GENERAL CHEMISTRY · SOLUTIONS

Track the solvent.
Fix the denominator.

Learn all five official topics through evidence ledgers, worked examples, retrieval prompts, and twenty original five-choice questions.

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5guided lessons

20practice questions

15study objectives

5choices per item

A repeatable solution routine

Components. Attractions. Property. Basis.

01Name components
Identify solvent, solute, and the particles that actually exist after mixing.
02Inventory attractions
Compare both starting particle sets with the new solute–solvent interactions.
03Classify the property
Separate effective particle-count effects from identity-dependent behavior.
04Fix the basis
Write the concentration denominator and model conditions before calculating.

Five prerequisite-aware lessons

Move from particles to evidence.

Each lesson uses two semantic study tables, one fully worked example, and closed-note retrieval before multiple-choice practice.

LESSON 1 · 18 MIN

Study + retrieve

Polarity from component roles to measured compatibility

Identify solution components, determine relevant polarity evidence, and make bounded compatibility predictions without turning ‘like dissolves like’ into an absolute rule.

ESSENTIAL QUESTIONWhich attractions can replace those in the separated solute and solvent?

Component and compatibility ledger

Evidence sequence for a polarity-based prediction
Question	Evidence	Boundary
Which is solvent?	Stated dissolving medium and preparation	Not physical state alone
Is the molecule polar?	Vector sum of bond dipoles in the geometry	Polar bonds can cancel
Can new attractions form?	Ion–dipole, dipole–dipole, hydrogen bond, dispersion	Name only structurally supported forces

What polarity matching can and cannot establish
Comparison	Qualitative support	Do not claim without data
Ionic or polar solute + polar solvent	Often favorable attraction matching	Complete or infinite solubility
Nonpolar solute + nonpolar solvent	Often favorable dispersion compatibility	An exact concentration
Mixed polar and nonpolar regions	Compare the whole controlled structure	One functional group always dominates

STUDY TABLE · CAPTION INCLUDED

Name the component roles

The solvent is the dissolving medium and the solute is dispersed within it. Phase or formula order alone does not determine the roles; use the preparation and stated medium.

Name the medium before predicting
A solution can be solid, liquid, or gas

Build the polarity evidence

Molecular polarity is the vector sum of bond dipoles in the three-dimensional structure. Ionic charge, a polar bond, and a net molecular dipole are different kinds of evidence.

Geometry can cancel bond dipoles
Charge is not a molecular dipole

Use compatibility as a heuristic

Polar and ionic particles are often stabilized by polar solvents, while nonpolar solutes are often more compatible with nonpolar solvents. Crystal energy, size, shape, temperature, and measured data can override a one-label shortcut.

Compare old and new attractions
Supplied measurements outrank a slogan

Worked example

A supplied table classifies methanol as completely miscible with water, 1-butanol as moderately soluble, and 1-octanol as only slightly soluble. What conclusion is supported?

1
Hold the polar alcohol functional group constant across the homologous series.
2
Identify the controlled structural change: the nonpolar carbon chain becomes longer.
3
Use the supplied observations, not an absolute polarity claim, to state the trend.

ConclusionWithin this stated series, a larger nonpolar region is associated with lower water solubility; the table does not prove a universal rule for every molecule.

Close the notes first

Retrieve the model.

01What evidence identifies the solvent when physical state is not enough?

The stated dissolving medium and preparation context.

A liquid is not automatically the solvent, and solutions can exist in every bulk phase.

02Can a molecule with polar bonds be nonpolar overall?

Yes, when the bond-dipole vectors cancel in the molecular geometry.

Molecular polarity is a vector result rather than a count of polar bonds.

03What does ‘like dissolves like’ actually provide?

A qualitative attraction-matching prediction.

It does not supply a numerical solubility or guarantee complete miscibility.

LESSON 2 · 20 MIN

Study + retrieve

Colligative properties from effective particle count

Use an explicit ideal particle model to compare vapor pressure, boiling point, freezing point, and osmotic pressure.

ESSENTIAL QUESTIONHow many dissolved particles does the stated model treat as effective?

Effective-particle ledger

Ideal colligative directions for a nonvolatile solute
Property	Direction from pure solvent	Particle model
Solvent vapor pressure	Lower	P_solvent = X_solventP°_solvent
Boiling temperature	Higher	ΔT_b = iK_bm
Freezing temperature	Lower	ΔT_f = iK_fm
Osmotic pressure	Greater with effective concentration	Π = iMRT

Particle-factor and membrane checks
Evidence	Use	Guardrail
Nonelectrolyte remains molecular	i = 1	Only under the stated model
Ideal complete dissociation	Use supplied or formula-based i	Real solutions can show ion pairing
Semipermeable membrane	Track which species can cross	Tonicity needs permeability evidence

STUDY TABLE · CAPTION INCLUDED

Count effective particles

Colligative magnitude follows effective dissolved-particle concentration. Multiply the stated concentration by a supplied or idealized van ’t Hoff factor; do not assume formula subscripts equal a measured factor in real concentrated solutions.

Effective concentration = i × concentration
Use supplied i when available

Track the direction first

For an ideal solution with a nonvolatile solute, solvent vapor pressure falls, boiling temperature rises, and freezing temperature falls. A direction check should precede arithmetic.

ΔTb = iKbm
ΔTf = iKfm

Read the membrane model

An ideal semipermeable membrane passes solvent but not the stated solute. Solvent moves toward greater effective solute concentration unless an opposing pressure balances the tendency.

Π = iMRT
Permeability controls biological interpretation

Worked example

Find the freezing-point depression for ideal 0.200 m CaCl₂ using i = 3.00 and Kf = 1.86 K·kg/mol.

1
Confirm that the prompt supplies the ideal particle factor i = 3.00.
2
Form the effective molality: (3.00)(0.200 m) = 0.600 m particles.
3
Multiply by Kf: ΔTf = (1.86)(0.600) = 1.116 K, reported as 1.12 K.

ConclusionThe solution freezes 1.12 K below the pure solvent under the stated ideal model.

Close the notes first

Retrieve the model.

01Which quantity distinguishes 0.10 m glucose from ideal 0.10 m CaCl₂ in a colligative comparison?

The effective particle concentration i × m.

Glucose remains molecular while the stated ideal CaCl₂ model produces three ions per formula unit.

02What happens to boiling and freezing temperatures after adding a nonvolatile solute?

Boiling temperature rises and freezing temperature falls.

The solvent vapor-pressure lowering shifts the two phase-equilibrium conditions in opposite temperature directions.

03In the ideal osmosis model, what crosses the membrane?

The solvent crosses; the stated impermeable solute does not.

Net solvent movement responds to the effective solute-concentration difference.

LESSON 3 · 19 MIN

Study + retrieve

Non-colligative properties from equilibrium and identity

Classify saturation, read supplied solubility evidence, apply a stated Henry convention, and connect conductivity to mobile ions rather than particle count alone.

ESSENTIAL QUESTIONDoes this property depend on equilibrium amount, particle identity, or both?

Equilibrium and identity ledger

Classifying dissolved amount against the equilibrium limit
Composition at stated conditions	State	Expected response
Below solubility limit	Unsaturated	More solute can dissolve
At solubility limit	Saturated	Dissolution–precipitation equilibrium
Above solubility limit while dissolved	Supersaturated	Metastable; precipitation can occur

Identity-dependent property evidence
Property	Required evidence	Common trap
Solid solubility versus temperature	Supplied curve or data	Assuming every solid increases
Gas concentration	Stated Henry form, partial pressure, temperature	Using the wrong constant convention
Conductivity	Mobile ion concentration and mobility	Confusing strength with concentration

STUDY TABLE · CAPTION INCLUDED

Locate the solubility limit

Unsaturated, saturated, and supersaturated compare dissolved amount with equilibrium solubility at the stated conditions. Concentrated and dilute do not name saturation states.

Below · at · above the limit
A supersaturated solution is metastable

Let conditions control the data

A supplied curve controls solid-solubility questions. For many nonreacting gases, C = kP under the stated convention and fixed temperature; gas solubility commonly decreases when temperature rises.

Check temperature before reading a curve
Henry constants are convention-specific

Count mobile charge carriers

Conductivity depends on charged-particle concentration, mobility, identity, temperature, and apparatus. Strong electrolyte describes extensive ion formation, not a high sample concentration.

Strong can still be dilute
Conductivity is not colligative

Worked example

A solution holds 80 g solute per stated solvent basis at high temperature. After cooling, the supplied equilibrium solubility is 30 g on the same basis. How much precipitates at equilibrium?

1
Confirm that both amounts use the same solvent basis.
2
The cooled solution can retain 30 g dissolved at equilibrium.
3
Subtract the new limit from the initially dissolved amount: 80 g − 30 g = 50 g.

ConclusionFifty grams precipitate if equilibrium is reached and no solvent is lost.

Close the notes first

Retrieve the model.

01Can a dilute solution be saturated?

Yes.

A low-solubility solute can reach its equilibrium limit at a small absolute concentration.

02If a Henry problem states C = kP, what pressure belongs in the equation?

The gas’s partial pressure under the stated conditions.

The dissolved concentration is tied to that gas component, not automatically to total pressure.

03Why is conductivity non-colligative?

It depends on charge, mobility, concentration, and identity—not particle count alone.

Neutral dissolved particles and mobile ions can have very different electrical behavior.

LESSON 4 · 18 MIN

Study + retrieve

Dissolution forces with a three-step energy ledger

Inventory the interactions broken and formed during solution preparation, calculate a stated dissolution enthalpy, and preserve the entropy boundary.

ESSENTIAL QUESTIONWhat must separate, what forms, and what conclusion does the energy evidence actually support?

Dissolution-energy ledger

Three signed steps for solution formation
Step	Interaction change	Enthalpy sign
Separate solute	Disrupt solute–solute attraction	Positive
Separate solvent	Disrupt solvent–solvent attraction	Positive
Solvate	Form solute–solvent attraction	Negative
Net	ΔH_solution = ΔH_solute + ΔH_solvent + ΔH_solvation

New attraction inventory and inference limits
Particle pair	Supported attraction	Boundary
Ion + polar molecule	Ion–dipole	The ion already carries charge
Compatible donor + acceptor	Hydrogen bond	Requires structural evidence
Two polar molecules	Dipole–dipole + dispersion	Does not alone set solubility
ΔH_solution known	Heat direction	Spontaneity also needs entropy

STUDY TABLE · CAPTION INCLUDED

Separate both starting particle sets

Separating solute particles and separating solvent particles require energy in the standard ledger. Ordinary molecular dissolution changes interparticle attractions rather than cleaving every intramolecular covalent bond.

Solute separation: positive
Solvent separation: positive

Form the new attractions

Solvation releases energy as new solute–solvent attractions form. Ion–dipole, hydrogen-bond, dipole–dipole, and dispersion interactions require their corresponding structural evidence.

Solvation contribution: negative
Dispersion is always present

Stop at the evidence boundary

ΔHsolution is the signed sum of the three steps, but enthalpy alone does not establish spontaneity or equilibrium solubility. The complete free-energy balance also includes entropy.

Preserve every sign
ΔH alone is not ΔG

Worked example

Solute separation requires +15 kJ/mol, solvent separation requires +10 kJ/mol, and solvation releases 30 kJ/mol. Find ΔHsolution.

1
Write the two separation terms as +15 and +10 kJ/mol.
2
Write the energy-releasing solvation term as −30 kJ/mol.
3
Add with signs preserved: +15 + 10 − 30 = −5 kJ/mol.

ConclusionThe stated dissolution is exothermic by 5 kJ/mol; that sign alone does not prove unlimited solubility or spontaneity at every condition.

Close the notes first

Retrieve the model.

01Which two dissolution steps require energy in the standard ledger?

Separating solute particles and separating solvent particles.

Existing attractions must be disrupted before the particle sets can mix.

02What new attraction stabilizes an ion in a polar solvent?

Ion–dipole attraction.

The solvent dipole orients around the existing charge.

03Can ΔHsolution alone establish spontaneity?

No.

The free-energy balance also includes temperature and entropy.

LESSON 5 · 21 MIN

Study + retrieve

Concentration calculations by denominator

Choose the concentration basis from the stated quantities, perform molarity and dilution calculations, and distinguish solution from solvent denominators.

ESSENTIAL QUESTIONWhat exactly belongs in the denominator of this concentration?

Concentration-basis ledger

Choose the denominator named by the concentration unit
Unit	Numerator	Denominator
Molarity, M	Moles solute	Liters total solution
Molality, m	Moles solute	Kilograms solvent
Mole fraction, X	Component moles	Total moles
Mass percent	Component mass	Total solution mass
ppm	Part amount × 10⁶	Total amount on stated basis

Preparation and conversion checks
Task	Relationship	Required check
Prepare molarity	M = n/V_solution	Use final total volume
Simple dilution	M₁V₁ = M₂V₂	Same nonreacting solute
Mass–volume bridge	Use supplied density	Do not assume 1.00 g/mL
Percent or parts	Part ÷ whole	Preserve the named basis

STUDY TABLE · CAPTION INCLUDED

Fix the basis before calculating

Molarity uses moles of solute per liter of total solution. Molality uses moles of solute per kilogram of solvent. Mole fraction uses component moles divided by total moles.

M: L solution
m: kg solvent

Conserve solute during dilution

For the same nonreacting solute, M₁V₁ = M₂V₂ follows from conserved solute amount. The target volume is final solution volume, not the amount of solvent added.

Stock aliquot first
Dilute to the mark

Keep percent and parts explicit

Mass percent divides solute mass by total solution mass. Volume percent and parts-based units require their stated basis, while density is needed to bridge mass and volume descriptions.

Part divided by whole
Do not assume density without permission

Worked example

What stock volume is needed to prepare 400 mL of 0.250 M solution from 2.00 M stock?

1
The solute is unchanged, so apply M₁V₁ = M₂V₂.
2
Solve V₁ = (0.250 M)(400 mL)/(2.00 M).
3
Cancel molarity and calculate V₁ = 50.0 mL.

ConclusionTransfer 50.0 mL of stock and add solvent until the total solution volume is 400 mL.

Close the notes first

Retrieve the model.

01What volume appears in the molarity denominator?

The final volume of the entire solution.

Molarity is not defined using only the solvent volume used during preparation.

02What mass appears in the molality denominator?

Kilograms of solvent.

Molality excludes solute mass from its denominator.

03What does V₂ mean in a simple dilution calculation?

The final total solution volume.

It is not automatically the volume of solvent added.

All twenty Solutions problems

Choose the valid comparison.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

20 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions have not been calibrated to the official score scale. Use explanations to repair the exact component, particle, property, force, or denominator decision, then use balanced practice to interleave all five topics.

DAT TRAIN does not claim topic quotas because the official manual does not publish them.