A solution is a type of mixture that is homogeneous—its macroscopic properties are the same everywhere. It has a solvent (major component) and one or more solutes (minor components), and can be solid, liquid, or gas. A general mixture can be either homogeneous (a solution) or heterogeneous—in a heterogeneous mixture the composition and properties vary with location (particles or phases are visibly distinct). On the AP Chem CED terms: solutions = homogeneous mixture; heterogeneous mixture = nonuniform composition. AP problems often ask you to use concentration measures for solutions (molarity: M = n_solute / L_solution) or to count solute particles, volumes, or perform dilutions (LO 3.7.A). For quick review, see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and practice questions (https://library.fiveable.me/practice/ap-chemistry).

Molarity (M) = moles of solute ÷ liters of solution. Step-by-step: 1. Identify the solute and its mass or moles. If given grams, convert to moles using molar mass: n = mass (g) / Molar mass (g·mol⁻¹). 2. Make sure the solution volume is in liters. If given mL, divide by 1000: L = mL / 1000. 3. Apply the formula: M = n_solute / L_solution. Include units (mol·L⁻¹). Example: 5.00 g NaCl in 250.0 mL solution. - Molar mass NaCl = 58.44 g·mol⁻¹ → n = 5.00 / 58.44 = 0.0856 mol. - Volume = 250.0 mL = 0.2500 L. - M = 0.0856 mol / 0.2500 L = 0.342 M. On the AP exam you’ll be expected to show these algebraic steps, correct units, and significant figures (see CED Topic 3.7: M = n_solute / L_solution). For more worked examples and topic review check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

A homogeneous mixture has the same macroscopic properties and composition everywhere in the sample—i.e., it’s uniform. Solutions are the classic example: salt dissolved in water, sugar in tea, or air (a gas solution). The AP CED even calls solutions “homogeneous mixtures” and emphasizes you’ll treat composition with measures like molarity M = n_solute / L_solution (Topic 3.7). A heterogeneous mixture has properties or composition that vary with location—you can see or separate different parts. Examples: oil + water layers, sand in water, or a salad dressing with suspended droplets. Colloids and suspensions fall between: particles may stay dispersed (colloid) or settle out (suspension). On the exam you should be able to identify which mixtures are uniform and apply concentration calculations for homogeneous solutions. For a focused review see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Molarity (M) is moles of solute divided by liters of solution—not liters of solvent. So M = n(solute) / L(solution), which means the final total volume after you’ve dissolved the solute. That’s the formula listed in the CED for Topic 3.7 (3.7.A.2). If you instead divide by kilograms of solvent you’re calculating molality (m), which is useful when temperature changes matter because molality doesn’t depend on volume. For lab prep: measure or make up to the final solution volume (e.g., dissolve solute, then dilute to the mark in a volumetric flask) to get the correct molarity. For more practice and review on solutions and concentration units, see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and extra problems at (https://library.fiveable.me/practice/ap-chemistry).

Molarity (M = n_solute / L_solution) is the lab favorite because it links directly to the amounts of reactants you use in common experiments. You measure moles of solute (from mass) and then dilute to a known solution volume—that volume is easy to control with volumetric flasks and pipets, so making and diluting solutions (and doing titrations) is straightforward. Molarity plugs directly into stoichiometry and rate/equilibrium expressions (concentrations in mol·L⁻¹), which is exactly what AP problems ask you to calculate in Topic 3.7. Other units exist: molality (moles per kg solvent) is temperature-independent and better for some colligative-property work, and mole fraction is useful for gases—but they’re less convenient for routine lab prep and titration. Note: molarity depends on solution volume, so it can change slightly with temperature. For more practice and examples on preparing/diluting molar solutions see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and extra problems (https://library.fiveable.me/practice/ap-chemistry).

Yes—solutions can be gases, not just liquids. The CED says a solution (a homogeneous mixture) can be a solid, liquid, or gas (3.7.A.1). Air is the classic gaseous solution: N2 is the major “solvent” and O2, CO2, and other gases are solutes. Gas mixtures are homogeneous when composition is uniform (contrast that with a heterogeneous mixture that varies by location). For AP work, remember molarity (M = n_solute / L_solution) is most often used for liquid solutions, since it uses solution volume, but for gases you often use mole fraction or partial pressure (Dalton’s law) to describe composition. Topic 3.7 questions on the exam can ask you to calculate number of solute particles, volume, or concentration—practice those in the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and more problems at (https://library.fiveable.me/practice/ap-chemistry).

Solute = the substance that gets dissolved (what’s present in smaller amount); solvent = the medium that does the dissolving (usually the larger amount). In a saltwater solution, NaCl is the solute and H2O is the solvent. In the AP CED language, a solution is a homogeneous mixture whose macroscopic properties don’t vary throughout the sample (3.7.A.1). Solutions can be solid, liquid, or gas (so “solute vs. solvent” applies to all phases). For concentration calculations you’ll often use molarity: M = n_solute / L_solution (3.7.A.2)—note that n refers to moles of solute and L is total solution volume. Remember: “solvent” is typically the majority component and determines the phase; “solute” changes properties like boiling point and freezing point (colligative effects). For quick review, check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Quick rule: a solution (homogeneous mixture) has the same macroscopic properties everywhere in the sample; a heterogeneous mixture’s properties depend on location (CED 3.7.A.1). Visual clues you can use: - Looks uniform/transparent at the scale you’re viewing it (e.g., salt dissolved in water) → likely homogeneous. - You can see different phases, chunks, layers, suspended particles, or cloudiness that scatters light → heterogeneous (colloid or suspension). - Tyndall effect (light beam visible through the sample) indicates particles large enough to scatter light → not a true solution. - If particles settle, can be filtered, or separate on standing → heterogeneous. - If unsure, sample different spots: if composition/appearance changes with location it’s heterogeneous; if it’s identical, it’s homogeneous. For lab confirmation: filtration/centrifugation, microscope, or letting it sit are practical tests. For more on AP-style definitions and examples, see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl). Practice problems are at (https://library.fiveable.me/practice/ap-chemistry).

Molarity M = moles of solute / liters of solution. You have 2.00 mol NaCl in 500 mL = 0.500 L, so M = 2.00 mol / 0.500 L = 4.00 M. Give the answer as 4.00 M (moles per liter). On the AP exam you should always write the formula M = nsolute / Lsolution, include units, and give an appropriate number of significant figures based on the data (here 3 sig figs). For a quick review of Topic 3.7 (solutions, molarity, and related calculations) see the Fiveable study guide for Unit 3 (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl). For extra practice problems, check Fiveable’s AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

Think of a solution (a homogeneous mixture) as completely uniform at the scale you can see or measure without special separation. “Macroscopic properties don’t vary throughout the sample” means properties you can measure—concentration, color, density, refractive index, boiling point, etc.—are the same at every location in the container. That happens because the solute particles are evenly dispersed among solvent molecules; molecular motion (diffusion) and mixing produce a single phase with no separate regions of different composition. By contrast, a heterogeneous mixture has distinct regions (like oil + water) so macroscopic properties depend on where you measure. For AP Chem, use this idea when classifying mixtures and when calculating molarity or colligative effects: molarity (M = nsolute / Lsolution) describes a uniform concentration for the whole solution (CED 3.7.A.1–A.2). If you want a quick refresher or practice problems on solutions and mixtures, check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and the AP practice bank (https://library.fiveable.me/practice/ap-chemistry).

Heterogeneous mixtures are ones whose macroscopic properties change with location (unlike solutions). Real-world examples you should know for Topic 3.7: oil + water (liquid-liquid immiscible layers), sand or gravel in water (suspension), salad dressing with herbs and oil, soil (solid particles + organic matter), concrete (cement + aggregate), granite (different mineral crystals), blood (cells + plasma—a suspension/colloid), smoke or dust in air (aerosol), and emulsions like mayonnaise (a dispersed phase in a continuous phase). Note: colloids (milk, fog) are a special type of heterogeneous mixture where particles are small enough to stay dispersed but not truly molecularly mixed. These examples connect to CED language: in heterogeneous mixtures the macroscopic properties depend on location. For extra review and AP-style practice on mixtures and concentrations (molarity, solute/solvent ideas), check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and Unit 3 resources (https://library.fiveable.me/ap-chemistry/unit-3); practice questions are at (https://library.fiveable.me/practice/ap-chemistry).

Molarity (M) = moles solute per liter of solution. Molality (m) = moles solute per kilogram of solvent. To convert you need the solution’s density (or enough info to get mass of 1.00 L of solution) and the solute’s molar mass. Quick recipe (M → m): 1. Start with 1.00 L of solution. M tells you moles solute = M × 1.00 L. 2. Find mass of that solute: moles × molar mass (g). 3. Get mass of 1.00 L of solution from density (g/mL × 1000 mL). 4. Mass of solvent = mass of solution − mass of solute (convert to kg). 5. Molality = moles solute / kg solvent. Reverse (m → M): 1. Choose 1.00 kg solvent. Molality gives moles solute = m × 1.00 kg. 2. Find total mass = mass solvent + mass solute (g). Use density to get volume of solution. 3. Molarity = moles solute / volume (L). These steps tie directly to AP Topic 3.7 (M = n_solute / L_solution). For worked examples and practice, check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and thousands of practice problems (https://library.fiveable.me/practice/ap-chemistry).

Molarity is defined as moles of solute per liter of solution (M = n_solute / L_solution) because concentration describes how many solute particles are present in a given volume that you actually have to work with—the final homogeneous mixture. When you dissolve a solute the total volume usually changes (solute can expand or contract the solution), so using the solution’s volume gives the true particle density. If you used liters of solvent instead, you’d ignore that volume change and get the wrong concentration for reactions, equilibrium, or titrations (all tested in Topic 3.7 / LO 3.7.A). Molality (moles per kg solvent) is the alternative when you want a temperature-independent measure (mass doesn’t change with T), which is why colligative-property problems sometimes use molality. For review and problems on this, see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and more practice (https://library.fiveable.me/practice/ap-chemistry).

Molarity (M) goes down when you dilute a solution with more water. Molarity is defined as M = n_solute / L_solution, and dilution adds solvent without changing the number of moles of solute (n stays the same) while increasing the solution volume (L goes up), so M decreases. Use the dilution formula for calculations: M1·V1 = M2·V2. Example: 1.0 L of 2.0 M solution diluted to 2.5 L gives M2 = (2.0·1.0)/2.5 = 0.80 M. This is exactly the kind of calculation AP CED Topic 3.7 expects (Learning Objective 3.7.A). For a quick review, check the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and practice lots of dilution problems at Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).

Yes—a solution can be a solid. The CED even says solutions can be solids, liquids, or gases: what matters is homogeneity. In a solid solution one component (solute) is uniformly dispersed in another solid (solvent) so macroscopic properties don’t vary. Common examples are alloys: brass (Cu + Zn) and stainless steel (Fe + C + other metals) are substitutional or interstitial solid solutions (atoms replace host atoms or sit in gaps). In semiconductors, “doping” (tiny amounts of P or B in Si) is another solid-solution idea that changes properties. Composition for solids is usually given as percent by mass, mole fraction, or ppm rather than molarity. This topic ties to AP keywords (homogeneous mixture, solute, solvent, percent mass). For a focused review, see the Topic 3.7 study guide (https://library.fiveable.me/ap-chemistry/unit-3/solutions-mixtures/study-guide/lmjYxgiQN5yljSWPxWUl) and unit overview (https://library.fiveable.me/ap-chemistry/unit-3). For extra practice, check the practice problem bank (https://library.fiveable.me/practice/ap-chemistry).