Theory — Ways to Measure Concentration

Concentration says how much solute is dissolved in how much solution or solvent. Several units are in common use, each suited to a different purpose, and being able to move between them is a core skill.

The main concentration units

Molarity and molalityMolarity M = moles of solute / liters of solution
Molality m = moles of solute / kilograms of solvent
Molarity uses solution volume, molality uses solvent mass; molality does not change with temperature
Mole fractionMole fraction X = moles of one component / total moles of all components
All the mole fractions in a mixture add up to 1
Percent concentrationsMass percent (w/w) = (mass solute / mass solution) × 100
Mass-volume percent (w/v) = (mass solute in grams / volume solution in mL) × 100
Volume percent (v/v) = (volume solute / volume solution) × 100
Mass solution = mass solute + mass solvent

To use any of these you first convert a mass of solute to moles with the molar mass, moles equals mass divided by molar mass, and a mass of solvent such as water to moles the same way. Very dilute solutions are often given in parts per million, which for water is close to milligrams of solute per liter of solution.

Preparing a solution by dilution

Adding solvent to a concentrated stock spreads the same amount of solute through a larger volume, so the concentration falls while the moles of solute stay the same. That conservation of solute gives the dilution relation.

DilutionC₁ V₁ = C₂ V₂
To make volume V₂ at concentration C₂ from a stock of concentration C₁, use V₁ = C₂ V₂ / C₁ of stock and add solvent up to V₂

Determining a concentration by titration

An unknown concentration can be found by reacting it with a solution of known concentration, the titrant, added until the reaction is just complete at the equivalence point, often shown by an indicator changing color. For a reaction in which one mole of titrant reacts with one mole of the unknown, the moles are equal at that point.

Titration (one-to-one reaction)moles titrant = moles unknown  →  C_t × V_t = C_u × V_u
so  C_u = C_t × V_t / V_u
Read the titrant volume at the endpoint, then solve for the unknown concentration

Molarity vs molality

Molarity is easy to measure with glassware; molality is preferred when temperature changes, since it does not depend on volume.

Dilution keeps moles fixed

Only solvent is added, so the amount of solute is unchanged and C₁V₁ = C₂V₂ follows directly.

Endpoint and equivalence

A good indicator changes color essentially at the equivalence point, so the endpoint volume gives the unknown concentration.

UnitDefinitionNeeds
Molarity (M)mol solute / L solutionmoles, solution volume
Molality (m)mol solute / kg solventmoles, solvent mass
Mole fraction (X)mol part / total molmoles of each component
Mass percentmass solute / mass solution × 100masses
Mass-volume percentg solute / mL solution × 100solute mass, solution volume

Apparatus

The equipment a real solution-preparation experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

exact volume
Volumetric flask
Prepares solutions of known molarity to an exact volume.
measures volume
Graduated cylinder
Measures solvent volumes for dilutions.
0.000 gmeasures mass
Analytical balance
Weighs the solute for molarity and mass-percent.
transfers fixed volume
Volumetric pipette
Transfers exact volumes for serial dilutions.
holds solutions
Beaker
Mixes and holds the prepared solutions.
delivers titrant
Burette
Used to standardize a solution by titration.

Instructions — Running the Virtual Experiment

This is a record, calculate, and compare lab. In each part you set up or measure the solution, calculate the requested quantity by hand, enter your value, and only then does the simulation reveal its own result so you can compare. Record every value in your worksheet.

Part A — Concentration Units (Solution Composition tab)
1
Open Simulation → Solution Composition. Choose a solute, then set the mass of solute, the mass of water, and the solution volume. The readout shows these values and the moles of solute and water.
2
Pick a target unit (molarity, molality, mole fraction, mass percent, or mass-volume percent), calculate it by hand, enter your value, and click Check. The simulation reveals its value so you can compare. Repeat for all five units of the same solution and record them.
Part B — Dilution and Preparation (Dilution tab)
1
Open Dilution. Set a stock concentration, a target concentration, and a target volume.
2
Calculate the volume of stock needed from V₁ = C₂ V₂ / C₁, enter it, and click Check. The simulation reveals the stock volume and the amount of solvent to add, so you can compare and describe how you would prepare the solution.
Part C — Determination by Titration (Titration tab)
1
Open Titration. A 25.0 mL sample of an unknown acid is titrated with 0.100 M base and an indicator. Add base in steps, taking smaller steps near the color change, until the indicator turns and stays pink. That endpoint volume is the equivalence volume.
2
Calculate the unknown concentration from C_u = C_t V_t / V_u, enter it, and click Calculate & check to compare with the true value.
For your reportInclude your data tables, your worked calculations for each part, screenshots, and a short discussion comparing your calculated values with the simulation and explaining how you would prepare one of the solutions.

Simulation — The Solutions Bench

Concentration Virtual LabBuild or measure, calculate, then reveal and compare
UnitYour valueActual
No rows yet — choose a unit, calculate it, enter it, and check.

Build the solution

Given values
Molar mass solute58.44 g/mol
Mass solute20 g
Mass water200 g
Solution volume250 mL
Moles solute—
Moles water—
Molarity—

Stock and target

Preparation
Stock C₁6.0 M
Target C₂1.5 M
Target V₂250 mL
Stock volume V₁—
Solvent to add—
The flask holds 25.0 mL of unknown acid with indicator. The titrant is 0.100 M base. Add until the pink color persists.

Add base titrant

Burette and flask
Acid volume V_u25.0 mL
Base conc. C_t0.100 M
Base added V_t0.0 mL
Indicatorcolorless
Unknown C_udetermine from endpoint

Team Questions

Question 1. What are the units of molarity? (write it as mol/L)
Question 2. How many moles are in 58.44 g of NaCl (molar mass 58.44 g/mol)? (one number)
Question 3. Dissolving 0.50 mol of solute in water to make 2.0 L of solution gives what molarity? (one number)
Question 4. Which unit uses kilograms of solvent rather than liters of solution? (one word)
Question 5. To make 100 mL of 1.0 M from a 5.0 M stock, what volume of stock is needed? (one number, in mL)
Question 6. In a titration, 20.0 mL of unknown acid needs 25.0 mL of 0.100 M base. What is the acid concentration? (one number, in M)
Question 7 — Challenge. Do the mole fractions of all components in a mixture add up to what number? (one number)

Example Lab Report

A worked example showing the expected format and the record, calculate, and compare workflow.

Dilution, Titration and Molar Concentration Units

Chemistry | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Objective

To express the concentration of a solution in several units, to prepare a solution by dilution, and to determine an unknown concentration by titration, comparing every calculated value with the simulation.

Part A — Concentration Units (worked example)

Solution: 20.0 g NaCl (molar mass 58.44) in 200 g water, total volume 250 mL. Moles solute = 20.0 / 58.44 = 0.342 mol; moles water = 200 / 18.02 = 11.10 mol.

UnitCalculationValue
Molarity0.342 / 0.250 L1.369 M
Molality0.342 / 0.200 kg1.711 m
Mole fraction0.342 / (0.342 + 11.10)0.0299
Mass percent20 / 220 × 1009.09 %
Mass-volume percent20 / 250 × 1008.00 %

Part B — Dilution (worked example)

To make 250 mL of 1.5 M from a 6.0 M stock: V₁ = C₂ V₂ / C₁ = (1.5 × 250) / 6.0 = 62.5 mL of stock, then add water to a total of 250 mL (about 187.5 mL of water). This matched the simulation.

Part C — Titration (worked example)

A 25.0 mL acid sample turned pink at 30.0 mL of 0.100 M base. C_u = C_t V_t / V_u = (0.100 × 30.0) / 25.0 = 0.120 M, matching the simulation.

Discussion and Conclusion

Every calculated value agreed with the simulation. The five units describe the same solution from different viewpoints, dilution preserved the moles of solute while lowering the concentration, and the titration recovered the unknown concentration from the endpoint volume.

Practice Questions

Question 1
Dissolving 9.0 g of glucose (molar mass 180.16) in water to a total volume of 500 mL gives what molarity?
Hint: moles = 9.0 / 180.16 = 0.0500 mol; M = 0.0500 / 0.500 = 0.100 M.
Question 2
A solution contains 15 g of salt in 135 g of water. Find the mass percent.
Hint: mass solution = 150 g; mass percent = 15 / 150 × 100 = 10 %.
Question 3
How would you prepare 500 mL of 0.50 M solution from a 4.0 M stock?
Hint: V₁ = (0.50 × 500) / 4.0 = 62.5 mL of stock, then add water to 500 mL.
Question 4
A 25.0 mL sample of acid needs 22.0 mL of 0.200 M base to reach the endpoint. Find the acid concentration.
Hint: C_u = (0.200 × 22.0) / 25.0 = 0.176 M.
Question 5 — Challenge
Explain why molality, not molarity, is used when an experiment is run over a range of temperatures.
Hint: volume expands or contracts with temperature, changing molarity, but mass does not, so molality stays fixed.