Solutions and mixtures — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: homogeneous/heterogeneous mixtures, solution terminology, mass percent, mole fraction, molarity, molality, ppm/ppb, concentration conversions, energetics of solution formation, factors affecting solubility, and Henry’s Law for AP Chemistry CED.

You should already know: types of intermolecular forces, basic mole and molar mass calculations, enthalpy change sign conventions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Solutions and mixtures?

A mixture is a physical combination of two or more pure substances, where no covalent/ionic chemical bonds are formed between components (though new intermolecular interactions do develop). Mixtures are divided into two broad categories: heterogeneous mixtures, which have non-uniform composition with distinct visible phases, and homogeneous mixtures, which have uniform composition throughout at the molecular level, which we call solutions. For AP Chemistry CED, this subtopic contributes approximately 3-5% of the total exam score, as part of Unit 3 (Intermolecular Forces and Properties), which accounts for 18-22% of the overall exam. Content from this topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with other solution-related concepts like colligative properties, acid-base equilibria, or solubility equilibria. Standard notation conventions for this topic are: $[X]$ for molar concentration of X, $M$ for molarity, $m$ for molality, ${\chi }_A$ for mole fraction of component A, and %(m/m) for mass percent. Unlike pure substances, solutions can have variable composition, which makes concentration calculations one of the core, recurring skills of this topic.

2. Concentration Units and Interconversions

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

1. Mass percent: Ratio of solute mass to total solution mass, multiplied by 100: $$\%\text{ mass}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times 100\%=\frac{m_{\text{solute}}}{m_{\text{solute}}+m_{\text{solvent}}}\times 100\%$$
2. Mole fraction: Ratio of moles of a component to total moles of all solution components: $${\chi }_A=\frac{n_A}{n_{\text{total}}}$$ The sum of all mole fractions in any solution equals 1, and mole fraction is temperature independent.
3. Molarity ($M$): Moles of solute per liter of total solution: $$M=\frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$$ Molarity is temperature dependent because liquid volume expands/contracts with temperature changes.
4. Molality ($m$): Moles of solute per kilogram of pure solvent: $$m=\frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$$ Molality is temperature independent, so it is the preferred unit for colligative property calculations where temperature changes occur.
5. Parts per million (ppm) / parts per billion (ppb): Used for very dilute solutions: $\text{ppm}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times 10^6$, $\text{ppb}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times 10^9$.

Worked Example

A solution is prepared by dissolving 12.0 g of urea ($C{H}_4{N}_{2}O$, 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: $180.\text{g}=0.180\text{kg}$. Then $m=\frac{0.200\text{mol}}{0.180\text{kg}}=1.11m$.
3. For (b) molarity: Convert total solution volume to L: $188\text{mL}=0.188\text{L}$. Then $M=\frac{0.200\text{mol}}{0.188\text{L}}=1.06M$.

Exam tip: Always explicitly label whether you are using solvent mass or total solution mass when starting a concentration calculation. Mixing these up is the most common error on this type of problem, and it is easily avoided with 10 extra seconds of labeling.

3. 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, for the same reason as Step 1).
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).

The total enthalpy of solution is: $$\Delta H_{\text{soln}}=\Delta H_1+\Delta H_2+\Delta H_3$$ 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: polar/ionic solutes dissolve in polar solvents because strong new solute-solvent attractions offset the energy required to separate solute and solvent particles, leading to a $\Delta H_{\text{soln}}$ small enough that entropy of mixing drives dissolution. Nonpolar solutes only dissolve in nonpolar solvents, because $\Delta H_1$ and $\Delta H_2$ are small (weak intermolecular forces), so $\Delta H_{\text{soln}}$ is close to zero, and entropy drives dissolution.

Worked Example

Predict the sign of $\Delta H_{\text{soln}}$ for octane (nonpolar $C_8{H}_{18}$) 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: $\Delta H_{\text{soln}}=(\text{small }+)+(\text{large }+)+(\text{small }-)=\text{large positive}$. $\Delta H_{\text{soln}}>0$ (endothermic), which is why octane does not dissolve in water.

Exam tip: 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.

4. Factors Affecting Solubility

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, described by Henry's Law: $$C=kP$$ 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.
• Temperature: For solid solutes in liquids, solubility usually increases with increasing temperature, because most solid dissolution is endothermic (Le Chatelier's principle: adding heat shifts equilibrium toward dissolved solute). Exceptions exist for exothermic dissolution, where solubility decreases with temperature. For gaseous solutes in liquids, solubility always decreases with increasing temperature: higher temperature gives gas molecules more kinetic energy to escape the solution by overcoming weak solute-solvent attractions.
• 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).

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. Use Henry's Law to calculate solubility: $C=kP=(1.4\times 10^{-3}{\text{M atm}}^{-1})(0.21\text{atm})=2.9\times 10^{-4}\text{M}$.
2. When water is heated to 80°C, temperature increases. For gaseous solutes, higher temperature increases the average kinetic energy of oxygen molecules, allowing more molecules to overcome the weak intermolecular attractions to water and escape the solution.
3. Solubility of oxygen will decrease at 80°C, to a value lower than $2.9\times 10^{-4}\text{M}$.

Exam tip: AP FRQ require a mechanistic explanation for temperature effects on gas solubility, not just a statement of the rule. Always link lower solubility at higher temperature to increased kinetic energy allowing gas escape to get full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using total mass of solution instead of mass of solvent when calculating molality. Why: 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 move: Write the formula for concentration explicitly before plugging in values, and highlight that molality requires mass of solvent, not solution.
• Wrong move: Claiming Step 1 or Step 2 of solution formation is exothermic. Why: Students misremember that "breaking bonds is exothermic", and incorrectly extend this to breaking intermolecular attractions between solute or solvent particles. Correct move: Memorize that separating any attracted particles requires energy, so Steps 1 and 2 are always endothermic, only the mixing Step 3 is exothermic.
• Wrong move: Applying Henry's Law to solid or liquid solutes. Why: Students associate Henry's Law with solubility, so they use it for any solute when pressure is mentioned. Correct move: Only use Henry's Law when the problem refers to a gaseous solute and its partial pressure above the solution.
• Wrong move: Assuming all solid solutes have increasing solubility with increasing temperature. Why: General rules lead students to assume this is always true, ignoring exceptions where dissolution is exothermic. Correct move: 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 move: Calculating ppm as mass of solute per mass of solvent, not per mass of solution. Why: 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 move: Use the same scaling as mass percent, just change the multiplier: ppm = (mass solute / mass solution) × 10^6.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

A solution of methanol ($C{H}_{3}OH$, 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? A) 0.10 m B) 6.2 m C) 1.0 m D) 5.6 m

Worked Solution: If the mole fraction of methanol is 0.10, the mole fraction of water is $1-0.10=0.90$. Assume 1 total mole of solution, so 0.10 mol methanol and 0.90 mol water. Mass of water (solvent) = $0.90\text{mol}\times 18.0\text{g/mol}=16.2\text{g}=0.0162\text{kg}$. Molality = $\frac{\text{moles methanol}}{\text{kg solvent}}=\frac{0.10\text{mol}}{0.0162\text{kg}}\approx 6.2\text{m}$. Common wrong answers come from confusing mole fraction with molality (A) or using total mass instead of solvent mass (C). The correct answer is B.

Question 2 (Free Response)

A 1.50 L aqueous solution of potassium chloride (KCl, molar mass 74.5 g/mol) has a molarity of 0.500 M. The density of the solution is 1.03 g/mL at 25°C. (a) Calculate the mass percent of KCl in the solution. (b) Calculate the molality of KCl in the solution. (c) The enthalpy of solution for KCl is +17.2 kJ/mol. If KCl is dissolved in room temperature water in an open beaker, will the beaker feel cooler or warmer than the surrounding air? Justify your answer.

Worked Solution: (a) Moles of KCl = $M\times V=0.500\text{mol/L}\times 1.50\text{L}=0.750\text{mol}$. Mass of KCl = $0.750\text{mol}\times 74.5\text{g/mol}=55.875\text{g}$. Total mass of solution = density × volume = $1.03\text{g/mL}\times 1500\text{mL}=1545\text{g}$. Mass percent = $\left(\frac{55.875\text{g}}{1545\text{g}}\right)\times 100\%=3.62\%$. (b) Mass of water (solvent) = total mass - mass of KCl = $1545\text{g}-55.875\text{g}=1489.125\text{g}=1.489\text{kg}$. Molality = $\frac{0.750\text{mol}}{1.489\text{kg}}=0.504m$. (c) The beaker will feel cooler. $\Delta H_{\text{soln}}$ is positive, so dissolution is endothermic. The process absorbs heat from the surrounding water and the glass beaker, lowering their temperature below room temperature.

Question 3 (Application / Real-World Style)

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 by the child.

Worked Solution: 15 ppb Pb means 15 g Pb per $10^9$ g solution, which equals 15 μg Pb per 1 kg solution (since 1 kg = $10^6$ mg = $10^9$ μg). Total mass of 1.5 L water = $1.00\text{kg/L}\times 1.5\text{L}=1.5\text{kg}$. Mass of Pb = $15\text{μg Pb/kg solution}\times 1.5\text{kg solution}=22.5\text{μg}\approx 23\text{μg}$. This means the child would consume 23 micrograms of lead per day, which is at the maximum allowed level for safe drinking water.

7. Quick Reference Cheatsheet

8. What's Next

This chapter 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. Follow-on topics to study next: Colligative Properties Intermolecular Forces Solubility Equilibria Acid-Base Equilibria