Molarity and Solution Concentration Explained
Molarity is the most widely used way to express the concentration of a solution in chemistry. Whether you are titrating an acid in a general chemistry lab, preparing reagents for a biochemistry experiment, or scaling up a pharmaceutical formulation, understanding molarity is essential. It connects the amount of dissolved substance to the volume of solution in a way that makes stoichiometric calculations straightforward and practical.
This guide explains what molarity means, how to calculate it from scratch, how to perform dilution calculations, and how to prepare solutions accurately in the laboratory.
What Is Molarity?
Molarity is a measure of solution concentration defined as the number of moles of solute dissolved in one liter of solution. It is denoted by a capital M and expressed in units of mol/L (moles per liter).
Solute
The substance being dissolved (e.g., NaCl, HCl, glucose)
Solvent
The liquid doing the dissolving (usually water)
Solution
The homogeneous mixture of solute + solvent
A 1 M (one molar) solution of sodium chloride contains exactly one mole of NaCl (58.44 grams) dissolved in enough water to make one liter of total solution. Notice the critical distinction: molarity is based on the total solution volume, not the volume of solvent added. This means you cannot simply add one mole of solute to one liter of water, because the dissolved solute will increase the total volume slightly beyond one liter.
The Molarity Formula
Molarity Formula
M = n / V
M = Molarity (mol/L)
n = Moles of solute (mol)
V = Volume of solution (L)
Since the number of moles is calculated from mass divided by molar mass, you can expand the formula:
Expanded Molarity Formula
M = mass of solute (g) / [molar mass (g/mol) x volume of solution (L)]
This expanded form is the most practical version because laboratory measurements typically involve weighing a solid solute on a balance (giving mass in grams) and measuring solution volume with a volumetric flask (giving volume in liters or milliliters).
How to Calculate Molarity Step by Step
- Identify the solute and find its molar mass. Use the periodic table to add up the atomic masses of all atoms in the chemical formula. For NaCl: Na (22.99) + Cl (35.45) = 58.44 g/mol.
- Determine the mass of solute dissolved. This comes from the problem statement or from weighing the solute on a balance.
- Calculate the moles of solute. Divide the mass by the molar mass: n = mass / molar mass.
- Determine the total solution volume in liters. Convert milliliters to liters if necessary by dividing by 1,000.
- Divide moles by volume. M = n / V gives you the molarity.
Calculation Example
Sophie dissolves 11.7 grams of NaCl in water to make 500 mL of solution. What is the molarity?
Step 1: Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
Step 2: Moles = 11.7 / 58.44 = 0.200 mol
Step 3: Volume = 500 mL = 0.500 L
Step 4: Molarity = 0.200 / 0.500 = 0.400 M
The solution is 0.400 M NaCl (or 400 mM).
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Use CalculatorDilution Calculations
Dilution is the process of reducing a solution's concentration by adding more solvent. The total moles of solute remain the same before and after dilution — only the volume changes. This principle gives us the dilution equation:
Dilution Equation
M1 x V1 = M2 x V2
M1 = Initial molarity (concentrated solution)
V1 = Volume of concentrated solution used
M2 = Final molarity (diluted solution)
V2 = Final total volume
Dilution Example
A lab needs 250 mL of 0.1 M HCl, but only has a 6 M HCl stock solution. How much stock solution is needed?
Using M1V1 = M2V2:
6 x V1 = 0.1 x 250
V1 = 25 / 6 = 4.17 mL
Measure 4.17 mL of 6 M HCl and add water to bring the total volume to 250 mL.
Safety note: When diluting strong acids, always add acid to water, never water to acid. The heat generated can cause the water to boil and splash concentrated acid.
Practical Examples
Example 1: Marcus, General Chemistry Student
Marcus needs to prepare 1.00 L of 0.250 M glucose (C6H12O6) solution for an osmosis experiment.
Step 1: Calculate the molar mass of glucose.
C6H12O6 = (6 x 12.01) + (12 x 1.008) + (6 x 16.00) = 72.06 + 12.10 + 96.00 = 180.16 g/mol
Step 2: Calculate the mass needed.
mass = M x V x molar mass = 0.250 x 1.00 x 180.16 = 45.04 grams
Marcus weighs 45.04 grams of glucose, dissolves it in water in a 1-liter volumetric flask, and adds water to the 1.00 L mark.
Example 2: Leah, Biochemistry Researcher
Leah needs to make a series of dilutions from a 2.0 M NaCl stock for an enzyme kinetics assay: 1.0 M, 0.5 M, 0.25 M, and 0.125 M, each in 100 mL total volume.
For 1.0 M: V1 = (1.0 x 100) / 2.0 = 50 mL of stock, add water to 100 mL
For 0.5 M: V1 = (0.5 x 100) / 2.0 = 25 mL of stock, add water to 100 mL
For 0.25 M: V1 = (0.25 x 100) / 2.0 = 12.5 mL of stock, add water to 100 mL
For 0.125 M: V1 = (0.125 x 100) / 2.0 = 6.25 mL of stock, add water to 100 mL
Alternatively, Leah could perform serial dilutions: make the 1.0 M first, then dilute that 1:1 to get 0.5 M, dilute that 1:1 to get 0.25 M, and so on. Serial dilutions from each previous solution reduce pipetting errors for very small volumes.
Example 3: James, Quality Control Analyst
James receives a solution of unknown NaOH concentration. He titrates 25.00 mL of the unknown solution with 0.100 M HCl, requiring 31.25 mL of HCl to reach the endpoint.
Step 1: Calculate moles of HCl used.
moles HCl = 0.100 x 0.03125 = 0.003125 mol
Step 2: Since NaOH + HCl react in a 1:1 ratio, moles NaOH = 0.003125 mol
Step 3: Calculate molarity of NaOH.
M = 0.003125 / 0.02500 = 0.125 M
The unknown NaOH solution is 0.125 M.
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Use CalculatorOther Concentration Units
While molarity is the most common, several other concentration units appear in chemistry. Understanding when to use each one prevents confusion:
| Unit | Symbol | Definition | Best Used For |
|---|---|---|---|
| Molarity | M | mol solute / L solution | Lab work, stoichiometry |
| Molality | m | mol solute / kg solvent | Colligative properties |
| Normality | N | equivalents / L solution | Acid-base titrations |
| Mass percent | % w/w | g solute / 100 g solution | Industrial formulations |
| Parts per million | ppm | mg solute / L solution | Trace analysis, water quality |
| Mole fraction | X | mol solute / total mol | Vapor pressure, gas mixtures |
Molarity is the standard for most laboratory stoichiometry because it directly connects volume measurements to the number of moles. Molality is preferred when temperature effects must be excluded since it is based on mass rather than volume. Parts per million is the standard for environmental chemistry and water quality testing where concentrations are extremely low.
Preparing Solutions in the Lab
Accurate solution preparation requires careful technique. Follow this protocol for preparing a solution from a solid solute:
- Calculate the required mass of solute using mass = M x V x molar mass.
- Weigh the solute accurately on an analytical balance. For solutions requiring high precision, use a balance that reads to 0.0001 g.
- Transfer the solute to a volumetric flask of the desired final volume. Use a funnel and rinse all residue from the weighing paper and funnel into the flask with small amounts of solvent.
- Add solvent to approximately 80% of the final volume and swirl or gently mix until the solute is completely dissolved.
- Bring the volume to the calibration mark by adding solvent dropwise with a wash bottle. Read the meniscus at eye level for accuracy.
- Mix thoroughly by inverting the stoppered flask several times to ensure homogeneity.
For dilutions from a liquid stock solution, use a pipette to transfer the calculated volume of stock solution to the volumetric flask, then fill to the mark with solvent. Always use calibrated volumetric glassware rather than beakers or graduated cylinders for accurate molarity preparation.
Tips for Working with Molarity
- Always use the molar mass of the actual solute form. If you are using a hydrated salt like CuSO4 · 5H2O, the molar mass includes the water of hydration (249.69 g/mol, not 159.61 g/mol for anhydrous CuSO4). Using the wrong molar mass will give you the wrong molarity.
- Convert milliliters to liters before calculating. The most common arithmetic error in molarity problems is forgetting to convert mL to L. Dividing by 1,000 first prevents this mistake.
- Label all solutions immediately. Every solution container should be labeled with the solute, molarity, date of preparation, and your initials. Unlabeled solutions are a safety hazard and a waste of reagents.
- Account for solute purity. Laboratory-grade reagents are not always 100% pure. If a reagent bottle says 98% purity, you need to weigh 2% more to compensate. Divide the calculated mass by the purity fraction: adjusted mass = calculated mass / 0.98.
- Use volumetric flasks for accuracy. A 250 mL volumetric flask is accurate to within 0.12 mL, while a 250 mL graduated cylinder may only be accurate to within 2 mL. For precise molarity, always use volumetric glassware.
Common Mistakes to Avoid
- Confusing volume of solvent with volume of solution. Molarity uses the total solution volume, not the volume of solvent added. If you add 58.44 g of NaCl to exactly 1,000 mL of water, the total solution volume will be slightly more than 1,000 mL, making the true molarity slightly less than 1.000 M.
- Using mass instead of moles in the formula. Plugging grams directly into M = n/V gives a meaningless number. You must first convert the mass to moles by dividing by the molar mass.
- Forgetting to convert units. If volume is given in milliliters and you need liters for the molarity formula, divide by 1,000. If mass is in milligrams and you need grams, divide by 1,000. Unit conversion errors are the leading cause of incorrect molarity calculations.
- Ignoring hydration in hydrated salts. Many laboratory chemicals are sold as hydrates. Using the anhydrous molar mass instead of the hydrated molar mass for compounds like MgSO4 · 7H2O produces a significant error because the water molecules represent nearly half the mass of each formula unit.
- Rounding too early in multi-step problems. Carry extra significant figures through intermediate steps and only round the final answer to the appropriate number of significant figures. Premature rounding compounds errors across sequential calculations.
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Use CalculatorFrequently Asked Questions
Molarity (M) measures moles of solute per liter of solution, while molality (m) measures moles of solute per kilogram of solvent. The key distinction is the denominator: molarity uses the total solution volume, while molality uses the mass of the solvent only. Molality does not change with temperature because mass is independent of temperature, whereas molarity changes slightly as the solution volume expands or contracts with temperature. Molality is preferred for precise physical chemistry calculations like boiling point elevation and freezing point depression.
Yes, temperature affects molarity because liquids expand when heated and contract when cooled. Since molarity is defined as moles per liter of solution, the same solution becomes slightly less concentrated when heated (because the volume increases) and slightly more concentrated when cooled. For most laboratory work at room temperature, this effect is negligible. However, for precise analytical chemistry or industrial processes operating at extreme temperatures, the temperature dependence of molarity must be accounted for.
To convert from percent concentration (weight/volume) to molarity, use the formula M = (10 x % concentration x density) / molecular weight. For example, a 10% NaCl solution with a density of 1.07 g/mL has a molarity of (10 x 10 x 1.07) / 58.44 = 1.83 M. To go the other direction, multiply molarity by molecular weight, divide by (10 x density), and the result is the weight/volume percentage.
Adding more solute to an existing solution increases its concentration. To calculate the new molarity, add the moles of new solute to the original moles in solution and divide by the total volume. Be aware that adding solid solute may slightly change the total volume of the solution, though for dilute solutions this effect is usually small enough to ignore. For concentrated solutions, remeasure the final volume after dissolving the additional solute for an accurate molarity.
Molarity is dominant in laboratory chemistry because it directly relates solution volume to the number of moles of solute, making stoichiometric calculations straightforward. When you know the molarity and volume of a solution, you immediately know how many moles of reactant you are adding to a reaction. Since laboratory measurements are typically volumetric (using graduated cylinders, pipettes, and burettes), molarity connects seamlessly to the way chemists actually dispense and measure reagents.
Yes, molarity can be much greater than 1 M. Concentrated stock solutions in the lab are routinely high molarity. Concentrated hydrochloric acid is approximately 12 M, concentrated sulfuric acid is about 18 M, and concentrated nitric acid is roughly 16 M. The maximum possible molarity depends on the solubility of the solute at a given temperature. At some point, no more solute can dissolve, and you reach a saturated solution at its maximum molarity for that temperature.
Use the dilution equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution. To find the final molarity after dilution, rearrange to M2 = M1V1 / V2. For example, if you dilute 50 mL of 6 M HCl to a final volume of 500 mL, the resulting molarity is (6 x 50) / 500 = 0.6 M.
Sources & References
- Chemistry LibreTexts — Concentration of solutions textbook chapter: chem.libretexts.org
- Chemistry LibreTexts — Analytical chemistry concentration methods: chem.libretexts.org
- Khan Academy — Molarity and solution stoichiometry article: khanacademy.org
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Last updated: February 23, 2026