Chemistry: Atoms First 2e · Composition of Substances and Solutions
Molarity
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In 30 seconds
A solution Homogeneous mixture of solute dissolved in solvent. Full entry → is a homogeneous mixture of a solute The substance being dissolved (usually the smaller amount). Full entry → (the substance dissolved) and a solvent The substance doing the dissolving (usually the larger amount). Full entry → (the substance doing the dissolving). To describe a solution quantitatively you need a concentration — how much solute per how much solution. The most widely used concentration unit in chemistry is molarity (M), the number of moles of solute per liter of solution:
M = moles of soluteliters of solution
The unit is mol/L, read "molar." A 1.0 M solution contains one mole of solute dissolved in enough solvent to make exactly one liter of solution (not one liter of solvent plus solute — the final volume is what matters). Molarity is the conversion factor that links solution volume to moles of solute, which makes it the workhorse unit for solution stoichiometry, titrations, dilutions, and virtually every lab preparation.
Why this matters
Molarity is how real chemistry is done. A hospital IV bag of 0.9% saline is a solution whose osmolarity must be tightly controlled; a pharmaceutical tablet's active ingredient is often assayed by dissolving it and titrating with a standard solution of known molarity. In the lab, "make 500 mL of 0.100 M NaCl" is a daily instruction — and getting the volume-to-solution distinction wrong (adding solute to 500 mL of water instead of diluting to 500 mL total) changes the concentration. Molarity also underlies the dilution Adding solvent to lower concentration while keeping solute moles constant. Full entry → math used everywhere from preparing buffers to diluting concentrated acid safely, and it is the bridge between the solution world and the reaction stoichiometry of the next chapter.
The college version
Core Concepts
Molarity is moles per liter of solution
M = nV
where n is moles of solute and V is liters of solution (final volume), not solvent. Rearranged, the same equation answers the two other questions that arise constantly:
n = M × V and V = nM
Note the lowercase convention: molarity is M (italic, the quantity); the unit is written mol/L or just "M" (e.g., 0.500 M NaCl). Molarity changes with temperature because volume changes with temperature, whereas molality (moles per kilogram of solvent, next topic) does not.
Preparing a solution from solid solute
To make a solution of known molarity from a solid: (1) calculate moles needed with n = M × V; (2) convert to mass with m = n × Mmolar; (3) dissolve the solid in a portion of solvent (less than the final volume) in a volumetric flask Glassware calibrated to hold one exact volume at the mark. Full entry →; (4) add solvent up to the calibration mark and mix. The final volume, not the volume of solvent added, defines the molarity.
Dilution: M1V1 = M2V2
Dilution adds solvent to a more concentrated solution. The moles of solute are unchanged, so the product of molarity and volume is constant:
M1 V1 = M2 V2
where subscript 1 is the concentrated stock solution and subscript 2 the diluted solution. This single equation handles the classic "how much stock do I need" calculation. (Note: diluting by adding water to a final volume is exact; adding a fixed amount of water is only approximate, because volumes are not strictly additive.)
Solution stoichiometry
Molarity converts volume → moles, which then feed the mole ratios of a balanced equation. The general path for any solution reaction:
volume × M⟶ moles of A mole ratio⟶ moles of B ÷ M or × Mmolar⟶ volume or mass of B
Titrations — measuring the volume of a standard solution needed to react completely with an unknown — are the classic application: at the equivalence point, the moles of titrant consumed match the unknown's moles according to the balanced equation.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| "1 L of solvent + solute" | 1 L of solution | Molarity is defined on final solution volume; dissolving solute in 1 L of water gives less than 1 L of solution and a different (lower) molarity. |
| Molarity M | Molality m | Molarity = mol solute/L solution (volume-based, temperature-dependent); molality = mol solute/kg solvent (mass-based, temperature-independent). |
| M1V1 = M2V2 | Using it when solute moles change | The dilution equation requires the same amount of solute before and after; it cannot be applied to a reaction where solute is consumed or produced. |
| "Dilution = adding a fixed volume of water" | Diluting to a final volume | Exact dilution means bringing the total volume to the target; adding a measured splash of water is approximate because volumes aren't additive. |
| Moles vs molarity | "Moles per liter" vs "moles" | 0.1 M is a concentration; 0.1 mol is an amount. Mixing them produces factor-of-volume errors. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Molarity is like the "sugariness" of lemonade: how many spoonfuls of sugar (solute) are in each big glass (liter) of the whole pitcher (solution). If you want sweeter lemonade, add more sugar to the pitcher, not more water. And if the pitcher is too sweet, you don't take sugar out — you add more water to make the whole pitcher bigger, spreading the same sugar through more glasses.
Worked example
Example 1: Calculating molarity from mass and volume
4.00 g of NaOH (molar mass 40.00 g/mol) is dissolved and diluted to 0.500 L of solution. What is the molarity?
First convert grams to moles:
n = mMmolar = 4.00 g40.00 g/mol = 0.100 mol
Then apply the definition:
M = nV = 0.100 mol0.500 L = 0.200 mol/L = 0.200 M
Dimensional analysis: g × (mol/g) = mol; mol ÷ L = mol/L ✓.
Example 2: Preparing a solution from solid solute
How many grams of NaCl (58.44 g/mol) are needed to prepare 250.0 mL of 0.100 M NaCl?
Convert mL to L, then moles, then mass:
n = M × V = 0.100 mol/L × 0.2500 L = 0.02500 mol
m = n × Mmolar = 0.02500 mol × 58.44 g/mol = 1.461 g
Procedure: weigh 1.461 g NaCl, dissolve in ~150 mL water in a 250-mL volumetric flask, then add water to the mark and mix. Dissolving in 250 mL of water directly would give a lower concentration than 0.100 M, because the salt adds volume to the solution.
Example 3: Dilution — M1V1 = M2V2
What volume of 6.00 M HCl is needed to prepare 250.0 mL of 0.500 M HCl?
M1 V1 = M2 V2
V1 = M2 V2M1 = 0.500 M × 0.2500 L6.00 M = 0.02083 L = 20.8 mL
Take 20.8 mL of the 6.00 M stock, add it to a 250-mL flask containing some water, then dilute to the mark. Safety note: always add acid to water, slowly and with stirring, to control the heat released — never the reverse.
Example 4: Solution stoichiometry (titration)
How many milliliters of 0.200 M NaOH are needed to neutralize 25.00 mL of 0.100 M HCl?
Balanced equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Moles of HCl:
nHCl = M × V = 0.100 mol/L × 0.02500 L = 0.002500 mol
Mole ratio 1:1, so nNaOH = 0.002500 mol. Volume of NaOH needed:
V = nM = 0.002500 mol0.200 mol/L = 0.0125 L = 12.5 mL
Key takeaways
- M = n/V with V = liters of solution; rearrange to n = M × V and V = n/M.
- Preparing a solution: dissolve solute in less than the final volume, then dilute to the mark — never add solute to the full solvent volume and call it done.
- Dilution: M1V1 = M2V2 because moles of solute are conserved.
- Dimensional analysis: L × (mol/L) = mol; mol ÷ (mol/L) = L.
- In solution stoichiometry, molarity is the volume→moles conversion factor; mole ratios come from the balanced equation.
- Molarity is temperature-dependent (volume-dependent); molality (next topic) is not.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
What is the molarity of a solution made by dissolving 0.250 mol of solute to a final volume of 0.500 L?
Show answer
M = 0.250/0.500 = 0.500 M.
How many grams of glucose (180.16 g/mol) are needed for 500.0 mL of 0.150 M solution?
Show answer
n = 0.150 × 0.5000 = 0.0750 mol; m = 0.0750 × 180.16 = 13.5 g.
To what final volume must you dilute 10.0 mL of 5.00 M stock to get a 0.250 M solution?
Show answer
V2 = M1V1/M2 = (5.00 × 0.0100)/0.250 = 0.200 L = 200 mL.
Why must you dilute "to the mark" in a volumetric flask instead of adding the full solvent volume first?
Show answer
The solute adds volume; molarity is defined on final solution volume, so you dissolve first and then bring the total to the exact mark — otherwise the concentration is lower than intended.
In the titration Adding standard solution until reaction with the unknown is complete. Full entry → HCl + NaOH → NaCl + H₂O, what volume of 0.100 M NaOH neutralizes 20.0 mL of 0.150 M HCl?
Show answer
nHCl = 0.150 × 0.0200 = 0.00300 mol = nNaOH (1:1); V = 0.00300/0.100 = 0.0300 L = 30.0 mL.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- solution
- Homogeneous mixture of solute dissolved in solvent.
- solute
- The substance being dissolved (usually the smaller amount).
- solvent
- The substance doing the dissolving (usually the larger amount).
- molarity, M
- Moles of solute per liter of solution, mol/L.
- dilution
- Adding solvent to lower concentration while keeping solute moles constant.
- volumetric flask
- Glassware calibrated to hold one exact volume at the mark.
- titration
- Adding standard solution until reaction with the unknown is complete.
Sources & references
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