Concentration terms

Open any bottle in a chemistry lab and the label says more than just "hydrochloric acid" — it says how much acid is dissolved in how much water. Reactions, dilutions, and titrations all depend on that "how much," and chemists have settled on a small, precise vocabulary to state it: mass per cent, mole fraction, molarity, molality, parts per million, and normality. Each term answers the same basic question — solute amount relative to solute-plus-solvent amount — but chooses different units (mass, moles, or volume) for numerator and denominator, and that choice quietly decides whether the number changes with temperature. This article, part of ISC Class 11's "Some Basic Concepts of Chemistry," builds directly on the mole concept and molar mass to define, calculate, and interconvert all six terms.

The core insight

Concentration terms are all ratios of solute amount to solute-plus-solvent amount, differing only in whether mass, moles, or volume is used for solute and for solution/solvent — and that choice determines whether the value depends on temperature.

The core idea

Mass per cent (% w/w) expresses the mass of solute per hundred units of mass of solution — purely a mass ratio, so it never changes with temperature.

\text{Mass \%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

where: mass of solute and mass of solution are both in the same unit (commonly grams), so the ratio is dimensionless before multiplying by 100.

Mole fraction ( $x_i$ ) is the ratio of moles of one component to the total moles of every component present in the mixture — a purely mole-based ratio, so it too is temperature-independent.

x_i = \frac{n_i}{\sum n}

where: $n_i$ is the moles of component $i$ and $\sum n$ is the sum of moles of all components. Mole fractions of every component in a mixture always add up to exactly 1 — a fast way to check any calculation.

Molarity ( $M$ ) is the number of moles of solute per litre of solution — note "solution," not "solvent." Because it uses volume, and liquids expand slightly on heating, molarity drifts a little with temperature even though the amount of solute is fixed.

M = \frac{n_{\text{solute}}}{V_{\text{solution}}\ (\text{L})}

where: $n_{\text{solute}}$ is moles of solute and $V_{\text{solution}}$ is the total volume of the made-up solution in litres, read off a volumetric flask or burette.

Molality ( $m$ ) is the number of moles of solute per kilogram of solvent — a mass-based denominator, so unlike molarity it does not shift with temperature.

m = \frac{n_{\text{solute}}}{w_{\text{solvent}}\ (\text{kg})}

where: $w_{\text{solvent}}$ is the mass of solvent alone (not the whole solution) in kilograms.

Parts per million (ppm) scales the mass-per-cent idea up for very dilute solutions, expressing mass of solute per million parts of solution — the standard way to report trace impurities or pollutants.

\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6

Normality ( $N$ ), retained as ISC-specific content, counts gram equivalents of solute per litre of solution rather than moles, and connects to molarity through the solute's n-factor (equivalents per mole — see the "Equivalent weight" article for the full n-factor rules):

N = M \times n\text{-factor}

The figure below groups the six terms by what they measure in the numerator and denominator, which is the fastest way to remember which ones are temperature-sensitive.

Figure

Six concentration terms grouped by numerator and denominator basis showing mass over mass mole over mole and mole over volume families

Molarity is the everyday choice for lab work because volumes are quick to measure with a burette or flask, but it is the one term that is not perfectly fixed once the temperature changes. Molality and mole fraction stay exactly constant regardless of temperature because they never reference a volume — a property that matters later for colligative properties (a Class 12 topic, out of scope here). Mass per cent and ppm are convenient whenever only masses, not volumes, are easy to weigh, and ppm specifically is the standard unit for very small trace amounts.

Derivation AConverting normality to molarity using the n-factor

The goal is to show precisely why $N = M \times n\text{-factor}$ holds, by comparing what a "mole" and a "gram equivalent" of solute actually mean, so that normality is never mistaken for just another name for molarity, and so that the n-factor is applied to the correct chemical role of the solute (acid, base, or redox species) rather than guessed.

Step 1. Recall the two definitions side by side. Molarity counts moles of solute per litre of solution: $M = n_{\text{solute}}/V$ . Normality counts gram equivalents of solute per litre of solution: $N = \text{(number of equivalents)}/V$ . Both share the same denominator, the volume of solution in litres, so any difference between $M$ and $N$ must come entirely from how "amount of solute" is counted in the numerator.

Step 2. Connect moles to equivalents. One mole of solute does not always correspond to one equivalent — it corresponds to as many equivalents as the solute's n-factor, which counts how many reactive units (replaceable hydrogen ions for an acid, replaceable hydroxide ions for a base, or electrons transferred for a redox species) one mole of that solute supplies. By definition, number of equivalents $= n_{\text{solute}} \times n\text{-factor}$ .

Step 3. Substitute into the normality formula. Since normality is equivalents per litre, $N = \dfrac{n_{\text{solute}} \times n\text{-factor}}{V} = \left(\dfrac{n_{\text{solute}}}{V}\right) \times n\text{-factor}$ . But the bracketed term is exactly the definition of molarity, $M$ , so $N = M \times n\text{-factor}$ falls out directly — no separate rule is needed, only the substitution.

Step 4. Read the result. Normality is always greater than or equal to molarity, and the two are equal only when the n-factor is 1 (for example, a monoprotic acid like $\text{HCl}$ ). For a diprotic acid like $\text{H}_2\text{SO}_4$ , whose n-factor is 2, the normality is exactly double the molarity of the same solution — which is precisely what the worked example below confirms numerically.

Worked example

$4.9\ \text{g}$ of sulphuric acid ( $\text{H}_2\text{SO}_4$ , molar mass $= 98\ \text{g mol}^{-1}$ ) is dissolved in water and made up to $500\ \text{mL}$ of solution. Find (a) the molarity, and (b) the normality of the solution, given the basicity of $\text{H}_2\text{SO}_4$ is 2.

Step 1. Find moles of $\text{H}_2\text{SO}_4$ from the given mass and molar mass:

n = \frac{m}{M} = \frac{4.9}{98} = 0.0500\ \text{mol}

Step 2. Convert the volume to litres, since molarity is defined per litre of solution:

500\ \text{mL} = 0.500\ \text{L}

Step 3. Calculate molarity by dividing moles by volume in litres:

M = \frac{n}{V} = \frac{0.0500}{0.500} = 0.100\ \text{mol L}^{-1}

Step 4. Calculate normality using $N = M \times n\text{-factor}$ , taking the n-factor of $\text{H}_2\text{SO}_4$ (its basicity) as 2:

N = 0.100 \times 2 = 0.200\ N

\boxed{M = 0.100\ \text{mol L}^{-1}, \quad N = 0.200\ N}

Competitive stretch

The same $500\ \text{mL}$ of $0.200\ N$ $\text{H}_2\text{SO}_4$ is exactly neutralised by $250\ \text{mL}$ of a $\text{NaOH}$ solution. Since equivalents of acid must equal equivalents of base at neutralisation, $N_{\text{acid}}V_{\text{acid}} = N_{\text{base}}V_{\text{base}}$ , so $0.200 \times 500 = N_{\text{base}} \times 250$ , giving $N_{\text{base}} = 0.400\ N$ . Because NaOH has an n-factor of 1, its normality equals its molarity: $M_{\text{NaOH}} = 0.400\ \text{mol L}^{-1}$ . This equivalence-based shortcut, using normality directly instead of converting through moles of each species, is exactly why ISC retains normality for titration calculations.

Common traps

Using the mass of solution instead of the mass of solvent in the molality formula — molality is strictly defined with solvent mass in kilograms.
Using the volume of solvent instead of the volume of the final solution in the molarity formula — dissolving a solute changes the total volume, so the two are never the same.
Forgetting that mole fractions of all components in a mixture must sum to exactly 1 — a quick sanity check on any mole-fraction calculation.
Applying $N = M \times n\text{-factor}$ with the wrong n-factor, such as treating a diprotic acid like $\text{H}_2\text{SO}_4$ as having an n-factor of 1 instead of 2.

Ask the tutor: want to practise converting between molarity, molality, mole fraction, and normality for the same solution, or need help setting up a mixed-concentration problem from scratch?

Why does ISC still teach normality when most modern chemistry uses molarity

Normality survives in the ISC syllabus mainly because it simplifies titration and volumetric calculations: at the exact neutralisation or equivalence point, equivalents of one reactant always equal equivalents of the other, so $N_1V_1 = N_2V_2$ works directly without first figuring out how many moles of acid react with how many moles of base. Molarity alone would require knowing the balanced equation's mole ratio first. Once you also know the n-factor, converting freely between $N$ and $M$ is straightforward, so learning both gives you the calculation shortcut without losing the more fundamental molarity.

If I mix two solutions do I just add their molarities to get the mixture concentration

No — molarity is not additive across different volumes because both moles and total volume change on mixing. Instead, find total moles of solute by adding $n_1 = M_1V_1$ and $n_2 = M_2V_2$ separately, add the two volumes to get the new total volume, and only then divide total moles by total volume to get the new molarity. The same care applies to mole fraction and mass per cent — always return to moles or mass before combining, then form the new ratio using the combined totals.

How do I quickly decide which concentration term to use in a given problem

Look at what data the problem gives you and what stays constant. If temperature is changing or will change (as in many practical or industrial contexts), prefer molality or mole fraction, since both are unaffected by volume changes. If you only have masses and no easy way to measure volume, mass per cent or ppm (for very dilute solutions) fits naturally. If the problem involves making up a solution to a fixed volume in the lab, molarity is standard. If it involves acid-base or redox titrations, normality via the n-factor is usually fastest.

Concentration terms

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads