Concentration & Molarity · Grade 11 Science
Chapter: Solutions & Dissolved ParticlesConcentration & 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.