Solutions and Solubility

🧪 Solution Terminology

Part 1 of 7 — Solutes, Solvents, and Solution Types

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 Solute and Solvent

A solution is a homogeneous mixture of two or more substances.

Key Points

• The solvent is usually present in the greater quantity
• Water is called the universal solvent because of its polarity and ability to dissolve many ionic and polar substances
• Solutions can exist in all phases: gas (air), liquid (saltwater), solid (alloys like brass)

Aqueous Solutions

When water is the solvent, we call it an aqueous solution, denoted by (aq) in chemical equations:

$\text{NaCl(s)} \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(\text{aq}) + \text{Cl}^-(\text{aq})$

🧊 Saturation States

The amount of solute that dissolves depends on the solubility of that solute in a given solvent at a specific temperature.

Three Saturation States

Making a Supersaturated Solution

1. Heat the solvent to increase solubility
2. Dissolve more solute than would normally dissolve at room temperature
3. Slowly cool the solution without disturbing it
4. The result: a supersaturated solution that can crystallize dramatically when a seed crystal is added

Solubility vs. Temperature

• Most solid solutes: solubility increases with temperature
• Gases: solubility decreases with temperature (think of a warm soda going flat)
• Pressure significantly affects gas solubility (Henry's Law) but has negligible effect on solids/liquids

Saturation Concept Check 🎯

💧 "Like Dissolves Like"

This is the most important rule for predicting solubility:

🔑 Key Concept: Polar solutes dissolve in polar solvents. Nonpolar solutes dissolve in nonpolar solvents.

Why?

Dissolving occurs when solute-solvent interactions are strong enough to overcome:

• Solute-solute attractions (breaking apart the solute)
• Solvent-solvent attractions (making room in the solvent)

The Dissolving Process for Ionic Compounds

When NaCl dissolves in water:

1. Water molecules surround ions — hydration (or solvation)
2. The partially negative oxygen of $H_{2}O$ attracts $Na^{+}$
3. The partially positive hydrogens attract $Cl^{-}$
4. Ion-dipole forces pull ions away from the crystal lattice

The energy released by hydration must be comparable to the lattice energy for dissolution to occur.

Predict the Solubility 🔽

Use the "like dissolves like" principle to predict whether each solute dissolves in the given solvent.

📏 AP Chemistry Solubility Rules (Aqueous Ionic Compounds)

For the AP exam, you need to know which ionic compounds are soluble in water:

Generally Soluble

Generally Insoluble

💡 Tip: Remember "NAG SAG" — Nitrates Always dissolve, Group 1 Salts Always Go into solution.

Solubility Rules Quiz 🎯

Exit Drill — Solution Terminology 🧮

1) In a solution of 5.0 g of sugar dissolved in 95.0 g of water, the solvent is water. What is the total mass of the solution in grams?

2) At 25°C, the solubility of NaCl is 36.0 g per 100 g of water. If 30.0 g of NaCl is added to 100 g of water at 25°C, how many grams of NaCl dissolve?

3) Using the same conditions as problem 2, how many grams of NaCl remain undissolved at the bottom?

Round all answers to 3 significant figures.

📏 Concentration Units

Part 2 of 7 — Molarity, Molality, Mass Percent, Mole Fraction, and ppm

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 Molarity ($M$)

Molarity is the most commonly used concentration unit in chemistry.

$\boxed{M = \frac{\text{moles of solute}}{\text{liters of solution}}}$

Units: mol/L (or simply $M$)

Example

Dissolve 4.00 g of NaOH ($M_{\text{NaOH}} = 40.00$ g/mol) in enough water to make 500.0 mL of solution.

$n = 4.00 \; \cancel{\text{g NaOH}} \times \frac{1 \text{ mol NaOH}}{40.00 \; \cancel{\text{g NaOH}}} = 0.100 \text{ mol NaOH}$

$M = \frac{0.100 \text{ mol}}{0.5000 \text{ L}} = 0.200 \text{ M}$

Key Points

• Molarity depends on volume of solution (not volume of solvent)
• Molarity changes with temperature because volume changes with temperature
• Most useful for stoichiometry in solution: moles = $M \times V$(in liters)

📌 Molality ($m$)

$\boxed{m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}}$

Units: mol/kg (or simply $m$)

Example

Dissolve 18.0 g of glucose ($M_{\text{glucose}} = 180.16$ g/mol) in 250.0 g of water.

$n = 18.0 \; \cancel{\text{g glucose}} \times \frac{1 \text{ mol glucose}}{180.16 \; \cancel{\text{g glucose}}} = 0.0999 \text{ mol glucose}$

$m = \frac{0.0999 \text{ mol}}{0.2500 \text{ kg}} = 0.400 \; m$

Why Molality?

• Molality depends on mass of solvent, which does not change with temperature
• This makes molality ideal for colligative property calculations (boiling point elevation, freezing point depression)
• On the AP exam, colligative property formulas use molality, not molarity

💡 Tip: Molality uses mass of solvent (kg), while molarity uses volume of solution (L). In dilute aqueous solutions they are approximately equal, but they diverge at higher concentrations.

📌 Other Concentration Units

Mass Percent (Weight Percent)

$\boxed{\text{mass \%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%}$

Example: 5.0 g NaCl in 95.0 g $H_{2}O$ → mass% = $\frac{5.0}{5.0 + 95.0} \times 100 = 5.0\%$

Mole Fraction ($\chi$)

$\boxed{\chi_A = \frac{n_A}{n_A + n_B + \cdots}}$

• Mole fraction is unitless and always between 0 and 1
• The sum of all mole fractions in a solution equals 1
• Used in Raoult's law and gas law calculations

Parts Per Million (ppm)

$\boxed{\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6}$

• Used for very dilute solutions (trace contaminants, water quality)
• 1 ppm = 1 mg solute per 1 kg solution (for dilute aqueous solutions, 1 ppm ≈ 1 mg/L)

Summary Table

Concentration Unit Concepts 🎯

Concentration Calculations 🧮

Given: Na = 22.99, Cl = 35.45, O = 16.00, H = 1.008, C = 12.01

1) What is the molarity of a solution prepared by dissolving 11.7 g of NaCl ($M = 58.44$ g/mol) in water to make 250.0 mL of solution? (to 3 significant figures, in mol/L)

2) What is the molality of a solution made by dissolving 36.0 g of glucose ($M = 180.16$ g/mol) in 500.0 g of water? (to 3 significant figures, in mol/kg)

3) What is the mass percent of a solution containing 8.0 g of NaOH in 200.0 g of solution? (to 3 significant figures)

🧪 Mole Fraction Worked Example

Problem: A solution contains 46.0 g of ethanol ($C_{2}H_{5}OH$, $M = 46.07$ g/mol) and 72.0 g of water ($M = 18.02$ g/mol). Calculate the mole fraction of ethanol.

Solution:

Step 1: Moles of ethanol: $n_{\text{eth}} = 46.0/46.07 = 0.9985$ mol

Step 2: Moles of water: $n_{\text{H}_2\text{O}} = 72.0/18.02 = 3.996$ mol

Step 3: Mole fraction of ethanol:

$\chi_{\text{eth}} = \frac{0.9985}{0.9985 + 3.996} = \frac{0.9985}{4.995} = 0.200$

Check: $\chi_{\text{H}_2\text{O}} = 3.996/4.995 = 0.800$. And $0.200 + 0.800 = 1.000$ ✓

Concentration Units — Identify and Apply 🔽

Exit Quiz — Concentration Units ✅

🔬 Dilution

Part 3 of 7 — $M_1V_1 = M_2V_2$, Preparing Solutions, and Serial Dilutions

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 The Dilution Equation

When you dilute a solution, you add more solvent. The amount of solute stays the same — only the volume changes.

Since moles of solute before = moles of solute after:

$\boxed{M_1V_1 = M_2V_2}$

🔑 Key Concept: The dilution equation works because moles of solute are conserved — only the volume changes. Since moles $= M \times V$, we get $M_1V_1 = M_2V_2$.

where:

• $M_1$ = initial molarity (concentrated)
• $V_1$ = initial volume
• $M_2$ = final molarity (diluted)
• $V_2$ = final volume

Why It Works

$\text{moles} = M \times V$

Since moles don't change: $M_1V_1 = M_2V_2$

Important Notes

• $V_1$ and $V_2$ must be in the same units (both mL or both L)
• $M_2 < M_1$ always (dilution lowers concentration)
• $V_2 > V_1$ always (you're adding solvent)
• The volume of solvent added = $V_2 - V_1$

🧪 Worked Example

Problem: How would you prepare 500.0 mL of 0.100 M HCl from a 12.0 M HCl stock solution?

Solution:

Step 1: Identify knowns.

• $M_1 = 12.0$ M, $M_2 = 0.100$ M, $V_2 = 500.0$ mL, $V_1 = ?$

Step 2: Solve for $V_1$.

$V_1 = \frac{M_2V_2}{M_1} = \frac{(0.100)(500.0)}{12.0} = 4.17 \text{ mL}$

Step 3: Procedure.

1. Measure 4.17 mL of 12.0 M HCl using a pipette
2. Add the acid to water (never water to acid — exothermic!)
3. Add water to bring the total volume to 500.0 mL in a volumetric flask
4. Mix thoroughly

⚠️ Warning: Always add acid to water ("Do as you oughta — add acid to water"). Adding water to concentrated acid can cause violent spattering due to the large heat of dilution.

Dilution Concept Check 🎯

📌 Serial Dilutions

A serial dilution is a series of sequential dilutions, each using the output of the previous step as input. This technique is used to:

• Create very low concentrations from stock solutions
• Prepare a set of standards for calibration curves
• Reduce concentration by orders of magnitude

How It Works

If each step dilutes by a factor of 10 (e.g., 1 mL into 9 mL):

General Formula

After $n$ serial dilutions, each with dilution factor $f$:

$\boxed{C_n = C_0 \times f^n}$

For a 1:10 dilution ($f = 0.1$), after 3 steps:

$C_3 = 1.0 \times (0.1)^3 = 0.001 \text{ M} = 1 \times 10^{-3} \text{ M}$

Dilution Calculations 🧮

1) What volume (in mL) of 16.0 M $HNO_{3}$ is needed to prepare 1.00 L of 0.400 M $HNO_{3}$?

2) If 25.0 mL of 0.800 M $CuSO_{4}$ is diluted to 100.0 mL, what is the final molarity? (to 3 significant figures)

3) A 1.00 M stock solution undergoes three serial 1:10 dilutions. What is the final concentration? (answer in M, use scientific notation as a × $10^{-3}$ — enter just the coefficient $a$)

🧪 Preparing Solutions in the Lab

From a Solid Solute

To prepare 250.0 mL of 0.200 M $Na_{2}CO_{3}$ ($M = 105.99$ g/mol):

1. Calculate mass needed: $m = n \times M_{\text{molar}} = (0.200 \times 0.2500) \times 105.99 = 5.30$ g
2. Weigh 5.30 g of $Na_{2}CO_{3}$ on an analytical balance
3. Transfer to a 250 mL volumetric flask
4. Add water to dissolve, then fill to the 250 mL mark
5. Mix by inverting several times

From a Stock Solution (Dilution)

To prepare 100.0 mL of 0.0500 M $Na_{2}CO_{3}$ from the 0.200 M solution:

$V_1 = \frac{(0.0500)(100.0)}{0.200} = 25.0 \text{ mL}$

Pipette 25.0 mL of stock into a 100 mL volumetric flask and add water to the mark.

Dilution and Preparation — Key Concepts 🔽

Exit Quiz — Dilution ✅

🌡️ Colligative Properties

Part 4 of 7 — Boiling Point Elevation and Freezing Point Depression

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🔬 What Are Colligative Properties?

The word "colligative" comes from Latin colligare meaning "to bind together." These properties depend on the collective number of dissolved particles, regardless of what those particles are.

🔑 Key Concept: Colligative properties depend only on the number of dissolved particles — not their chemical identity. More particles → greater effect.

The Four Colligative Properties

Why Do They Occur?

When solute particles are added to a solvent:

• They disrupt the orderly arrangement needed for freezing → lower freezing point
• They lower the vapor pressure → solvent must be heated to a higher temperature to boil
• More particles = greater effect

Key Distinction

• Nonelectrolytes (glucose, sucrose): dissolve as intact molecules → 1 particle per formula unit
• Electrolytes (NaCl, $CaCl_{2}$): dissociate into ions → multiple particles per formula unit

📌 Boiling Point Elevation

$\boxed{\Delta T_b = iK_bm}$

where:

• $\Delta T_b$ = change in boiling point (°C)
• $i$ = van't Hoff factor (number of particles per formula unit)
• $K_b$ = ebullioscopic constant of the solvent (°C/m)
• $m$ = molality of the solution (mol solute / kg solvent)

The New Boiling Point

$T_{b,\text{solution}} = T_{b,\text{pure}} + \Delta T_b$

The boiling point increases (we add $\Delta T_b$).

For Water

$K_b = 0.512$ °C/m and $T_{b,\text{pure}} = 100.0$ °C

Worked Example

Problem: What is the boiling point of a solution of 1.00 mol of glucose (nonelectrolyte, $i = 1$) in 1.00 kg of water?

Solution:

$\Delta T_b = (1)(0.512)(1.00) = 0.512 \text{ °C}$

$T_b = 100.0 + 0.512 = 100.5 \text{ °C}$

📌 Freezing Point Depression

$\boxed{\Delta T_f = iK_fm}$

where:

• $\Delta T_f$ = change in freezing point (°C)
• $i$ = van't Hoff factor
• $K_f$ = cryoscopic constant of the solvent (°C/m)
• $m$ = molality

The New Freezing Point

$T_{f,\text{solution}} = T_{f,\text{pure}} - \Delta T_f$

The freezing point decreases (we subtract $\Delta T_f$).

For Water

$K_f = 1.86$ °C/m and $T_{f,\text{pure}} = 0.0$ °C

Worked Example

Problem: What is the freezing point of a solution of 0.500 mol of NaCl ($i = 2$) in 1.00 kg of water?

Solution:

$\Delta T_f = (2)(1.86)(0.500) = 1.86 \text{ °C}$

$T_f = 0.0 - 1.86 = -1.86 \text{ °C}$

Why Does NaCl Have $i = 2$?

NaCl dissociates into 2 ions: $Na^{+}$ + $Cl^{-}$. Each formula unit produces 2 particles:

$\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$

📊 The van't Hoff Factor ($i$)

The van't Hoff factor tells you how many particles one formula unit produces in solution.

Ideal vs. Actual $i$

In real solutions, ion pairing can reduce the effective $i$:

• NaCl: theoretical $i = 2$, actual $i \approx 1.9$ in moderate concentrations
• $CaCl_{2}$: theoretical $i = 3$, actual $i \approx 2.7$

For AP calculations, use the theoretical (ideal) $i$ unless told otherwise.

⚠️ Warning: For electrolytes, always multiply by the van't Hoff factor $i$. Forgetting $i$ is the most common mistake on colligative property problems — NaCl gives $i = 2$, $CaCl_{2}$ gives $i = 3$, etc.

Colligative Properties Quiz 🎯

Colligative Property Calculations 🧮

Given: $K_b = 0.512$ °C/m, $K_f = 1.86$ °C/m for water.

1) Calculate the boiling point (in °C) of a solution containing 0.300 mol of KBr ($i = 2$) in 500.0 g of water. (to 3 significant figures)

2) Calculate the freezing point (in °C) of a solution of 58.44 g of NaCl ($M = 58.44$ g/mol, $i = 2$) in 2.00 kg of water. (to 3 significant figures)

3) An antifreeze solution is 2.50 m ethylene glycol ($i = 1$). What is its freezing point in °C? (to 3 significant figures)

Colligative Properties — Conceptual 🔽

Exit Quiz — Colligative Properties ✅

🔬 Osmotic Pressure and the van't Hoff Factor

Part 5 of 7 — $\Pi = iMRT$, Electrolytes vs. Nonelectrolytes

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 Osmosis

Osmosis is the net movement of solvent (usually water) through a semipermeable membrane from a region of lower solute concentration to higher solute concentration.

Semipermeable Membrane

A membrane that allows solvent molecules to pass through but blocks solute particles (ions or large molecules).

Driving Force

The solvent naturally moves to dilute the more concentrated side — this is an entropy-driven process.

Direction Rules

Biological Importance

• Red blood cells placed in a hypotonic solution swell and may burst (lysis)
• In a hypertonic solution, they shrink (crenation)
• IV fluids must be isotonic (typically 0.9% NaCl — "normal saline")

💨 Osmotic Pressure ($\Pi$)

Osmotic pressure is the minimum pressure that must be applied to the solution side to prevent osmosis.

$\boxed{\Pi = iMRT}$

where:

• $\Pi$ = osmotic pressure (atm)
• $i$ = van't Hoff factor
• $M$ = molarity of the solution (mol/L)
• $R$ = gas constant = $0.08206$ L·atm/(mol·K)
• $T$ = temperature in Kelvin

This looks like the ideal gas law!

$\Pi V = nRT \quad \Rightarrow \quad \Pi = \frac{n}{V}RT = MRT$

With the van't Hoff factor for electrolytes: $\Pi = iMRT$

Worked Example

Problem: Calculate the osmotic pressure at 25°C of a 0.100 M NaCl ($i = 2$) solution.

Solution:

$T = 25 + 273 = 298 \text{ K}$

$\Pi = (2)(0.100)(0.08206)(298) = 4.89 \text{ atm}$

That is a surprisingly large pressure — about 72 psi — from a relatively dilute solution!

🔋 Electrolytes vs. Nonelectrolytes

Strong Electrolytes

• Completely dissociate into ions in solution
• Examples: NaCl, KBr, HCl, NaOH, $CaCl_{2}$
• Conduct electricity strongly
• $i$ equals the total number of ions produced

Weak Electrolytes

• Partially dissociate in solution
• Examples: $CH_{3}COOH$ (acetic acid), $NH_{3}$, HF
• Conduct electricity weakly
• $i$ is between 1 and the theoretical maximum (closer to 1 for weak electrolytes)

Nonelectrolytes

• Do not dissociate — dissolve as whole molecules
• Examples: glucose $(C_{6}H_{12}O_{6})$, sucrose, ethanol, urea
• Do not conduct electricity
• $i = 1$ always

Impact on Colligative Properties

For 0.10 m solutions in water ($K_f = 1.86$ °C/m):

Osmotic Pressure Concepts 🎯

Osmotic Pressure Calculations 🧮

Given: $R = 0.08206$ L·atm/(mol·K)

1) Calculate the osmotic pressure (in atm) of a 0.200 M glucose ($i = 1$) solution at 37°C (body temperature). (to 3 significant figures)

2) A 0.150 M $CaCl_{2}$ ($i = 3$) solution at 25°C has what osmotic pressure? (in atm, to 3 significant figures)

3) A protein solution has an osmotic pressure of 0.0821 atm at 25°C. What is the molarity of the protein? ($i = 1$; answer in M to 3 significant figures)

📌 Reverse Osmosis

If you apply pressure greater than the osmotic pressure to the concentrated side, you can force solvent to flow backward — from high concentration to low concentration.

Applications

• Water desalination — removing salt from seawater (seawater $\Pi \approx 27$ atm, so applied pressure must exceed this)
• Water purification — removing contaminants
• Maple syrup production — concentrating sap

Reverse Osmosis vs. Osmosis

Electrolytes and Osmosis — Key Concepts 🔽

Exit Quiz — Osmotic Pressure ✅

🧮 Problem-Solving Workshop

Part 6 of 7 — Mixed Concentration and Colligative Property Calculations

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

🛠️ Problem-Solving Strategy

Step-by-Step Approach

1. Read the problem — identify what is given and what is asked
2. List the relevant formula(s)
3. Convert units if needed (g → mol, mL → L, °C → K)
4. Determine the van't Hoff factor $i$ (does the solute dissociate?)
5. Plug in and solve
6. Check — does the answer make physical sense?

Key Formulas Reference

Constants for Water

• $K_b = 0.512$ °C/m
• $K_f = 1.86$ °C/m
• $T_b = 100.0$ °C, $T_f = 0.0$ °C
• $R = 0.08206$ L·atm/(mol·K)

🧪 Worked Example 1: Concentration Conversion

Problem: A solution is prepared by dissolving 34.2 g of sucrose ($C_{12}H_{22}O_{11}$, $M = 342.30$ g/mol) in 200.0 g of water. The density of the resulting solution is 1.024 g/mL. Calculate molality, molarity, mass percent, and mole fraction.

Given

Step-by-Step Solution

🔑 Tip: Notice how each concentration unit uses different denominators: molality uses kg of solvent, molarity uses L of solution, mass percent uses g of solution, and mole fraction uses total moles.

Concentration Conversion Quiz 🎯

Mixed Calculations — Set 1 🧮

Given: $K_f = 1.86$ °C/m, $K_b = 0.512$ °C/m, $R = 0.08206$ L·atm/(mol·K)

1) A solution of 9.00 g of glucose ($M = 180.16$ g/mol, $i = 1$) in 250.0 g of water has what freezing point? (in °C, to 3 significant figures)

2) What is the boiling point (°C) of a solution made by dissolving 14.6 g of NaCl ($M = 58.44$ g/mol, $i = 2$) in 500.0 g of water? (to 3 significant figures)

3) What is the osmotic pressure (atm) of a 0.0500 M KBr ($i = 2$) solution at 25°C? (to 3 significant figures)

🧪 Worked Example 2: Finding Molar Mass from Colligative Data

Problem: A solution of 5.00 g of an unknown nonelectrolyte in 100.0 g of water freezes at $-0.930$ °C. Find the molar mass of the unknown.

Given

Step-by-Step Solution

🔑 AP Application: This technique is used in the lab to identify unknown compounds — a classic AP free-response question topic.

Mixed Calculations — Set 2 🧮

1) An unknown nonelectrolyte ($i = 1$) dissolves: 3.50 g in 50.0 g of water, and the solution freezes at $-1.86$ °C. What is the molar mass (g/mol) of the unknown? (to 3 significant figures)

2) What mass (in grams) of ethylene glycol ($M = 62.07$ g/mol, $i = 1$) must be added to 4.00 kg of water to lower the freezing point to $-10.0$ °C? (to nearest whole number)

3) A 0.200 M solution of an electrolyte at 25°C has an osmotic pressure of 14.6 atm. What is the van't Hoff factor $i$? (to 3 significant figures)

Problem-Solving Strategies 🔽

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Connecting Solubility Rules, Concentration, and Colligative Properties

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

📌 The Big Picture

How It All Connects

$\text{Solubility Rules} \rightarrow \text{Does it dissolve?} \rightarrow \text{How much? (Concentration)} \rightarrow \text{What effect? (Colligative Properties)}$

Unit Summary

AP Exam Tips

• Always check whether the solute is an electrolyte or nonelectrolyte → determines $i$
• Watch units: molarity (mol/L of solution) vs. molality (mol/kg of solvent)
• Colligative properties: more particles = greater effect
• For free-response: show all work, label formulas, include units

AP-Style Multiple Choice — Set 1 🎯

AP-Style Multiple Choice — Set 2 🎯

Comprehensive Calculations 🧮

Given: $K_f = 1.86$ °C/m, $K_b = 0.512$ °C/m, $R = 0.08206$ L·atm/(mol·K)

1) An aqueous solution of $MgCl_{2}$ ($i = 3$, $M = 95.21$ g/mol) is made by dissolving 9.52 g in 200.0 g of water. What is the freezing point of the solution? (in °C, to 3 significant figures)

2) How many mL of 12.0 M HCl must be diluted to prepare 500.0 mL of 0.600 M HCl? (to 3 significant figures)

3) A 0.0250 M solution of an unknown electrolyte at 25°C has an osmotic pressure of 1.83 atm. What is the van't Hoff factor? (to 3 significant figures)

AP Conceptual Review 🔽

✏️ AP Free-Response Practice Problem

Problem: A student performs a freezing point depression experiment. She dissolves 2.56 g of an unknown compound in 50.0 g of water. The solution freezes at $-0.744$ °C.

Solution:

(a) Calculate the molality of the solution.

$m = \frac{\Delta T_f}{iK_f} = \frac{0.744}{(1)(1.86)} = 0.400 \; m$

(Assuming $i = 1$ initially)

(b) Calculate the moles of solute.

$n = m \times \text{kg solvent} = 0.400 \times 0.0500 = 0.0200 \text{ mol}$

(c) Calculate the apparent molar mass.

$M = \frac{2.56}{0.0200} = 128 \text{ g/mol}$

(d) The compound is known to be naphthalene ($C_{10}H_{8}$, actual $M = 128.17$ g/mol). Is this consistent? Yes — the experimental value matches the actual molar mass, confirming $i = 1$ (naphthalene is a nonelectrolyte).

(e) If the compound were actually an electrolyte with $i = 2$, what would the molar mass be?

$m = \frac{0.744}{(2)(1.86)} = 0.200 \; m, \quad n = 0.200 \times 0.0500 = 0.0100, \quad M = \frac{2.56}{0.0100} = 256 \text{ g/mol}$

Final Exit Quiz — AP Synthesis ✅