Molarity from dissolved solute:
formula and worked example

Molarity as the working unit of solution chemistry

When reagents are in solution, moles are measured by volume rather than by mass. Molarity is the unit that makes this possible: it states how many moles of solute sit in each litre of solution. Titrations, buffer preparation, reaction-rate work and most wet-lab protocols are written in terms of it.

Because it is a ratio, the same molarity can describe a few millilitres or a full tank, and the amount of solute scales with whatever volume you take. A 0.5 mol/L solution contains half a mole in each litre, so measuring 200 mL gives you 0.1 mol, without having to know anything about the bulk batch.

It also gives a direct route to the amount in any portion. Because molarity is a per-litre quantity, multiplying it by a volume in litres recovers the moles in that portion. That reverse relationship is what lets you pipette a measured volume from a stock and know exactly how much solute you have taken, which is the basis of titration and serial dilution.

The formula, its inputs and units

Molarity is the amount of dissolved solute divided by the volume of the whole solution.

• Moles: the amount of solute, typically a mass divided by its molar mass.
• Volume l: the final volume of solution in litres. Convert millilitres by dividing by 1000.
• Result: molarity in mol/L, often written M and read as molar.

Notice that the denominator is solution volume. Solute plus solvent after mixing is what the volumetric flask holds, and that is what must go in the formula.

Worked example: a quarter mole in half a litre

1. 1Moles of solute: 0.25 mol.
2. 2Final volume of solution: 0.5 L, which is 500 mL.
3. 3Divide: 0.25 / 0.5 = 0.5.
4. 4State the unit: 0.5 mol/L, or 0.5 M.

Dividing by a number smaller than one makes the result larger than the moles, a point that surprises people the first time. Any volume under one litre increases the molarity relative to the number of moles.

What a molarity value implies

Concentration only tells you how crowded the solute is. Most routine laboratory solutions fall roughly between millimolar and a few molar, and anything far outside that is either a stock solution to be diluted or a very dilute analytical standard. Solubility sets the ceiling: you cannot make a solution more concentrated than the solute allows at the working temperature.

To prepare a different strength from a stock, the relationship C1 x V1 = C2 x V2 holds, because moles are conserved on dilution. A 0.5 mol/L stock diluted to 100 mL at 0.1 mol/L needs 20 mL of stock. This is why the molarity of a stock must be correct before any dilution is done.

Keep in mind that molarity is tied to volume, and volume changes with temperature. A solution made up at one temperature will have a very slightly different molarity at another.

How moles and volume trade off

Adding 10% more solute at the same volume lifts the molarity from 0.5 to 0.55 mol/L, a proportional change. Volume works the other way: doubling it while keeping the moles fixed halves the concentration. These two levers are all there is, which is why a concentration problem can usually be solved by asking whether the moles or the volume changed. Note that in the real lab, adding a large amount of solute also changes the final volume, so scale-up should be checked against a volumetric measurement.

Typical molarity errors

• Dividing by the volume of solvent added rather than the final solution volume, which overstates the concentration.
• Leaving the volume in millilitres. 0.25 mol in 500 mL is 0.5 mol/L, not 0.0005.
• Computing moles from mass using the anhydrous molar mass when the solid was a hydrate.
• Confusing molarity with molality. Molarity uses litres of solution and changes with temperature; molality uses kilograms of solvent and does not.

A last pitfall is reading the concentration of an ion when the solute is a salt. A 0.5 mol/L solution of calcium chloride has 0.5 mol/L of calcium ions but 1.0 mol/L of chloride, since each formula unit releases two. State clearly whether you mean the compound or one of its ions when you quote the figure.

Making a solution of a given molarity

Calculate the moles required, convert to a mass using the molar mass, dissolve the solid in less than the final volume of solvent, transfer to a volumetric flask and make up to the mark. Working in that order guarantees that the final volume is the number you use in the calculation.

As an illustration, a student dissolves 0.25 mol of solute and dilutes to the 500 mL mark, obtaining 0.5 mol/L. A later experiment needs 50 mL of that solution. Multiplying 0.5 mol/L by 0.050 L gives 0.025 mol, one tenth of the original amount. Tracking the moles through each step like this makes it obvious where a concentration or volume went wrong if a result looks off.