Solutions & Dissolved Particles — Print · Physical Science: Chemistry, Grade 11
Study saturation, concentration, conductivity, and particle-count effects.
Printable chapter lessons and review.
Solutions & Solubility
Learning goals
- I can identify the solute and solvent in a solution and describe how dissolving happens at the particle level.
- I can explain what solubility means and predict how temperature and pressure affect it.
Readings for this unit
Before this unit
Mixtures, polarity, and saturation.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Does faster dissolving mean a solvent can hold more solute at equilibrium?
Illustrative evidence case. A stirred and an unstirred sample reach the same final dissolved mass at matched conditions.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Does faster dissolving mean a solvent can hold more solute at equilibrium?
Your first model
Draw time-dependent approach separately from the final equilibrium amount.
Build the language from the idea
- solution
- A homogeneous mixture in which one substance is dissolved evenly throughout another, uniform right down to the particle level.
- solute
- The substance that is dissolved in a solution, usually present in the smaller amount.
- solvent
- The substance that does the dissolving in a solution, usually present in the larger amount; water is the most common one.
- solubility
- The maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature, often given in grams per 100 g of water.
- saturated solution
- A solution that holds all the solute it can at a given temperature; any extra solute added will simply not dissolve.
A solution is a special kind of mixture — one so thoroughly blended that it looks completely uniform, right down to the level of individual particles. Stir a spoonful of sugar into hot tea and it seems to vanish; what has really happened is that the sugar has broken apart into separate molecules and spread evenly throughout the liquid. Every sip tastes equally sweet because the sugar is distributed the same everywhere. Chemists split a solution into two roles: the solute is the substance being dissolved (the sugar), and the solvent is the substance doing the dissolving (the water). Usually the solvent is simply whichever substance is present in the larger amount.
Water is a polar solvent. Its interactions with solute particles help explain why many ionic or polar substances dissolve, but polarity alone does not guarantee high solubility. Dissolution also involves separating solute particles and rearranging solvent interactions, with energy and entropy both contributing. The phrase like dissolves like is a useful first prediction rather than a law. Many oils have low water solubility; some ionic solids also have low water solubility because their overall balance differs.
When sodium chloride dissolves, water molecules interact with surface ions and hydrate ions that enter solution. The oxygen end of water tends to orient toward cations, and hydrogen ends toward anions. Dissolved ions remain charged and surrounded by interacting water molecules; they are not bare particles in empty space. Whether a whole crystal dissolves depends on its amount, solvent amount, temperature, and solubility. At saturation, further net dissolution is limited even though microscopic exchanges can continue.
There's a limit, though. At a given temperature, only so much solute will dissolve in a given amount of solvent — that maximum is the substance's solubility, usually stated as grams of solute per 100 grams of water. Keep adding solute past that point and the extra simply piles up undissolved at the bottom. A solution holding all it can is saturated; one that could still dissolve more is unsaturated. Under careful conditions you can even coax a solution into holding more than its normal limit — a delicate supersaturated state that is just waiting for a nudge to crystallize.
Temperature can change equilibrium solubility as well as dissolution rate. Many solids become more soluble as temperature rises, but use the supplied curve because some do not. At a specified gas partial pressure, many gases become less soluble as water warms; increasing the gas partial pressure often increases its dissolved concentration in a suitable dilute model. Stirring or grinding usually helps a solid reach equilibrium sooner without changing its bulk equilibrium solubility at fixed temperature. Heating must not be placed in that rate-only category.
Recognizing solutions and their limits is genuinely practical chemistry. It explains why the ocean can hold only so much dissolved oxygen for fish to breathe, why sprinkling salt melts the ice on a winter road, and why a doctor must dissolve a medicine to exactly the right strength before it enters your bloodstream. Once you can picture solute particles spreading among solvent molecules, a huge amount of everyday chemistry suddenly clicks into place.
Model check: Dissolving faster is not the same as having a larger equilibrium solubility. It can speed reaching equilibrium but does not generally change the equilibrium solubility.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Describe dissolving without losing species
A solution is a homogeneous mixture at the relevant scale. The solute is the dissolved component and the solvent is the medium, often but not always the larger amount. A molecular solute can remain molecular, while an ionic solid can dissociate into hydrated ions. Dissolving does not mean the solute loses mass or disappears from the species inventory. Some solutes also react with the solvent, so the actual dissolved species require a chemical model rather than simply shrinking a solid picture into invisible pieces.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can two clear solutions contain very different kinds of dissolved entities?
Visual reasoning — after your attempt
One can contain mainly intact solute molecules and another separated ions; similar appearance does not establish the same composition.
Dissolution disperses solute among solvent particles. Some solutes retain molecules and others dissociate into ions. A homogeneous solution is still a mixture. Dissolved substances do not disappear, and a clear appearance does not establish chemical purity.
Sugar molecules and salt ions require different particle drawings in water. Use actual solute identity before deciding how many dissolved particles are represented.
Pause and explain
Does dissolving always create ions?
Check your reasoning after answering
No. Molecular solutes can disperse as molecules, while suitable electrolytes produce ions. The particle model depends on the substance.
Part 2 · Balance competing interactions
To explain dissolution, consider interactions disrupted in the solute and solvent, interactions formed between them, and the entropy contribution under the stated conditions. Solubility is an equilibrium amount for a specified solvent, temperature and pressure, not simply how fast a stirred sample seems to disappear. Polar and ionic interactions can help water stabilize some solutes, but a strong crystal lattice can oppose dissolution. “Like dissolves like” is a useful heuristic whose limits are tested by this fuller account.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why does a strong attraction between water and ions not guarantee every ionic solid is highly soluble?
Visual reasoning — after your attempt
Separating the crystal and reorganizing the solvent also contribute to the full energy/free-energy balance; lattice stabilization can oppose dissolution.
Dissolving changes solute-solute, solvent-solvent, and solute-solvent interactions; energy and entropy both matter. Like dissolves like is a qualitative tendency rather than an absolute law. A simple polarity comparison does not supply an exact solubility value or account for chemical reaction with the solvent.
Water can stabilize suitable ions, but a strong crystal lattice can still limit a salt’s solubility. Consider both starting and final interactions.
Pause and explain
Why are some ionic compounds poorly soluble in water?
Check your reasoning after answering
Favorable ion-water interactions may not sufficiently offset the solid lattice and other thermodynamic contributions under the conditions.
Part 3 · Distinguish rate from equilibrium capacity
Dissolution rate is how quickly solute enters solution, while equilibrium solubility specifies a final capacity under stated conditions. Stirring and smaller particles can speed approach to saturation by improving transport or exposed area without changing the final equilibrium amount in the selected model. Temperature can affect both rate and equilibrium, with the direction depending on the system. A faster curve reaching the same plateau supports a rate difference, not a claim that stirring created extra solvent capacity.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What is the main difference between the two selected curves, and what stays the same?
Visual reasoning — after your attempt
The stirred model approaches the common 15 g plateau faster. The selected final equilibrium capacity is unchanged.
Stirring and reducing particle size often speed dissolution without increasing equilibrium solubility at the same temperature. Temperature effects depend on the substance; not all solids become more soluble when heated. Gas solubility also depends on pressure and often temperature.
A stirred sample can reach the same saturated amount sooner. Faster disappearance is rate evidence, not proof of a higher maximum dissolved amount.
Pause and explain
Does faster dissolution necessarily mean higher solubility?
Check your reasoning after answering
No. Rate concerns how quickly dissolution proceeds; solubility concerns equilibrium capacity under stated conditions.
Part 4 · Interpret saturation dynamically
A saturated solution is at the relevant solute equilibrium under stated conditions; with an excess solid phase present, dissolution and crystallization can continue at equal rates. An unsaturated solution can dissolve more under those conditions, while a supersaturated solution is metastable relative to the equilibrium and may crystallize after a disturbance. A clear solution’s status cannot be determined from appearance alone. State temperature, solvent amount and remaining solid before comparing a measured mass with a solubility reference.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does unchanged visible crystal mass prove that no solute particles move between solid and solution?
Visual reasoning — after your attempt
No. Equal dissolution and crystallization rates can keep the visible amount steady while exchange continues.
A saturated solution is in equilibrium with undissolved solute under suitable conditions. Dissolution and crystallization can continue at equal average rates. Supersaturation is a metastable excess dissolved amount that can crystallize when triggered; it is not the same as visible undissolved solid.
Use a supplied solubility curve to predict crystallization on cooling while stating complete equilibration and unchanged solvent assumptions.
Pause and explain
Why can a saturated solution remain dynamic?
Check your reasoning after answering
Particles can dissolve and crystallize continuously while their average rates balance and the dissolved amount stays steady.
Apply the whole unit · Use a supplied solubility change
Illustrative solubility of a named model solid is 40 g per 100 g water at 60°C and 25 g per 100 g water at 20°C. Initially 40 g is dissolved in 100 g water; neglect evaporation.
Your task
Predict solid formation after equilibrium cooling, describe particle exchange, and explain why stirring affects timing rather than this supplied final capacity.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
At 20°C the 100 g water can retain 25 g dissolved in the supplied model.
Excess=40−25=15 g, predicted to crystallize at equilibrium.
Keep solvent mass constant and distinguish final equilibrium from a possible temporarily supersaturated state.
The equilibrium prediction is 15 g crystallized and 25 g dissolved. A cooling sample may temporarily remain supersaturated, so an immediate observation need not match the final prediction. Stirring or seeding can affect timing but does not replace the supplied temperature-dependent equilibrium capacity. Evaporation or chemical change would require a new account.
Scientific references and source-model foundations
Parent / teacher teaching notes
Keep rate and capacity in separate columns. Ask students to state whether data are per mass of water, per mass of solution, or another basis. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
A solution forms when a solute spreads evenly through a solvent; solubility is the maximum that will dissolve at a given temperature, and it shifts with temperature, pressure, and the polarity of the substances.
Worked examples
Example 1
- The solvent is the substance present in the larger amount and does the dissolving — here that is the 250 g of water.
- The solute is the substance being dissolved, present in the smaller amount — here that is the 5 g of drink mix.
- The solution is the uniform mixture that results — the sweetened, colored water.
Example 2 · Worked application
- Capacity at that temperature is 30 g.
- 40−30=10 g exceeds that amount.
- Assume no reaction or water loss and enough time for equilibrium.
Solubility is temperature-specific.
Example 3
- Solubility of 64 g per 100 g of water is a ratio you can scale up.
- 250 g of water is 2.5 times as much as 100 g (250 ÷ 100 = 2.5).
- Multiply: 64 g × 2.5 = 160 g of KNO₃ will dissolve.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Solubility is temperature-specific.
- Stirring often changes dissolution rate; equilibrium solubility is set by substance and conditions.
Solubility is temperature-specific.
Example 5 · Reason from the visual model
- Dissolution rate is how quickly solute enters solution, while equilibrium solubility specifies a final capacity under stated conditions. Stirring and smaller particles can speed approach to saturation by improving transport or exposed area without changing the final equilibrium amount in the selected model. Temperature can affect both rate and equilibrium, with the direction depending on the system. A faster curve reaching the same plateau supports a rate difference, not a claim that stirring created extra solvent capacity.
- The stirred model approaches the common 15 g plateau faster. The selected final equilibrium capacity is unchanged.
Original illustrative approach-to-equilibrium records; the curves are not experimental fits and apply to the stated fixed-condition model.
Avoid these mistakes
- Dissolving faster is not the same as having a larger equilibrium solubility.
Check your understanding
Would stirring necessarily increase grams dissolved at saturation and fixed T?
Discuss the answer
It can speed reaching equilibrium but does not generally change the equilibrium solubility.
Return to the phenomenon
Return to the initial case. Does faster dissolving mean a solvent can hold more solute at equilibrium? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The curves can differ in rate while sharing the same capacity. Temperature, composition and remaining phases determine the equilibrium comparison.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Does faster dissolving mean a solvent can hold more solute at equilibrium? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The curves can differ in rate while sharing the same capacity. Temperature, composition and remaining phases determine the equilibrium comparison.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
At a supplied temperature, solubility is 20.0 g solute per 100 g water. Add 30.0 g to 100 g water and allow equilibrium with excess solid. Find dissolved and undissolved masses and explain what stirring changes.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Under the stated model, 20.0 g dissolves and 10.0 g remains solid. Stirring can speed equilibration but does not change this selected equilibrium capacity at fixed conditions.
Words to know
- solution
- A homogeneous mixture in which one substance is dissolved evenly throughout another, uniform right down to the particle level.
- solute
- The substance that is dissolved in a solution, usually present in the smaller amount.
- solvent
- The substance that does the dissolving in a solution, usually present in the larger amount; water is the most common one.
- solubility
- The maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature, often given in grams per 100 g of water.
- saturated solution
- A solution that holds all the solute it can at a given temperature; any extra solute added will simply not dissolve.
Solubility-product reasoning
For AgCl(s)⇌Ag⁺+Cl⁻, Ksp=[Ag⁺][Cl⁻] in a dilute concentration model. The solid is omitted while that phase is present. Compare the current ion product with Ksp: a larger product favors precipitation; a smaller one permits dissolution if solid is available. If only pure AgCl dissolves into pure water with equal ion concentration s, Ksp=s², so s=√Ksp. Adding a common ion changes that simple equality and usually lowers additional solubility. Temperature and activity matter; use supplied constants rather than assuming one value applies everywhere.
In the real world
Soft-drink makers dissolve carbon dioxide under high pressure to carbonate soda; when you open the can the pressure drops, the gas becomes far less soluble, and it fizzes back out of solution.
Exit check
Would stirring necessarily increase grams dissolved at saturation and fixed T?
Check your explanation after answering
It can speed reaching equilibrium but does not generally change the equilibrium solubility.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Concentration & Molarity
Learning goals
- I can calculate the molarity of a solution from moles (or grams) of solute and the volume of solution.
- I can use M₁V₁ = M₂V₂ to find the concentration or volume of a diluted solution.
Readings for this unit
Before this unit
Moles, liters, and solution versus solvent.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Can concentration fall while the amount of solute stays unchanged?
Illustrative evidence case. A selected solution is diluted to a larger final volume with no reaction or material loss.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Can concentration fall while the amount of solute stays unchanged?
Your first model
Track solute amount separately from the denominator used for concentration.
Build the language from the idea
- concentration
- A measure of how much solute is dissolved in a given amount of solution.
- molarity (M)
- Concentration expressed as the number of moles of solute per liter of solution (M = mol ÷ L).
- mole
- The SI unit of amount of substance, containing exactly 6.02214076×10²³ specified entities per mole.
- molar mass
- The mass of one mole of a substance in grams, found by adding the atomic masses from the periodic table.
- dilution
- Adding solvent to a solution to lower its concentration; the moles of solute stay the same, so M₁V₁ = M₂V₂.
Knowing that something is dissolved in water only tells you part of the story — the crucial follow-up question is *how much*. Concentration is the general word for how much solute is packed into a given amount of solution. A few crystals of drink mix in a full pitcher give a pale, weak drink; the same pitcher with ten scoops gives an intense, syrupy one. Same solute, same solvent — wildly different concentrations, and chemists need a precise way to capture that difference.
The concentration unit chemists reach for most is molarity, symbolized with a capital M. Molarity is defined as the number of moles of solute divided by the number of liters of solution: M = moles ÷ liters. A '1.0 M' solution ('one molar') contains one mole of solute in every liter of solution. Molarity is popular because reactions happen between *particles*, and a mole is really just a chemist's counting unit for particles — one mole is 6.022×10²³ of them — so molarity tells you how crowded the reacting particles are.
To use molarity you often have to start from a mass on a balance, not a number of moles, so you need one extra step: convert grams to moles by dividing by the substance's molar mass (the mass of one mole, read off the periodic table). For example, one mole of sodium chloride weighs 58.5 grams, so 117 grams of it is 117 ÷ 58.5 = 2.0 moles. Dissolve those 2.0 moles in enough water to make 1.0 liter of solution and you have a 2.0 M solution. Notice the careful wording — 'enough water to make 1.0 liter of *solution*,' not 'added to 1.0 liter of water.' The volume that matters is the final volume of the whole solution.
That distinction is exactly why chemists prepare precise solutions in a volumetric flask, a special glass flask with a single etched line on its neck. You weigh out the solute, dissolve it in some water, pour it into the flask, and then add water slowly until the liquid level reaches the line — guaranteeing an exact final volume and therefore an exact molarity. Eyeballing it in a random cup would never be accurate enough for real chemistry.
Often you don't need to build a solution from scratch because you already have a concentrated stock solution and simply need a weaker one. Adding solvent to lower a solution's concentration is called dilution, and it follows a wonderfully simple rule: M₁V₁ = M₂V₂. Because you are only adding water, the number of moles of solute never changes — so its concentration times its volume stays constant before and after. Dilute 50 mL of 6.0 M acid up to 300 mL and the new concentration is (6.0 × 50) ÷ 300 = 1.0 M. The very same equation works in reverse when you boil water *off* to concentrate a solution.
Getting concentration right isn't just lab housekeeping — it can be a matter of life and death. A hospital IV bag of saline is prepared to almost exactly 0.15 M sodium chloride so that it matches the concentration of fluid inside your cells; too strong or too weak and it would damage them. From brewing and cooking to medicine and water treatment, molarity is the language people use whenever 'how much is dissolved' truly matters.
Model check: Molarity uses final solution volume, not the volume of water added. M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Use solution volume in molarity
Molarity is mole amount of a specified dissolved solute divided by liters of final solution. It is not grams per liter or moles per liter of solvent before mixing. Dissolution can change volume, so preparation uses a defined final volume after the solute dissolves and temperature is appropriate. The same molarity can represent different species inventories if formulas dissociate differently. Write solute identity and units whenever stating a concentration rather than treating the symbol M as a generic amount.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can dissolving 0.100 mol in 0.500 L water fail to produce exactly 0.200 M solution?
Visual reasoning — after your attempt
The final solution volume need not equal the initial solvent volume. Molarity requires the actual final solution volume.
Molarity C=n/V uses moles of specified solute per liter of final solution. It does not use the initial solvent volume unless that equals the stated final volume. A concentration describes a ratio and is not itself total amount. State the species being counted.
0.100 mol in a final 0.500 L solution is 0.200 M. The same molarity in two liters contains four times the amount present in half a liter.
Pause and explain
Why is concentration not the same as total moles?
Check your reasoning after answering
Molarity is amount per volume. Total amount also depends on how much solution is present.
Part 2 · Convert mass and count before finding concentration
To obtain molarity from a weighed sample, divide its target-solute mass by molar mass, then divide the resulting amount by final solution liters. Include purity or hydration if those alter the weighed composition. For a selected NaCl model with complete dissociation, sodium and chloride each have the formula concentration, while total dissolved-ion concentration is twice that value. Activities and real nonideality are different from these ideal concentration counts. Keep formula amount, individual-ion amount and combined particle amount separate.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Is total ideal ion concentration in 0.400 M NaCl equal to 0.400 M?
Visual reasoning — after your attempt
No. The selected dissociation gives approximately 0.400 M Na⁺ plus 0.400 M Cl⁻, or 0.800 mol/L total ions.
If mass is supplied, divide by molar mass to find moles, then divide by solution liters. Electrolyte formula amount and ion amounts can differ by stoichiometric dissociation. Use the supplied ideal-dissociation approximation rather than assuming every solute remains one particle.
0.10 M CaCl₂ corresponds ideally to 0.10 M Ca²⁺ and 0.20 M Cl⁻ if fully dissociated and no other processes matter.
Pause and explain
Does 0.10 M CaCl₂ mean 0.10 M of every ion?
Check your reasoning after answering
No. One formula unit gives one calcium ion and two chloride ions in the full-dissociation model.
Part 3 · Derive dilution from conserved solute amount
Dilution adds solvent while retaining the same amount of the specified solute, assuming no reaction, precipitation or loss. From n=cV, this gives c₁V₁=c₂V₂. The final volume V₂ is the total diluted solution volume, not merely the added solvent volume. The same expression should not be used automatically for reacting acid and base solutions, because the named solute amount can change. Draw a before-and-after entity inventory to identify what is actually conserved.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why does doubling final volume halve concentration in this selected dilution?
Visual reasoning — after your attempt
The unchanged solute amount is distributed through twice the solution volume, so n/V halves.
For dilution without reaction or solute loss, initial and final moles are equal: C₁V₁=C₂V₂. Final volume includes the solution after dilution, not merely the added solvent. This relationship is an amount account and differs from stoichiometric reaction mixing.
Diluting 25.0 mL of 0.400 M stock to 100.0 mL gives 0.100 M. The solute amount stays 0.0100 mol.
Pause and explain
Why use final solution volume rather than added water volume?
Check your reasoning after answering
Concentration refers to total final solution volume. Added solvent volume is not automatically identical to it.
Part 4 · Distinguish concentration conventions and conditions
Molality uses moles of solute per kilogram of solvent, while molarity uses liters of solution. Mass fraction uses solute mass per total solution mass. These cannot be exchanged by renaming a unit; conversion can require solution density and composition. Temperature changes volume and therefore can alter molarity for the same material inventory, while mass-based definitions behave differently if nothing is lost. State the convention in a calculation, especially for colligative models that require molality rather than molarity.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What extra information commonly helps convert between mass-based and volume-based solution concentrations?
Visual reasoning — after your attempt
A suitable solution density, along with composition and temperature, connects mass and volume.
Mass percent, molarity, and molality use different denominators. Temperature can change volume and therefore molarity without changing solute amount. A solution recipe should specify the convention and conditions rather than use an unlabeled percent or a volume assumption.
10% by mass means 10 g solute per 100 g solution, not per 100 g water. Molality uses kilograms of solvent and will be used in later colligative models.
Pause and explain
What must accompany a percentage concentration label?
Check your reasoning after answering
State the numerator and denominator convention, such as mass/mass, volume/volume, or another defined basis.
Apply the whole unit · Dilute and account for ions
Illustrative stock: 0.400 M CaCl₂, aliquot 25.0 mL, diluted without reaction to final 100.0 mL. Assume ideal full dissociation.
Your task
Calculate retained formula moles, final formula molarity, and ideal calcium and chloride concentrations. Explain which quantity stays constant and which changes during dilution.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
n(CaCl₂)=0.400×0.0250=0.0100 mol.
Cfinal=0.0100/0.1000=0.100 M.
Ideal ion concentrations: Ca²⁺0.100 M,Cl⁻0.200 M.
The retained solute amount remains 0.0100 mol while concentration decreases as final solution volume increases. Ion concentrations follow formula ratios under the supplied full-dissociation approximation. Real ion interactions and temperature can complicate precise solution behavior; use the actual final solution volume rather than simply an assumed amount of added water.
Scientific references and source-model foundations
Parent / teacher teaching notes
Write amount and concentration on different lines. Require final-volume and dissociation assumptions before comparing formula and ion concentrations. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Molarity (M) measures concentration as moles of solute per liter of solution (M = mol ÷ L); to dilute a solution, use M₁V₁ = M₂V₂ because the moles of solute stay the same.
Molarity
M = moles of solute ÷ liters of solution
M is the concentration in mol/L (molar). To use a mass instead of moles, first convert grams to moles by dividing by the substance's molar mass.
Dilution
M₁V₁ = M₂V₂
M₁ and V₁ are the concentration and volume before dilution; M₂ and V₂ are after. It works because adding solvent changes the volume but never the number of moles of solute.
Worked examples
Example 1
- Molarity = moles of solute ÷ liters of solution.
- M = 2.0 mol ÷ 4.0 L.
- M = 0.50 M.
Example 2 · Worked application
- Use M₁V₁=M₂V₂ for conserved solute during dilution.
- V₁=0.20×0.50/1.0=0.10 L.
- Dilute to a final total volume of 0.50 L.
Final solution volume is not necessarily the volume of added water.
Example 3
- First convert grams to moles: moles = mass ÷ molar mass = 40.0 g ÷ 40.0 g/mol = 1.00 mol.
- Now apply molarity = moles ÷ liters = 1.00 mol ÷ 0.500 L.
- M = 2.00 M.
Example 4
- Use the dilution equation M₁V₁ = M₂V₂, solving for M₂ = M₁V₁ ÷ V₂.
- M₂ = (6.0 M × 50.0 mL) ÷ 300 mL = 300 ÷ 300.
- M₂ = 1.0 M.
Example 5 · Find and fix the mistake
- Test the claim against the stated conditions: Final solution volume is not necessarily the volume of added water.
- The target is total solution volume; solute and stock already occupy part of it.
Final solution volume is not necessarily the volume of added water.
Example 6 · Reason from the visual model
- Dilution adds solvent while retaining the same amount of the specified solute, assuming no reaction, precipitation or loss. From n=cV, this gives c₁V₁=c₂V₂. The final volume V₂ is the total diluted solution volume, not merely the added solvent volume. The same expression should not be used automatically for reacting acid and base solutions, because the named solute amount can change. Draw a before-and-after entity inventory to identify what is actually conserved.
- The unchanged solute amount is distributed through twice the solution volume, so n/V halves.
Original equal-count dilution model; dot spacing and solvent volume are schematic rather than a molecular-scale simulation.
Avoid these mistakes
- Molarity uses final solution volume, not the volume of water added.
Check your understanding
Find M for.20 mol dissolved to a final.50L.
Discuss the answer
M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs.
Return to the phenomenon
Return to the initial case. Can concentration fall while the amount of solute stays unchanged? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Dilution preserves the named solute amount but increases final solution volume. Concentration falls because n/V decreases, not because solute particles vanish.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Can concentration fall while the amount of solute stays unchanged? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Dilution preserves the named solute amount but increases final solution volume. Concentration falls because n/V decreases, not because solute particles vanish.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
Prepare 250. mL of 0.200 M solution by diluting a 1.00 M stock without reaction or loss. Find stock volume and explain whether 250. mL is added water volume.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.
Words to know
- concentration
- A measure of how much solute is dissolved in a given amount of solution.
- molarity (M)
- Concentration expressed as the number of moles of solute per liter of solution (M = mol ÷ L).
- mole
- The SI unit of amount of substance, containing exactly 6.02214076×10²³ specified entities per mole.
- molar mass
- The mass of one mole of a substance in grams, found by adding the atomic masses from the periodic table.
- dilution
- Adding solvent to a solution to lower its concentration; the moles of solute stay the same, so M₁V₁ = M₂V₂.
The mole: chemistry's dozen, but astronomically bigger
A mole is simply a counting word, like 'dozen' or 'pair,' but for an unimaginably large number of particles: 6.022 × 10²³, known as Avogadro's number. Chemists need such a huge count because atoms are so tiny — a single drop of water contains more molecules than there are stars in the observable universe. The mole is defined so that one mole of a substance has a mass in grams equal to its atomic or molecular mass in atomic mass units, which is the bridge that lets you weigh out a known number of particles on an ordinary balance. Molarity uses moles precisely because reactions happen particle-by-particle.
Other ways to say 'how much': molality, ppm, and percent
Molarity is the most common concentration unit, but it isn't the only one. Molality measures moles of solute per kilogram of solvent (not solution) — useful because, unlike volume, mass doesn't change with temperature. Percent by mass states grams of solute per 100 g of solution, common on food and cleaning labels. And parts per million (ppm) or parts per billion (ppb) describe extremely dilute solutions, such as pollutants or fluoride in drinking water. Each unit exists because different situations care about different things — temperature stability, everyday convenience, or vanishingly small trace amounts.
The volumetric flask and the art of a precise solution
Preparing a solution of an exact molarity is a genuine laboratory skill. Chemists use a volumetric flask, a pear-shaped glass vessel with a single etched calibration line on its narrow neck. You weigh the solute precisely, dissolve it in a little water, transfer it to the flask, and then add water drop by drop until the bottom of the curved liquid surface — the meniscus — just touches the line at eye level. Reading the meniscus from the wrong angle introduces 'parallax error.' This care matters because in analytical chemistry, medicine, and manufacturing, being off by even a percent can ruin an experiment or a dose.
Dilution everywhere: from IV bags to frozen juice
The equation M₁V₁ = M₂V₂ works because dilution only adds solvent — the number of moles of solute never changes, so concentration times volume stays constant. This one relationship quietly runs a huge amount of daily life. A hospital pharmacist dilutes a concentrated stock drug to a safe deliverable dose; a swimming-pool technician dilutes concentrated chlorine; and when you add three cans of water to frozen juice concentrate, you're performing a dilution calculation without knowing it. Learn the equation once and you can predict all of them.
Parts per million and the vanishingly small: lead in water
Some substances matter at concentrations far too small for molarity to describe conveniently, so scientists switch to parts per million (ppm) — one milligram of solute in one kilogram of solution, roughly one drop in a large bathtub. Safety standards for contaminants live in this realm: the action level for lead in drinking water is measured in parts per billion, and even those trace amounts can harm developing brains. The lesson is that 'a tiny concentration' is not the same as 'harmless' — chemistry cares about the identity of the solute as much as the amount, a theme that returns with strong versus weak acids.
In the real world
Pharmacists and nurses mix medicines and IV fluids to an exact molarity — a saline drip is close to 0.15 M salt so that it matches the fluid inside your cells and doesn't damage them.
Exit check
Find M for.20 mol dissolved to a final.50L.
Check your explanation after answering
M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Electrolytes & Colligative Properties
Learning goals
- I can distinguish molality from molarity and ions from neutral solute particles.
- I can calculate ideal phase-point changes and state the model’s limits.
Readings for this unit
Before this unit
Moles, concentration, phase changes, and ions.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why can equal mole amounts of two solutes shift freezing differently?
Illustrative evidence case. One selected solute remains molecular while another dissociates into several ions.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why can equal mole amounts of two solutes shift freezing differently?
Your first model
Count effective dissolved entities and identify the solvent-mass denominator.
Build the language from the idea
- molality
- Moles solute per kilogram of solvent.
- particle factor i
- Effective number of dissolved particles per solute formula unit in a model.
- osmosis
- Net solvent flow across a selectively permeable membrane.
A solution’s behavior depends on both particle identity and particle count. Ionic solutes may dissociate into ions; molecular solutes such as sugar can remain neutral molecules. A strong electrolyte produces many mobile ions and conducts; a weak electrolyte ionizes partially. Solubility and electrical conductivity answer different questions.
Molality b is moles of solute per kilogram of solvent, unlike molarity which uses liters of solution. Molality is useful when studying temperature-dependent properties because solvent mass does not expand with temperature. Convert grams of solvent to kilograms before dividing.
In a dilute ideal model, adding nonvolatile solute lowers vapor pressure and changes freezing and boiling points. ΔTf = iKf b and ΔTb = iKb b; i is the effective dissolved-particle factor. Sugar has i about 1, whereas complete ideal dissociation of NaCl gives about 2. Real solutions can depart from ideal values.
The freezing point is the pure-solvent freezing temperature minus ΔTf, while boiling point is pure-solvent boiling temperature plus ΔTb. For water near ordinary pressure, Kf ≈1.86°C·kg/mol and Kb≈0.512°C·kg/mol. State pressure, dilute-solution assumptions, and solute choice.
Osmosis is net solvent movement across a selectively permeable barrier toward the side with lower solvent chemical potential, often the more concentrated solute side in a simple comparison. Osmotic pressure in an ideal dilute solution is π=iMRT. Use consistent pressure units with R and kelvin temperature.
Road salt works only within practical temperature and concentration limits and can damage soils and waterways. Do not infer that twice the nominal salt always doubles the real effect. Safe paper calculations explain particle-count models; measurements should compare conditions and include limitations.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Count independent dissolved particles
An electrolyte produces mobile ions in a suitable solution. Introductory colligative properties depend on effective dissolved-particle amount under an approximate solution model rather than only the solute’s formula mass. The van’t Hoff factor i expresses an effective particle ratio: ideal complete NaCl dissociation suggests two, and CaCl₂ three. Ion association and nonideal interactions can make measured effective values differ, so formula dissociation counts are a starting approximation, not exact experimental constants.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can one mole of CaCl₂ have a larger ideal particle effect than one mole of glucose?
Visual reasoning — after your attempt
The selected complete dissociation produces three moles of ions per mole of CaCl₂, while glucose stays as one mole of intact molecules.
Colligative effects depend on dissolved-particle amount in suitable dilute ideal models. Electrolyte dissociation can increase particle count; association and nonideality can make the effective van ’t Hoff factor differ from an integer. Specify whether formula units or effective dissolved particles are being counted.
Ideal NaCl gives approximately two ions per formula unit; glucose remains one molecule. This does not mean every real NaCl solution has exactly i=2.
Pause and explain
Why can a measured electrolyte factor differ from the ideal count?
Check your reasoning after answering
Ion interactions, association, and nonideal behavior can change the effective particle contribution relative to complete independent dissociation.
Part 2 · Use molality for temperature-shift models
For a dilute idealized solution with the specified nonvolatile solute behavior, freezing-point depression is approximated by ΔTf=iKf b, where b is molality and Kf is a solvent-specific constant. Boiling-point elevation uses a corresponding Kb under its own conditions. A positive depression magnitude means the freezing temperature is lower than the pure solvent’s reference. The model does not mean every solute produces the same effect at equal grams or that concentrated mixtures follow the dilute equation exactly.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why would substituting 0.200 mol/L directly for b be unjustified without more information?
Visual reasoning — after your attempt
The required b is mol/kg solvent. Converting from molarity needs density/composition information rather than a changed unit label.
Molality is moles solute per kilogram solvent. Introductory freezing depression and boiling elevation use ΔTf=iKfm and ΔTb=iKbm with supplied solvent constants. These differ from molarity because volume is not the denominator. State dilute, nonvolatile-solute assumptions where required.
0.10 mol solute in 0.50 kg water gives 0.20 mol/kg, not 0.10 M unless a final solution volume is supplied.
Pause and explain
Why cannot molarity be substituted automatically for molality?
Check your reasoning after answering
Their denominators differ: liters of solution versus kilograms of solvent. They are only approximately related under suitable conditions.
Part 3 · Explain effects through phase equilibrium
A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What does the slope describe in the selected linear model, and why should it not be extrapolated without limit?
Visual reasoning — after your attempt
It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.
A suitable nonvolatile solute lowers solvent vapor pressure in an ideal mixture and changes phase-equilibrium temperatures. The molecular explanation concerns solvent chemical potential and particle mixing, not heavier water molecules or a permanent change in the solvent identity. Exact behavior needs a suitable solution model.
A freezing-point depression is a change in when solvent-rich solid can equilibrate with solution. Solute concentration can change as freezing proceeds, so one initial shift is not an entire freezing history.
Pause and explain
Does solute make each water molecule heavier?
Check your reasoning after answering
No. It changes the solution’s thermodynamic and particle environment, not the mass or identity of each solvent molecule.
Part 4 · Keep osmotic pressure and membrane conditions explicit
Osmosis is net solvent movement across a selectively permeable membrane in response to chemical-potential differences. Under equal-pressure simple conditions with effectively retained solute, solvent tends toward the higher effective solute side. An opposing pressure can balance that tendency; dilute ideal osmotic pressure is approximated by π=i cRT with appropriate concentration and temperature units. A membrane permeable to the solute changes the long-term account, so “water always moves toward more solute” is incomplete without pressure and permeability conditions.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What can stop net solvent flow even when the two sides have different retained-solute concentrations?
Visual reasoning — after your attempt
A suitable opposing pressure difference can balance the solvent chemical-potential difference; membrane permeability and conditions must be stated.
For a dilute ideal solution, osmotic pressure can be modeled as π=iCRT with consistent units. Osmosis involves a selectively permeable membrane and solvent transfer. Without a suitable membrane or concentration difference, the simple osmotic setup is not present.
Use a supplied membrane that passes solvent but not the solute. A membrane allowing both freely requires a different account.
Pause and explain
Why does membrane selectivity matter?
Check your reasoning after answering
The osmotic model requires different permeability for solvent and relevant solute. Free passage of both changes the conditions for sustained osmotic behavior.
Apply the whole unit · Compare ideal freezing-point depressions
Illustrative solutions each have molality 0.100 mol/kg in water. A is glucose with i=1; B is NaCl with supplied ideal i=2. Use Kf=1.86°C·kg/mol.
Your task
Calculate depression for each, predict initial freezing temperatures relative to pure water at 0°C, and explain why the comparison is an ideal particle model rather than an exact guarantee for all concentrations.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
A: ΔTf=1×1.86×0.100=0.186°C.
B: ΔTf=2×1.86×0.100=0.372°C.
Initial predicted freezing temperatures are −0.186°C and −0.372°C.
The ideal NaCl prediction has twice the depression because its effective particle factor is supplied as two. Real association and concentration effects can alter that factor, and continuing freezing changes the liquid concentration. State molality and solvent conditions; do not compare molarity records as though their denominators were identical.
Scientific references and source-model foundations
Parent / teacher teaching notes
Ask students to count particles and identify the denominator before substituting. Separate a temperature shift from the direction of heat transfer. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Colligative changes depend approximately on the number of dissolved particles, not merely solute grams.
Worked examples
Example 1
- b=.10/.50=.20 mol/kg.
- Sugar i≈1.
- ΔTf=1×1.86×.20.
Example 2 · Worked application
- Colligative change scales with particle molality i×m.
- Glucose gives 0.10 particle-molality units; NaCl gives 0.20.
- The ideal NaCl effect is twice as large, with real nonideality possible.
Use molality, not molarity, for the standard freezing-point relation.
Example 3
- Sugar gives one particle per molecule.
- NaCl gives about two ions per formula unit.
- Use ΔTf proportional to i.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Use molality, not molarity, for the standard freezing-point relation.
- Dissociation changes the number of dissolved particles; use the supplied van't Hoff factor and model limits.
Use molality, not molarity, for the standard freezing-point relation.
Example 5 · Reason from the visual model
- A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.
- It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.
Avoid these mistakes
- Molarity and molality are different; ΔTf is a change, not the final freezing temperature.
Check your understanding
Why compare equal mole amounts rather than equal grams for colligative effects?
Discuss the answer
Different molar masses and dissociation factors produce different dissolved-particle counts per gram.
Return to the phenomenon
Return to the initial case. Why can equal mole amounts of two solutes shift freezing differently? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The dilute colligative model uses effective particle amount, not just formula moles or grams. Dissociation, association and phase-equilibrium conditions determine its useful scope.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why can equal mole amounts of two solutes shift freezing differently? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The dilute colligative model uses effective particle amount, not just formula moles or grams. Dissociation, association and phase-equilibrium conditions determine its useful scope.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
Using Kf=1.86 K·kg/mol, compare ideal freezing depressions for 0.100 mol glucose and 0.100 mol CaCl₂, each in 0.500 kg water. Assume i=1 and 3 respectively.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.
Words to know
- molality
- Moles solute per kilogram of solvent.
- particle factor i
- Effective number of dissolved particles per solute formula unit in a model.
- osmosis
- Net solvent flow across a selectively permeable membrane.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
Why compare equal mole amounts rather than equal grams for colligative effects?
Check your explanation after answering
Different molar masses and dissociation factors produce different dissolved-particle counts per gram.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
Solute = sugar; solvent = water.
The solvent is the substance present in the larger amount that does the dissolving — the 200 g of water. The solute is the substance being dissolved, present in the smaller amount — the 8 g of sugar.
Problem 2
Answer and explanation
96 g
150 g of water is 1.5 times 100 g (150 ÷ 100 = 1.5). Multiply the solubility: 64 g × 1.5 = 96 g of KNO₃ will dissolve.
Problem 3
Answer and explanation
14 g
Only 36 g of NaCl can dissolve in 100 g of water at this temperature (the solution becomes saturated). So 50 g − 36 g = 14 g of NaCl remains undissolved.
Problem 4
Answer and explanation
0.25 M
Molarity = moles of solute ÷ liters of solution = 0.50 mol ÷ 2.0 L = 0.25 M.
Problem 5
Answer and explanation
1.0 M
Use the dilution equation M₁V₁ = M₂V₂, solving for M₂ = M₁V₁ ÷ V₂ = (4.0 M × 25.0 mL) ÷ 100.0 mL = 100 ÷ 100 = 1.0 M.
Problem 6
Answer and explanation
2.00 M
First convert grams to moles: moles = mass ÷ molar mass = 20.0 g ÷ 40.0 g/mol = 0.500 mol. Then molarity = moles ÷ liters = 0.500 mol ÷ 0.250 L = 2.00 M.
Problem 7
Answer and explanation
Final volume = 150 mL, so you must add 100 mL of water.
Use M₁V₁ = M₂V₂ to find the final volume: V₂ = M₁V₁ ÷ M₂ = (6.0 M × 50.0 mL) ÷ 2.0 M = 300 ÷ 2.0 = 150 mL. Since you started with 50.0 mL, the water added is 150 mL − 50 mL = 100 mL.
Problem 8
Answer and explanation
It can speed reaching equilibrium but does not generally change the equilibrium solubility.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 9
Answer and explanation
Under the stated model, 20.0 g dissolves and 10.0 g remains solid. Stirring can speed equilibration but does not change this selected equilibrium capacity at fixed conditions.
Under the stated model, 20.0 g dissolves and 10.0 g remains solid. Stirring can speed equilibration but does not change this selected equilibrium capacity at fixed conditions.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 10
Answer and explanation
M=n/V=.40 mol/L. Dilution lowers M without changing solute amount if no reaction or loss occurs.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 11
Answer and explanation
V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.
V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 12
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
0.372 °C
2×1.86×.10.
Problem 13
Answer and explanation
Molarity uses solution volume; molality uses solvent mass in kilograms.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 14
Answer and explanation
Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.
Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.