Lesson 9

Solutions and Concentration

Learn about solution preparation, molarity, dilution, and concentration units used in chemistry.

11 learning objectivesquantitative

The Dissolving Process and Like Dissolves Like

A solution is a homogeneous mixture of a solute (the substance dissolved) in a solvent (the substance doing the dissolving). At the molecular level, dissolving occurs when solute–solvent interactions are strong enough to overcome the solute–solute and solvent–solvent attractions.

The guiding principle is “like dissolves like”:

  • Polar solvents (like water) dissolve polar and ionic solutes because they can form ion-dipole or dipole-dipole interactions. NaCl dissolves in water; sugar (polar) dissolves in water.
  • Nonpolar solvents (like hexane) dissolve nonpolar solutes. Grease dissolves in oil but not in water.

When an ionic compound dissolves, it dissociates into its constituent ions, which are stabilized by a shell of water molecules (hydration). When a molecular compound dissolves, individual molecules disperse among solvent molecules.

Miscibility. Some liquids may be mixed in any proportions to yield solutions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol are completely miscible with water; two-cycle motor oil is miscible with gasoline. Two liquids that do not mix to an appreciable extent are called immiscible: gasoline, oil, benzene, carbon tetrachloride, and other nonpolar liquids are immiscible with water (separate layers form when poured into the same container). Two liquids of moderate mutual solubility are said to be partially miscible: bromine and water are a classic example. When the two are shaken and allowed to settle, the upper layer is water saturated with bromine (faintly orange), and the lower layer is bromine saturated with water. Each layer is itself a saturated solution. The distinction between miscibility and immiscibility is really one of extent — miscible pairs have infinite mutual solubility, immiscible pairs have very low (though not zero) mutual solubility, and partially miscible pairs sit between.

Go deeperWhy this matters: vitamins obey like-dissolves-like

Vitamins A, D, E, and K are largely nonpolar molecules: they dissolve in fat, accumulate in your body’s fatty tissue, and can build to harmful levels if chronically over-supplemented. Most B vitamins and vitamin C are polar: they dissolve in water, circulate in blood plasma, and a surplus of the readily excreted ones (vitamin C especially) leaves in the urine rather than building up in tissue. Water solubility is not a guarantee against storage, though: vitamin B12 is the notable exception, stashed in the liver in quantities that last years. Same rule, your kitchen sink: plain water slides off a greasy pan (polar solvent, nonpolar grime), while soap molecules bridge the two worlds with a polar head and a nonpolar tail. “Like dissolves like” is one of the few chemistry rules your body and your dishwashing both demonstrate daily.

Saturation and the Effect of Temperature

Solutions are classified by how much solute they contain relative to the maximum possible:

  • Unsaturated — contains less solute than the solubility limit. More solute can dissolve.
  • Saturated — contains the maximum amount of dissolved solute at a given temperature. Additional solute remains undissolved.
  • Supersaturated — temporarily contains more dissolved solute than the saturation limit. These are unstable; a seed crystal or disturbance can trigger rapid crystallization.

Temperature effects on solid solutes: For most solid solutes, solubility in water increases with temperature.

Temperature effects on gas solutes: Gas solubility in water decreases as temperature rises (which is why warm soda goes flat faster than cold soda, and why thermal pollution — warm-water discharge from power plants — reduces dissolved oxygen and stresses cold-water fish like trout).

Pressure effects on gas solutes — Henry’s law. Gas solubility increases as the partial pressure of the gas above the liquid increases. Carbonated beverages illustrate the relationship: the bottle is sealed under a relatively high pressure of CO2, saturating the beverage with dissolved CO2 at that pressure. When the bottle is opened the CO2 pressure drops, the solubility drops with it, and dissolved CO2 escapes as bubbles until the beverage equilibrates with atmospheric CO2 pressure (which is why open soda goes flat). For many gaseous solutes the relationship is directly proportional: Cg = kPg, where k depends on the gas–solvent pair and on temperature. This is Henry’s law: the quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.

Go deeperWhy this matters: supersaturation in your pantry

Honey is a naturally supersaturated sugar solution, which is why a jar left in the cupboard eventually crystallizes: one stray seed crystal or scratch gives the excess glucose its excuse to come out of solution. Reusable hand warmers exploit the same instability on demand: they hold supersaturated sodium acetate, and clicking the metal disc provides the nucleation trigger that crystallizes the whole pack, releasing heat. Gentle warming redissolves the solute and resets the trap. Supersaturation is not an exam curiosity; it is a storable, triggerable state of matter.

Molarity: The Primary Concentration Unit

Molarity (M) is the most commonly used concentration unit in chemistry. It is defined as the number of moles of solute per liter of solution:

M = mol solute ÷ L solution

Units: mol/L (often written as M). A 0.50 M NaCl solution contains 0.50 mol of NaCl per liter of total solution.

To calculate moles from molarity: mol = M × V (where V is in liters). Example: How many moles of HCl are in 250 mL of 0.10 M HCl? mol = 0.10 × 0.250 = 0.025 mol.

To prepare a solution of a given molarity: (1) Calculate the mass of solute needed using mol = M × V and mass = mol × molar mass. (2) Weigh the solute and transfer to a volumetric flask. (3) Add solvent to the mark to achieve the exact final volume.

Go deeperCommon mistake: per liter of solution, not of water

Molarity’s denominator is the final solution volume, not the volume of water you added. Dissolving 1 mol of NaCl in 1 L of water does not give a 1 M solution: the salt itself occupies space, so the final volume exceeds 1 L and the true concentration lands below 1 M.

This is exactly why volumetric flasks exist and why the lab procedure is dissolve first, then fill: add the solute to a partly filled flask, dissolve it completely, and only then add solvent up to the calibration line. The line, not the water you poured, defines the liter.

Go deeperTry it: prepare half a liter of 0.100 M saline

How many grams of NaCl (58.44 g/mol) do you need for 500.0 mL of 0.100 M solution? Work it, then check below.

Answer: moles needed = M × V = 0.100 mol/L × 0.5000 L = 0.0500 mol; mass = 0.0500 mol × 58.44 g/mol = 2.92 g. Weigh it, dissolve in less than 500 mL of water, transfer to a 500 mL volumetric flask, and fill to the line. Note the order of operations mirrors the definition: moles first (from M × V), grams second (from molar mass).

Dilution Calculations

Dilution is the process of adding more solvent to a solution to decrease its concentration. The amount of solute does not change — only the volume increases. This gives the dilution equation:

M⁠1V⁠1 = M⁠2V⁠2

where M⁠1 and V⁠1 are the initial molarity and volume, and M⁠2 and V⁠2 are the final molarity and volume. Example: What volume of 6.0 M HCl is needed to prepare 500 mL of 0.10 M HCl? V⁠1 = M⁠2V⁠2/M⁠1 = (0.10)(0.500)/(6.0) = 0.0083 L = 8.3 mL.

Serial dilutions involve repeated dilution steps. If you dilute by a factor of 10 three times in succession, the overall dilution factor is 10 × 10 × 10 = 1000. Serial dilutions are used when very low concentrations are needed and a single dilution would require impractically small volumes.

Go deeperWhy this matters: dilution is a daily act

Juice concentrate, cleaning products marked “dilute 1:10,” an americano (espresso stretched with hot water): each is M⁠1V⁠1 = M⁠2V⁠2 in disguise, the same solute redistributed through a larger volume. Labs run the equation constantly to make working solutions from concentrated stock.

One safety rule rides along with it: when diluting a concentrated acid, always add the acid to the water, never water to the acid. Dilution of concentrated sulfuric acid releases enough heat to flash-boil the first drops of added water and spatter acid; pouring acid slowly into water lets the larger volume absorb the heat safely.

Percent Composition and Parts per Million/Billion

Several concentration units express the ratio of solute to solution (or solvent) on a mass or volume basis:

  • Weight/weight percent (w/w%) = (mass solute ÷ mass solution) × 100%. Used for solid mixtures.
  • Volume/volume percent (v/v%) = (volume solute ÷ volume solution) × 100%. Used for liquid–liquid mixtures (e.g., alcohol in beverages).
  • Weight/volume percent (w/v%) = (mass solute in g ÷ volume solution in mL) × 100%. Common in medicine and biology.

For very dilute solutions:

  • Parts per million (ppm) = mg solute per kg solution (or mg/L for aqueous solutions since density ≈ 1 g/mL).
  • Parts per billion (ppb) = μg solute per kg solution. Used for trace contaminants in water and environmental analysis.
Go deeperWhy this matters: reading ppm and ppb in the news

These units are how public health speaks. Drinking-water regulators set lead action levels in the parts-per-billion range (on the order of 10 ppb); recommended fluoride sits near 0.7 ppm (0.7 mg per liter); a blood-alcohol reading of 0.08% is a weight/volume percent, 0.08 g of ethanol per 100 mL of blood. Grasping the scale matters: 1 ppb is one second in about 32 years. Trace amounts can still matter enormously (lead has no safe level), but a headline number means nothing until you check the unit and the basis behind it.

Molality and Mole Fraction

Two additional concentration units are useful when temperature-independent values are needed (since they are based on mass, not volume):

Molality (m) = moles of solute ÷ kilograms of solvent. Note: the denominator is mass of solvent, not total solution. Units: mol/kg or simply m.

Example: 1.5 mol NaCl dissolved in 2.0 kg of water gives m = 1.5/2.0 = 0.75 m.

Mole fraction (χ) = moles of one component ÷ total moles of all components. Mole fractions are dimensionless and always sum to 1. For a two-component solution: χsolute = nsolute/(nsolute + nsolvent).

Molality is especially important for colligative property calculations (freezing-point depression, boiling-point elevation), which depend on the number of dissolved particles rather than concentration by volume.

Go deeperWhy chemists keep a temperature-proof unit

Molarity has a hidden flaw: it is built on volume, and volume drifts with temperature. Warm a solution and it expands; the moles of solute have not changed, but the molarity has quietly fallen. For routine benchwork the drift is negligible, but for precise physical measurements it is a real error source.

Molality dodges the problem entirely: kilograms of solvent do not expand or contract. That temperature-independence is why molality, not molarity, anchors the freezing-point and boiling-point calculations in the colligative-properties topic ahead: those experiments exist precisely to change the temperature.

Converting Between Concentration Units

Converting between molarity, molality, mole fraction, and percent requires knowledge of solution density and/or molar masses. The general strategy:

  1. Assume a convenient amount — typically 1 L of solution (for molarity) or 1 kg of solvent (for molality).
  2. Calculate the mass of solute and solvent from the given unit.
  3. Use density to convert between mass and volume as needed.
  4. Compute the desired unit from the new quantities.

Molarity ↔ molality: Starting from 1 L of solution with known molarity M and density d: mass of solution = d × 1000 g. Mass of solute = M × molar mass. Mass of solvent = mass of solution − mass of solute. Then m = M ÷ (mass of solvent in kg).

The conversion is straightforward when density is known, but impossible without it (since volume depends on temperature).

Concentration Calculations: Method Selection and Common Mistakes

Choose methods by problem type instead of memorizing isolated formulas:

  • Moles from solution volume: use n = M·V (V in liters).
  • Simple dilution: use M1V1 = M2V2.
  • Serial dilution: apply dilution stepwise; each stage uses the previous stage as its new initial condition.
  • Composition units (ppm/ppb, %): confirm whether basis is mass/mass, volume/volume, or mass/volume.

Lab pitfalls that change answers: using mL directly in molarity equations without converting to liters, assuming additive volumes for non-dilute mixtures, and mixing temperature-dependent densities with room-temperature assumptions without stating it. Best practice: write units at every line and include one sentence about physical reasonableness (e.g., diluted solution must have lower concentration than stock).

Key Equations

Molarity
M = mol solute / L solution
Units: mol/L
Dilution
M⁠1V⁠1 = M⁠2V⁠2
Volumes can be in any unit as long as both match
Molality
m = mol solute / kg solvent
Temperature-independent; used for colligative properties
Mole Fraction
XA = nA / (nA + nB + …)
Dimensionless; all mole fractions sum to 1

Learning Objectives

After studying this topic, you should be able to:

  1. Define molarity and calculate the molar concentration of a solution
  2. Perform dilution calculations using M₁V₁ = M₂V₂
  3. Interconvert among concentration units (molarity, molality, mole fraction, mass percent)
  4. Describe how to prepare a standard solution of a given molarity from a solid solute
  5. Express solution concentration in mass percent (w/w, w/v) and volume percent (v/v)
  6. Describe the effect of temperature on the solubility of solids and gases
  7. Distinguish between saturated, unsaturated, and supersaturated solutions
  8. Calculate molality from mass of solvent and moles of solute
  9. Describe the dissolving process at the molecular level in terms of solute-solvent interactions
  10. Apply the "like dissolves like" principle to predict solubility of solutes in various solvents
  11. Express concentrations in parts per million (ppm) and parts per billion (ppb)

How-To Procedure

How to Prepare a Solution of Known Molarity

  1. Determine the target molarity and final volume of solution you need.
  2. Calculate moles of solute required: mol = M × V (with V in liters).
  3. Convert moles to grams using the molar mass of the solute.
  4. Weigh the calculated mass of solute on an analytical balance.
  5. Transfer the solute to a volumetric flask of the correct size.
  6. Add distilled water, swirl to dissolve, then fill to the calibration mark on the flask.
  7. Mix thoroughly by inverting the flask several times.

Worked Example

Dilution Calculation: Preparing a Standard Solution

Problem

You need 250.0 mL of 0.200 M NaOH. Your stock solution is 6.00 M NaOH. What volume of stock solution must you dilute?

Solution
  1. Identify knowns: M⁠1 = 6.00 M, V⁠1 = ? , M⁠2 = 0.200 M, V⁠2 = 250.0 mL = 0.2500 L.
  2. Apply the dilution equation: M⁠1V⁠1 = M⁠2V⁠2.
  3. Solve for V⁠1: V⁠1 = M⁠2V⁠2 / M⁠1 = (0.200 M)(0.2500 L) / (6.00 M) = 0.00833 L = 8.33 mL.
  4. Procedure: Measure 8.33 mL of 6.00 M NaOH with a pipet, transfer to a 250.0 mL volumetric flask, then add distilled water to the 250.0 mL mark.
Answer

Dilute 8.33 mL of 6.00 M NaOH to a total volume of 250.0 mL to obtain 0.200 M NaOH.

Test Your Understanding

A student prepares a 1.00 M NaCl solution by dissolving 58.44 g of NaCl in 1.00 L of water. Will the actual molarity be higher, lower, or exactly 1.00 M? Explain.

Practice Problems

calculation

What volume of 12.0 M HCl stock solution is needed to prepare 250.0 mL of 0.500 M HCl?

calculation

A solution is prepared by dissolving 20.0 g of glucose (C6H12O6, molar mass 180.16 g/mol) in 500.0 g of water. Calculate the molality of the solution.

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Self-Study Questions

What is a solution and what are its components?

What is molarity and how is it calculated?

What is the difference between a saturated, unsaturated, and supersaturated solution?

Hint: Think about the amount of dissolved solute relative to solubility.

What does “like dissolves like” mean?

What is molality and how does it differ from molarity?

What is mass percent composition of a solution?

What is dilution and what quantity stays constant during dilution?

What is a stock solution?

What is mole fraction?

How does temperature generally affect the solubility of solids versus gases?

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