Molarity:
how much solute is in each litre of solution

The formula and its three forms

A solution with a molarity of 0.50 mol/L contains half a mole of dissolved substance in every litre. If 0.75 mol of solute ends up in 1.5 L of solution, dividing gives exactly that: 0.50 mol/L, often written 0.50 M.

Molarity is the working unit of most laboratory chemistry. Reaction equations speak in moles, and pipettes and flasks measure volume, so molarity is the bridge that lets you say how many moles are in a measured scoop of liquid.

Rearranged, the same relation gives moles from a volume and concentration (n = M × V) or the volume needed to carry a given number of moles (V = n ÷ M). Keep the volume in litres; millilitres must be divided by 1,000 first.

From grams on the balance to a molarity

Most recipes are written in grams, so the usual route has one extra step: convert mass to moles with the molar mass.

In practice, you weigh the solid, dissolve it in less than the target volume of water, then top up with solvent to the calibration mark of a volumetric flask. That is why the final volume, not the initial water volume, is the one in the formula.

Diluting a stock solution

When you dilute, adding solvent does not change the number of moles; it only spreads them over more volume. That gives the dilution relation M1 × V1 = M2 × V2.

Say you need 250 mL of a 0.50 M solution and your stock is 2.0 M. Then V1 = (0.50 × 250) ÷ 2.0 = 62.5 mL. You would measure 62.5 mL of stock into a flask and top up with water to 250 mL, which means adding about 187.5 mL of water.

Molarity beside its neighbours

Molarity depends on temperature because liquids expand when heated, so the same solution is slightly less concentrated on a warm day. For routine bench work that effect is small, but for precise work chemists switch to molality.

From a percentage label to molarity

Reagent bottles list concentration as a mass percentage and a density, not as moles per litre. To get molarity, multiply the density in g/mL by 1,000 to find grams of solution per litre, multiply by the mass fraction to get grams of solute per litre, then divide by the molar mass.

Commercial concentrated hydrochloric acid is about 37% HCl by mass with a density near 1.18 g/mL. That is 1,180 g of solution per litre, of which 0.37 × 1,180 = 436.6 g is HCl. With a molar mass of 36.46 g/mol the concentration is 436.6 ÷ 36.46 = 12.0 mol/L. Check the actual label of your bottle, because assay and density vary slightly by supplier and batch.

Once you know the stock is about 12 M, the dilution rule gives the volume to measure for any weaker solution. Making 500 mL of 1.0 M needs 1.0 × 500 ÷ 12.0 = 41.7 mL of stock, always added to water.

Practical flask technique

The calculation only means something if the solution is actually prepared at the stated final volume. A volumetric flask is calibrated to hold an exact volume at a stated temperature, commonly 20 °C, up to its etched mark. A beaker's graduations are rough guides and a measuring cylinder is better but still less precise than a flask.

1. 1Weigh the solid and dissolve it in roughly half the final volume of solvent.
2. 2Transfer the solution to the flask, rinsing the beaker into it so no solute is lost.
3. 3Add solvent until the level is just below the mark, then finish dropwise until the bottom of the meniscus rests on it.
4. 4Stopper and invert several times to mix; the concentration is only uniform after mixing.

Reading concentrations in practice

Solutions of different strengths get used for different jobs. Physiological saline is about 0.154 M sodium chloride, equal to 0.9 g per 100 mL, which matches the salt level of body fluids. Household vinegar is roughly 0.8 M acetic acid, and a typical lab stock acid is several molar. Moving between these scales is mostly a matter of keeping track of whether you are talking about moles or grams.

Millimolar and micromolar are common in biology, where 1 mM is 0.001 mol/L and 1 µM is 0.000001 mol/L. Cell-culture additives and drug assays are often described that way. To prepare 10 mL of a 5 mM solution from a 1 M stock, V₁ = 0.005 × 10 ÷ 1 = 0.05 mL, which is 50 µL, a volume small enough that serial dilution in two steps would be more accurate than measuring it once.

The same rule M₁V₁ = M₂V₂ applies at every scale. Only the care needed in measuring changes.

Mistakes that spoil the answer

• Using the volume of water you started with instead of the final solution volume.
• Forgetting to convert mL to L.
• Dividing grams by volume and calling the result molarity; grams per litre is a different quantity.
• Ignoring that hydrated salts carry water in their formula, which changes the molar mass.
• Applying M₁V₁ = M₂V₂ with mismatched volume units.