Molarity counts moles of solute per liter of solution. It is easy to measure out, but its volume changes a little with temperature, and it is not the natural unit for every job. Three other units fill the gaps.
Key idea
Molality, $m = \dfrac{\text{mol solute}}{\text{kg solvent}}$ (unit mol/kg). Mole fraction, $\chi_A = \dfrac{\text{mol A}}{\text{total mol}}$ (no unit; all the fractions add to 1). Mass percent $= \dfrac{\text{mass solute}}{\text{mass solution}} \times 100$. Parts per million, ppm $= \dfrac{\text{mass solute}}{\text{mass solution}} \times 10^{6}$ (for dilute water solutions, 1 ppm is about 1 mg per liter).
Why it works
Each unit suits a job. Molality uses masses, which don't change when the solution warms or cools, so it is the unit for freezing and boiling points. Mole fraction counts particles of every kind, which is what decides a vapor pressure or a partial pressure. Mass percent is what you get by weighing. Parts per million keeps tiny amounts readable: "0.0002% lead" becomes "2 ppm". Note what each one divides by: molality by the solvent alone, in kilograms; mass percent and ppm by the whole solution.
Worked example: Molality
$10.0\ \text{g}$ of NaCl ($58.44\ \text{g/mol}$) is dissolved in $250.0\ \text{g}$ of water. Find the molality.
- Moles of NaCl: $10.0/58.44 = 0.1711\ \text{mol}$.
- Kilograms of solvent: $250.0\ \text{g} = 0.2500\ \text{kg}$ (the water only, not the water plus the salt).
- $m = 0.1711/0.2500 = 0.684\ \text{mol/kg}$.
Answer: $0.684\ \text{mol/kg}$
Worked example: Mole fraction
A solution contains $46.08\ \text{g}$ of ethanol ($46.08\ \text{g/mol}$) and $90.10\ \text{g}$ of water ($18.02\ \text{g/mol}$). Find the mole fraction of ethanol.
- Moles: ethanol $1.000$; water $90.10/18.02 = 5.000$.
- $\chi_\text{ethanol} = \dfrac{1.000}{1.000 + 5.000} = 0.1667$.
Answer: $0.167$
Worked example: Parts per million
A $500.0\ \text{g}$ water sample contains $1.2\ \text{mg}$ of dissolved iron. Express the iron concentration in ppm.
- $1.2\ \text{mg} = 1.2 \times 10^{-3}\ \text{g}$.
- ppm $= \dfrac{1.2 \times 10^{-3}}{500.0} \times 10^{6} = 2.4\ \text{ppm}$.
Answer: $2.4\ \text{ppm}$
For a dilute water solution, a liter of solution is very nearly a kilogram of water, so molarity and molality are almost equal there. They differ more and more as the solution gets concentrated, which the next topic works out with densities.
Common mistake
For molality, dividing by liters of solution (that is molarity) or by the mass of the whole solution. Molality's denominator is the mass of the solvent only, in kilograms. Check: write the units out, mol/kg of solvent, and make sure the number under the bar excludes the solute's mass.