CHEM UNIT 1

Absolutely — these three slide decks give us a very clear picture of what your Chemistry Exam 1 is centered on. I’m going to separate your prep into two study guides, because the way you study the concepts should be different from the way you prepare for the numerical problems.

Your slides cover Chapter 11: Liquids & Phase Changes, Chapter 12: Solids & Solid-State Materials, and Chapter 13: Solutions & Their Properties.

🧪 CHEMISTRY EXAM 1 — STUDY GUIDE 1

Concepts, Relationships & Application Questions

CHAPTER 11 — LIQUIDS AND PHASE CHANGES

1. Bond Polarity vs. Molecular Polarity

This distinction is extremely important because molecular polarity determines the intermolecular forces (IMFs) a substance experiences.

Bond polarity

A bond becomes polar when electrons are shared unequally.

The more electronegative atom pulls electron density toward itself.

We represent this using partial charges:

δ+ → δ−

The δ− side is the more electronegative side.

Molecular polarity

Having polar bonds does NOT automatically make a molecule polar.

You must consider:

1. Are any bonds polar?

2. What is the molecular geometry?

3. Do the bond dipoles cancel?

Examples from your slides:

CO₂

O=C=O

The C–O bonds are polar, but the molecule is linear.

The dipoles point in opposite directions and cancel.

→ Nonpolar molecule

BF₃

Trigonal planar and symmetrical.

The B–F bonds are polar, but all three dipoles cancel.

→ Nonpolar

H₂O

Bent geometry.

The O–H dipoles do NOT cancel.

→ Polar

Exam strategy

When asked whether something is polar:

Lewis structure → geometry → bond dipoles → cancellation

Don't decide based only on whether the bonds are polar.

2. Intermolecular Forces

Intermolecular forces are attractions between different particles.

They are NOT the same as the covalent or ionic bonds holding atoms together within a compound.

Your slides focus on:

London dispersion forces
Dipole-dipole forces
Hydrogen bonding

London Dispersion Forces

Dispersion forces occur because electrons are constantly moving.

For a brief moment, electrons may be distributed unevenly, creating an:

instantaneous dipole

That instantaneous dipole can induce a dipole in a neighboring molecule.

KEY FACT

ALL atoms and molecules have London dispersion forces.

Even nonpolar substances.

What makes dispersion stronger?

Primarily:

more electrons → greater polarizability → stronger dispersion

Larger/heavier molecules therefore often have stronger dispersion forces.

But shape also matters.

Long, spread-out molecules can have more surface contact than compact molecules.

More contact → stronger dispersion → usually higher boiling point.

3. Dipole-Dipole Forces

These occur between polar molecules.

The δ+ end of one molecule attracts the δ− end of another.

Therefore:

polar molecule → dipole-dipole attractions

Remember that the molecule also has dispersion forces.

4. Hydrogen Bonding

Hydrogen bonding is a particularly strong type of dipole-dipole interaction.

Your slides specify that H must be bonded to:

N, O, or F

Think:

H–N
H–O
H–F

Examples:

CH₃OH → YES

H₂O → YES

NH₃ → YES

HF → YES

But:

HCl → NO hydrogen bonding

CH₄ → NO

Major exam trap

Having H somewhere in the formula is not enough.

It must be bonded directly to N, O, or F.

5. Determining the IMFs of a Molecule

Use this decision process:

Step 1

Every molecule has:

London dispersion

Step 2

Is the molecule polar?

YES → add dipole-dipole

Step 3

Does it contain H directly bonded to N/O/F?

YES → add hydrogen bonding

Example:

CH₃OH

Dispersion ✔
Polar ✔
Dipole-dipole ✔
O–H ✔
Hydrogen bonding ✔

6. IMFs and Physical Properties

This is one of the most important relationship sets to memorize.

Stronger intermolecular forces cause:

↑ boiling point

↑ viscosity

↑ surface tension

↓ vapor pressure

↓ volatility

The opposite is true for weaker IMFs.

Why?

Strong IMFs make molecules harder to separate.

Therefore they require more energy to escape the liquid.

7. Boiling Point

Boiling occurs when:

Vapor pressure = external atmospheric pressure

This definition is essential.

If IMFs are strong:

molecules have difficulty escaping → vapor pressure decreases → more heating is necessary → boiling point increases

Therefore:

Strong IMF → HIGH boiling point

Atmospheric pressure and boiling point

Lower atmospheric pressure means the liquid doesn't need to develop as much vapor pressure before boiling.

Therefore:

Lower atmospheric pressure → lower boiling point

This explains why liquids boil at lower temperatures at high altitude.

8. Vapor Pressure

Vapor pressure is the pressure of a vapor that is in equilibrium with its liquid.

Temperature relationship

↑ Temperature → ↑ Vapor pressure

because more molecules have enough kinetic energy to escape.

IMF relationship

↑ IMF → ↓ Vapor pressure

because molecules have greater difficulty escaping.

Therefore:

weak IMF → high vapor pressure → volatile

9. Volatility

A volatile substance evaporates readily.

High volatility means:

weak IMF
high vapor pressure
usually lower boiling point

Keep this relationship together rather than memorizing each one independently.

10. Viscosity

Viscosity = resistance to flow.

Think:

Water → relatively low viscosity

Syrup → high viscosity

Stronger IMFs → higher viscosity

because molecules resist moving past one another.

11. Surface Tension

Surface tension is the liquid's resistance to increasing its surface area/spreading out.

Molecules inside a liquid experience attractions in all directions.

Surface molecules experience a net inward attraction.

Strong IMF → high surface tension

12. Phase Changes

Know all six.

Energy absorbed

Solid → Liquid → Gas

These processes require energy.

Melting = endothermic
Vaporization = endothermic
Sublimation = endothermic

Energy released

Gas → Liquid → Solid

Condensation = exothermic
Freezing = exothermic
Deposition = exothermic

13. Enthalpies of Phase Changes

ΔHfusion

Energy required to convert:

1 mol solid → liquid

ΔHvap

Energy required to convert:

1 mol liquid → gas

ΔHsub

Energy required to convert:

1 mol solid → gas

Because sublimation can conceptually occur through:

solid → liquid → gas

Hess's Law gives:

ΔHsub = ΔHfus + ΔHvap

14. Gibbs Free Energy and Phase Changes

Your slides give:

ΔG = ΔH − TΔS

where:

ΔG = Gibbs free-energy change
ΔH = enthalpy change
T = temperature in Kelvin
ΔS = entropy change

At phase equilibrium:

ΔG = 0

Therefore:

0 = ΔH − TΔS

so:

ΔS = ΔH/T

This can appear as a calculation question.

CHAPTER 12 — SOLIDS

15. Crystalline vs. Amorphous Solids

Crystalline

Particles have:

rigid, long-range repeating order

Example structures form crystal lattices.

Amorphous

No regular three-dimensional long-range arrangement.

Examples from your slides:

glass
rubber

16. Four Types of Crystalline Solids

You need to distinguish:

Ionic solids

Made of ions.

Attraction:

electrostatic attraction between cations and anions

Molecular solids

Lattice points contain molecules.

Held together by:

dispersion
dipole-dipole
and/or hydrogen bonding

Covalent network solids

Atoms form an enormous network connected by:

covalent bonds

These are not individual molecules packed together.

Metallic solids

Metal atoms held together through delocalized electrons — the electron sea idea.

This explains properties such as:

good conductivity
malleability

17. Unit Cells

A unit cell is:

the basic repeating structural unit of a crystalline solid.

Your main cubic structures are:

Simple Cubic (SC)
Body-Centered Cubic (BCC)
Face-Centered Cubic (FCC)

18. Unit Cell Facts — MEMORIZE THIS TABLE

Also:

HCP

Coordination number = 12

Packing efficiency = 74%

FCC corresponds to cubic closest packing in your slides.

19. Counting Atoms in Unit Cells

This is both conceptual AND mathematical.

Corner atom

Shared between 8 cells.

Contribution:

1/8 atom

Therefore:

8 corners × 1/8 = 1 atom

Face atom

Shared between 2 cells.

Contribution:

1/2 atom

Six faces:

6 × 1/2 = 3 atoms

Body-center atom

Entirely inside one cell.

Contribution:

1 atom

Therefore:

SC

8(1/8)

= 1 atom

BCC

8(1/8) + 1

= 2 atoms

FCC

8(1/8) + 6(1/2)

= 1 + 3

= 4 atoms

20. Coordination Number

Coordination number = number of nearest neighboring atoms.

Memorize:

SC = 6
BCC = 8
FCC = 12
HCP = 12

Higher coordination number means particles are more tightly packed.

CHAPTER 13 — SOLUTIONS

21. What Is a Solution?

A solution is a:

homogeneous mixture

Its composition is uniform throughout.

Solvent

Component present to the greater extent.

Solute

Component present to the lesser extent.

There can be more than one solute.

22. Miscibility

Two liquids are miscible if they dissolve completely in each other in all proportions.

Example from the slides:

CCl₄ + benzene

Both are nonpolar.

This connects to the common idea:

substances with compatible intermolecular forces tend to mix.

23. Solution Formation and Energy

When dissolving something, intermolecular attractions must be disrupted and new ones formed.

Your slides connect this to Hess's Law:

energy associated with breaking existing attractions + energy associated with forming new attractions → overall ΔHsolution

Solutions can therefore form:

endothermically OR exothermically

Example:

NH₄Cl dissolving → endothermic → solution becomes cold

AlCl₃ dissolving → exothermic → solution becomes hot

24. Entropy and Dissolution

Enthalpy isn't the only factor.

Entropy also affects whether dissolution is favored.

Particles generally have more freedom of movement in solution than in an ordered crystalline solid.

Thus dissolution can sometimes occur even when it is endothermic.

25. Saturated vs. Unsaturated vs. Supersaturated

Saturated

Contains the maximum equilibrium amount of dissolved solute.

At equilibrium:

rate of dissolving = rate of crystallization

Unsaturated

Can still dissolve additional solute.

Supersaturated

Contains more dissolved solute than the equilibrium amount.

26. Temperature and Solubility

A major rule from your slides:

Gas solubility decreases as temperature increases.

So:

↑ temperature → ↓ gas solubility

Think about warm soda losing CO₂ more readily.

27. Pressure and Gas Solubility

For gases:

↑ pressure → ↑ gas solubility

Henry's Law describes this mathematically.

This is why CO₂ can remain dissolved in a sealed soda under pressure.

Open it:

pressure ↓ → CO₂ solubility ↓ → gas escapes.

28. Colligative Properties

This is a HUGE Chapter 13 topic.

Colligative properties depend on:

NUMBER of dissolved particles

not their chemical identity.

Four major properties:

vapor-pressure lowering
boiling-point elevation
freezing-point depression
osmotic pressure

29. van't Hoff Factor

Symbol:

i

It accounts for how many dissolved particles an electrolyte produces.

For an idealized example:

NaCl → Na⁺ + Cl⁻

Therefore:

i ≈ 2

A nonelectrolyte that stays intact:

glucose → glucose

i = 1

Your slides also note that real electrolyte solutions can deviate because of ion pairing, particularly at higher concentrations.

30. What Adding a Nonvolatile Solute Does

This relationship is extremely important.

Add a nonvolatile solute:

Vapor pressure ↓

Boiling point ↑

Freezing point ↓

Remember:

VP ↓
BP ↑
FP ↓

31. Osmosis

Osmosis is movement of:

solvent through a semipermeable membrane from the less concentrated side toward the more concentrated side.

Pay attention:

It's the solvent moving.

Not the solute.

🔢 STUDY GUIDE 2 — ALL THE MATH/CALCULATION QUESTIONS

These are the calculations I would prioritize heavily from your decks.

1. Heating/Cooling Calculations

Use:

q = mcΔT

where:

q = heat
m = mass
c = specific heat
ΔT = Tf − Ti

CRITICAL

Use this equation only while a substance stays in one phase.

Example:

25°C liquid water → 80°C liquid water

q = mcΔT

But if water boils, you need an additional phase-change calculation.

2. Phase-Change Energy

Use:

q = nΔH

where:

n = moles

ΔH = molar enthalpy of the phase change

For vaporization:

q = nΔHvap

For fusion:

q = nΔHfus

3. Multi-Step Heating Problems

Your Chapter 11 slides specifically contain a problem asking how much energy is needed to heat water from 25°C to 120°C.

This is a classic exam problem.

You CANNOT just do:

q = mcΔT ❌

because water changes phase.

Instead:

STEP 1 — Heat liquid

25°C → 100°C

q₁ = mcΔT

STEP 2 — Vaporize

100°C liquid → 100°C gas

q₂ = nΔHvap

STEP 3 — Heat gas

100°C → 120°C

q₃ = mcΔT

STEP 4

Add everything:

qtotal = q₁ + q₂ + q₃

This problem type is VERY worth practicing.

4. Clausius-Clapeyron Equation

Used to relate:

vapor pressure and temperature

Your slides apply it to finding octane's boiling point at a lower atmospheric pressure.

The two-point form is:

ln(P₂/P₁) = −ΔHvap/R(1/T₂ − 1/T₁)

Variables

P₁ = initial vapor pressure
P₂ = final vapor pressure
T₁ = initial temperature
T₂ = final temperature
ΔHvap = enthalpy of vaporization
R = gas constant

VERY IMPORTANT

Temperature must be:

Kelvin

and your energy units for ΔHvap and R must agree.

If R is in J:

convert:

kJ → J

5. Molality

m = moles solute / kg solvent

This is NOT molarity.

Biggest trap

Denominator is:

kg of SOLVENT

not kg solution.

Workflow:

grams solute
↓ ÷ molar mass
moles solute

grams solvent
↓ ÷1000
kg solvent

Then:

moles/kg.

6. Percent by Mass

mass % = (mass solute / mass solution) × 100

Remember:

mass solution = mass solute + mass solvent

Example:

20 g solute + 80 g water

solution mass = 100 g

Mass % =

20/100 × 100

= 20%

7. Henry's Law

Your slides use:

c = kP

where:

c = concentration of dissolved gas
k = Henry's Law constant
P = gas pressure

Therefore:

higher pressure → higher concentration.

Your slides' CO₂ example:

At 1.0 atm:

c = 3.2 × 10⁻² M

Therefore:

k = c/P

= 3.2 × 10⁻² M/atm

At 2.5 atm:

c = kP

= (3.2 × 10⁻²)(2.5)

c = 0.080 M

8. Mole Fraction

You need this for Raoult's Law.

Xₐ = nₐ / ntotal

For two components:

Xₐ = nₐ/(nₐ+nᵦ)

Mole fractions should add to:

Xₐ + Xᵦ = 1

9. Raoult's Law — Nonvolatile Solute

Your slides give:

Psolution = Xsolvent P°solvent

This means adding nonvolatile solute reduces the mole fraction of solvent.

Therefore:

Xsolvent ↓

causes:

Psolution ↓

Typical workflow

Given grams:

grams solvent → moles solvent

grams solute → moles solute

Then:

Xsolvent = nsolvent/(nsolvent + nsolute)

Then:

Psolution = XsolventP°solvent

10. Raoult's Law — BOTH Components Volatile

Your slides also cover this.

Now BOTH substances contribute vapor.

Component A

Pₐ = XₐP°ₐ

Component B

Pᵦ = XᵦP°ᵦ

Then Dalton's Law:

Ptotal = Pₐ + Pᵦ

So your workflow becomes:

moles A
↓
moles B
↓
mole fractions
↓
partial pressures
↓
add

11. Boiling-Point Elevation

For the ideal colligative-property calculations represented in your slides:

ΔTb = iKb m

where:

i = van't Hoff factor
Kb = boiling-point elevation constant
m = molality

Then:

Tb,new = Tb,pure + ΔTb

Remember:

ADD for boiling.

12. Freezing-Point Depression

ΔTf = iKf m

Then:

Tf,new = Tf,pure − ΔTf

Remember:

SUBTRACT for freezing.

An easy memory trick:

Boiling goes UP.
Freezing goes DOWN.

13. Osmotic Pressure

Your slides give:

Π = MRT

For electrolyte calculations, particle effects may be represented using the van't Hoff factor:

Π = iMRT

where:

Π = osmotic pressure
M = molarity
R = 0.08206 L·atm/(mol·K)
T = Kelvin

Your slides ask:

What is the molarity of solute in human blood if its osmotic pressure is 7.7 atm at 37°C?

Start:

Π = MRT

Solve:

M = Π/RT

First convert temperature:

37 + 273.15 = 310.15 K

Then substitute.

14. Finding Molar Mass from Freezing-Point Depression

Your final Chapter 13 example does exactly this.

You're essentially working backward.

Given:

ΔTf
Kf
grams solvent
grams solute

Start:

ΔTf = Kf m

Solve:

m = ΔTf/Kf

Then:

m = mol solute/kg solvent

Therefore:

mol solute = m × kg solvent

Finally:

molar mass = grams solute / moles solute

This is an excellent multi-step exam problem.

15. Finding Molar Mass from Osmotic Pressure

Another possible application:

Π = MRT

Solve:

M = Π/RT

But:

M = mol/L

Therefore:

moles = M × L

Then:

molar mass = grams/moles

16. Gibbs Free Energy / Entropy

From Chapter 11:

ΔG = ΔH − TΔS

At phase equilibrium:

ΔG = 0

Therefore:

0 = ΔH − TΔS

and:

ΔS = ΔH/T

Again:

T MUST be Kelvin.

⭐ YOUR EXAM 1 MASTER FORMULA SHEET

I would memorize/know how to use these:

🚨 EXAM TRAPS TO WATCH FOR

These are the mistakes I especially don't want you making:

1. °C in an equation requiring Kelvin

K = °C + 273.15

2. Grams instead of moles in mole fraction

Convert everything to moles FIRST.

3. Using solution mass instead of solvent mass for molality

Molality denominator = kg solvent.

4. Forgetting the van't Hoff factor

Electrolytes produce multiple particles.

5. Adding ΔTf

Freezing-point depression means:

Tf,new = Tf,pure − ΔTf

6. Subtracting ΔTb

Boiling-point elevation means:

Tb,new = Tb,pure + ΔTb

7. Using q = mcΔT during a phase change

During a phase change use:

q = nΔH.

8. Calling every molecule containing H hydrogen-bonding

H must be bonded to N, O, or F.

9. Assuming polar bonds automatically mean a polar molecule

Geometry matters.

10. Forgetting that all molecules have dispersion forces

They do.

11. Comparing IMFs without looking at molecular size/shape

Especially for dispersion-force questions.

12. Mixing up vapor pressure and boiling point

Strong IMF:

VP ↓ but BP ↑

🎯 What I would focus on most

Based specifically on how much emphasis and how many worked/application examples appear in your slides, I would make sure you can do these almost automatically:

Concept side: molecular polarity → identify IMF → compare boiling point/vapor pressure/viscosity/surface tension; phase changes; unit-cell structures and counting; solution terminology; gas solubility; colligative-property relationships.

Math side: multi-step heating curves, Clausius-Clapeyron, unit-cell counting, molality, mass %, Henry's Law, mole fractions, Raoult's Law with one vs. two volatile components, boiling-point elevation, freezing-point depression, osmotic pressure, and finding molar mass from colligative properties.

And I wouldn't stop at memorizing the equations. Your slides contain several problems where the hardest part is figuring out which equation(s) to use, so that's exactly how we should practice.

Next, I can make you a Chem Exam 1 practice set just for the math, starting easier and progressing to exam-level multi-step questions. I can give them one at a time without showing the answer, have you solve each one, and then correct your work before moving on.