need to be broken.
2. Step 2 (separate water molecules): $\Delta H_2$ is large positive, because strong hydrogen bonds between water molecules must be broken to make space for octane.
3. Step 3 (mix octane and water): $\Delta H_3$ is small negative, because only weak London dispersion forces form between octane and water.
4. Sum the enthalpy changes for the total enthalpy of solution:

   $$\Delta H_{\text{soln}} = (\text{small }+) + (\text{large }+) + (\text{small }-) = \text{large positive}$$
5. Conclusion: $\Delta H_{\text{soln}} > 0$ (endothermic), which is why octane does not dissolve in water.

> **warning**
>
> Never confuse breaking intermolecular attractions with breaking chemical bonds — both are endothermic. Students often incorrectly mark Step 1 or Step 2 as exothermic, which is a common MCQ trap.

> **Exam tip:** Breaking any attraction (intermolecular or covalent/ionic) is always endothermic

## Factors Affecting Solubility and Henry's Law

Solubility is the maximum concentration of a solute that can dissolve in a solvent at a given temperature and pressure. Key factors affecting solubility are pressure, temperature, and intermolecular interactions:

- **Pressure**: Only affects solubility of gaseous solutes in liquid solvents. Higher partial pressure of the gas above the solution increases solubility.
- **Temperature**: For solid solutes in liquids, solubility usually increases with increasing temperature (most dissolution is endothermic, per Le Chatelier's principle). For gaseous solutes in liquids, solubility *always* decreases with increasing temperature.
- **Intermolecular interactions**: Solubility is highest when solute-solvent intermolecular forces are similar in strength to solute-solute and solvent-solvent forces (the 'like dissolves like' rule).

$$C = kP$$

This is Henry's Law, where $C$ is the molar solubility of the gas, $k$ is the Henry's Law constant (specific to the gas-solvent pair and temperature), and $P$ is the partial pressure of the gas above the solution.

**Worked example:** The Henry's Law constant for oxygen gas in water at 20°C is $1.4 \times 10^{-3}\ \text{M atm}^{-1}$. Partial pressure of oxygen in the atmosphere is 0.21 atm. Calculate the solubility of oxygen in water at 20°C, and explain how this value changes if the water is heated to 80°C.

1. Calculate solubility using Henry's Law:

   $$C = kP = (1.4 \times 10^{-3}\ \text{M atm}^{-1})(0.21\ \text{atm}) = 2.9 \times 10^{-4}\ \text{M}$$
2. When heated to 80°C, increased temperature raises the average kinetic energy of oxygen molecules, allowing more molecules to overcome weak intermolecular attractions to water and escape the solution.
3. Conclusion: Solubility of oxygen will decrease at 80°C, to a value lower than $2.9 \times 10^{-4}\ \text{M}$.

**Worked example:** The legal limit for lead (Pb) in drinking water is 15 ppb Pb by mass. Tap water has a density of 1.00 kg/L. A child drinking 1.5 L of water per day consumes water at the legal limit. Calculate the mass of lead (in μg) consumed per day.

1. 15 ppb Pb means 15 g Pb per $10^9$ g solution. Since 1 kg = $10^9$ μg, this equals 15 μg Pb per 1 kg solution.
2. Calculate total mass of 1.5 L of water:

   $$1.00\ \text{kg/L} \times 1.5\ \text{L} = 1.5\ \text{kg}$$
3. Calculate total mass of lead consumed:

   $$15\ \text{μg Pb/kg solution} \times 1.5\ \text{kg solution} = 22.5\ \text{μg} ≈ 23\ \text{μg}$$

> **Exam tip:** FRQ require mechanistic explanations for solubility trends, not just rules

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using total mass of solution instead of mass of solvent when calculating molality.
  - Why it fails: Students confuse molality (solvent mass) with molarity (total solution volume) and mass percent (total solution mass), so they plug in the wrong value by default.
  - Correct: Write the formula for concentration explicitly before plugging in values, and highlight that molality requires mass of solvent, not solution.
- **Wrong:** Claiming Step 1 or Step 2 of solution formation is exothermic.
  - Why it fails: Students misremember that 'breaking bonds is exothermic', and incorrectly extend this to breaking intermolecular attractions between solute or solvent particles.
  - Correct: Memorize that separating any attracted particles requires energy, so Steps 1 and 2 are always endothermic, only the mixing Step 3 is exothermic.
- **Wrong:** Applying Henry's Law to solid or liquid solutes.
  - Why it fails: Students associate Henry's Law with solubility, so they use it for any solute when pressure is mentioned.
  - Correct: Only use Henry's Law when the problem refers to a gaseous solute and its partial pressure above the solution.
- **Wrong:** Assuming all solid solutes have increasing solubility with increasing temperature.
  - Why it fails: General rules lead students to assume this is always true, ignoring exceptions where dissolution is exothermic.
  - Correct: Always link temperature effect to the sign of $\Delta H_{\text{soln}}$: endothermic dissolution → higher T = higher solubility; exothermic dissolution → higher T = lower solubility.
- **Wrong:** Calculating ppm as mass of solute per mass of solvent, not per mass of solution.
  - Why it fails: For very dilute aqueous solutions, solvent mass ≈ solution mass, so students get away with this for dilute problems, and carry the error over to more concentrated solutions.
  - Correct: Use the same scaling as mass percent, just change the multiplier: ppm = (mass solute / mass solution) × 10^6.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Mass Percent | \% \text{ mass} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100\% | $m_{\text{solution}} = m_{\text{solute}} + m_{\text{solvent}}$ |
| Mole Fraction | \chi_A = \frac{n_A}{\sum n_i} | Sum of all $\chi$ = 1; temperature independent |
| Molarity | $M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$ | Temperature dependent; used for solution stoichiometry |
| Molality | $m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$ | Temperature independent; used for colligative properties |
| Parts per Million | $\text{ppm} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^6$ | 1 ppm ≈ 1 mg/L for dilute aqueous solutions |
| Parts per Billion | $\text{ppb} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^9$ | Used for trace contaminants |
| Enthalpy of Solution | $\Delta H_{\text{soln}} = \Delta H_1 + \Delta H_2 + \Delta H_3$ | $\Delta H_1, \Delta H_2 > 0$; $\Delta H_3 < 0$ |
| Henry's Law | $C = kP$ | Only applies to gaseous solutes; $k$ is temperature-dependent |

## What's next

This subtopic lays the foundational concentration calculation and solution chemistry skills you need for all subsequent solution-focused topics in AP Chemistry. Immediately next in Unit 3, you will study colligative properties, which relies entirely on the molality and mole fraction skills you mastered here. Without being able to reliably interconvert between concentration units, you cannot correctly calculate boiling point elevation, freezing point depression, or osmotic pressure, which are common high-weight FRQ topics on the AP exam. Beyond Unit 3, solutions and mixtures are core to acid-base equilibria, solubility product equilibria, reaction kinetics, and electrochemistry, all of which require consistent, accurate concentration calculations.

- [Intermolecular Forces](https://www.owlsprep.com/study/ap-chemistry-u3-intermolecular-forces/)
- [Solubility Equilibria](https://www.owlsprep.com/study/ap-chemistry-u7-solubility-equilibria/)
- [Ideal Gas Law](https://www.owlsprep.com/study/ap-chemistry-u3-ideal-gas-law/)

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