Concentration of Solutions

A pinch of salt stirred into a glass of water, an IV drip in a hospital ward, the acid in a car battery — all of them are solutions, and none of them can be discussed precisely without a number for "how much solute per how much solvent (or solution)." That number is what concentration terms give chemistry: a common language so that a lab technician in Delhi and one in Tokyo, reading "0.20 M NaCl," prepare exactly the same solution. This article takes the four concentration terms of the Class 11 syllabus — mass percentage, mole fraction, molarity and molality — builds the molarity formula from first principles, works a full molarity-then-dilution problem through to a numeric answer, and flags the traps that cost the most marks: mixing up solution volume with solvent volume, and confusing molarity with molality.

The core insight

Concentration terms — mass percentage, mole fraction, molarity and molality — are different ratios of solute to solvent or solution, chosen to match what is easiest to measure or most useful in a given situation, with molarity ( $M = n_{\text{solute}}/V_{\text{solution}}$ ) the workhorse of laboratory volumetric work.

The core idea

A solution is a homogeneous mixture of two or more substances; the solute is present in the smaller amount and the solvent in the larger amount. Because "how much" can be expressed as mass, moles or volume, and because the denominator can be the solute, the solvent, or the whole solution, chemists have settled on a small family of standard ratios rather than one universal number.

Mass percentage expresses the solute as a fraction of the total mass of solution, per hundred:

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

where: the mass of solution is the mass of solute plus the mass of solvent. This is the term you see on commercial labels — "5% by mass" — because masses are what a simple weighing balance measures directly, with no need for a thermometer or a volumetric flask.

Mole fraction ( $x_i$ ) compares moles of one component to the total moles of every component present:

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 in the solution. Because it is a ratio of particle counts rather than masses or volumes, mole fraction has no unit, and the mole fractions of all components in a solution always add up to 1: $\sum x_i = 1$ .

Molarity ( $M$ ) is moles of solute per litre of the final solution:

M = \frac{n_{\text{solute}}}{V_{\text{solution}}}

where: $n_{\text{solute}}$ is in moles and $V_{\text{solution}}$ is in litres. Because volumes expand and contract slightly with temperature, molarity is temperature-dependent — the same bottle of solution has a marginally different molarity on a hot day than on a cold one, even though nothing was added or removed.

Molality ( $m$ ), defined here only for its role as a temperature-independent lead-in term, is moles of solute per kilogram of solvent:

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

where: the denominator is solvent mass only, never solution mass. Because mass does not change with temperature, molality is temperature-independent — this is exactly why it becomes the preferred term once a calculation must survive a change of temperature, a use developed fully in the Class 12 Solutions chapter with colligative properties. At Class 11 level you are expected to know its definition and do simple mole/mass calculations with it, not its full quantitative application. The figure below makes the molarity-versus-molality distinction visual: molarity divides by the volume of the whole solution after solute and solvent have been mixed together, while molality divides by the mass of the solvent alone, before mixing.

Figure

A beaker of solute dissolving in solvent to form a solution, with molarity dividing moles of solute by the volume of the whole solution and molality dividing moles of solute by the mass of the solvent alone

Derivation Abuilding the molarity formula

The goal is to show, from the plain meaning of "moles per litre," exactly why molarity is defined as $M = n_{\text{solute}}/V_{\text{solution}}$ with the solution's volume in the denominator rather than the solvent's, and to carry a real substance through the formula so the definition becomes a number: given $5.85\ \text{g}$ of sodium chloride dissolved to make $500\ \text{mL}$ of final solution, what is the molarity?

Step 1. Start from the plain-language definition. Concentration in general answers "how much solute per fixed amount of solution?" For molarity, the "fixed amount" is chosen to be one litre of the finished solution, because a burette or volumetric flask measures out solution volume directly and instantly — no weighing, no evaporation. So molarity must be built from two measured quantities: the amount of solute (in moles, since chemical reactions occur mole-for-mole) and the volume of the solution that results once solute and solvent are mixed together.

Step 2. Convert the given mass of solute to moles, using molar mass $M_{\text{molar}}$ as the mass-to-mole converter: $n = \dfrac{\text{mass}}{M_{\text{molar}}}$ . For NaCl, $M_{\text{molar}} = 58.5\ \text{g mol}^{-1}$ , so $n(\text{NaCl}) = \dfrac{5.85\ \text{g}}{58.5\ \text{g mol}^{-1}} = 0.10\ \text{mol}$ .

Step 3. Express the volume in litres, since molarity is defined per litre by convention: $500\ \text{mL} = 0.500\ \text{L}$ . This is the volume of the solution after mixing, not the volume of water used before the solute was added — dissolving NaCl in water changes the total volume slightly, and it is the final, made-up volume that the formula requires.

Step 4. Divide moles of solute by litres of solution: $M = \dfrac{n_{\text{solute}}}{V_{\text{solution}}} = \dfrac{0.10\ \text{mol}}{0.500\ \text{L}} = 0.20\ \text{mol L}^{-1}$ . The unit $\text{mol L}^{-1}$ , also written $M$ , is molarity's signature: moles of solute for every litre of the made-up solution. Check: the definition used only moles (an amount independent of temperature and volume) divided by the solution's own volume, which is exactly why molarity — unlike molality — shifts slightly if the solution is warmed or cooled: the volume in the denominator expands or contracts while the moles of solute stay fixed.

Worked example

Problem. The $0.20\ \text{mol L}^{-1}$ NaCl solution from the derivation above is used to prepare a more dilute solution: $100\ \text{mL}$ of it is taken and diluted with water to a final volume of $250\ \text{mL}$ . What is the molarity of the diluted solution?

Step 1. Identify what stays constant on dilution. Adding water does not add or remove any NaCl — the moles of solute in the $100\ \text{mL}$ portion before dilution equal the moles of solute in the full $250\ \text{mL}$ after dilution. This constancy is captured by the dilution relation $M_1V_1 = M_2V_2$ , which is simply "moles before = moles after" written using $n = MV$ on each side.

Step 2. Find the moles of NaCl taken out in the $100\ \text{mL}$ portion, using the original molarity: $n = M_1 V_1 = 0.20\ \text{mol L}^{-1} \times 0.100\ \text{L} = 0.020\ \text{mol}$ .

Step 3. Apply the same $0.020\ \text{mol}$ to the new, larger volume: $M_2 = \dfrac{n}{V_2} = \dfrac{0.020\ \text{mol}}{0.250\ \text{L}}$ .

Step 4. Compute: $M_2 = 0.080\ \text{mol L}^{-1}$ .

The diluted solution has a molarity of $0.080\ \text{M}$ — smaller than the original $0.20\ \text{M}$ because the same $0.020\ \text{mol}$ of NaCl is now spread through a larger $250\ \text{mL}$ of solution.

Competitive stretch

If instead $0.10\ \text{mol}$ of ethanol were mixed with the $0.020\ \text{mol}$ of NaCl already present in $250\ \text{mL}$ of the diluted solution above, what would be the mole fraction of ethanol, taking the water present as $8.0\ \text{mol}$ (as in the classic ethanol–water mixture)? Total moles $= 0.10 + 0.020 + 8.0 = 8.12\ \text{mol}$ , so $x_{\text{ethanol}} = 0.10/8.12 \approx 0.012$ — a reminder that mole fraction always divides by every component present, not just the two most obvious ones.

Common traps

Using the volume of solvent (the water poured in) instead of the volume of the final solution (after mixing and making up to the mark) when calculating molarity.
Mixing up molarity ( $M$ , per litre of solution, changes with temperature) with molality ( $m$ , per kilogram of solvent, does not change with temperature).
Forgetting that mole fraction divides by the total moles of all components present, not just solute and one solvent, when more than two substances are mixed.
Reaching for normality or equivalent weight in a Class 11 theory answer — these belong to titration practicals only and are not examined as concentration terms in this chapter.

Ask the tutor: if a problem gives you a volume and you are not sure whether it is the solvent poured in or the final solution volume, ask the AI tutor to help you parse the wording before you compute molarity.

Why does molarity change with temperature but molality does not?

Molarity is moles of solute divided by the volume of the whole solution, and volume expands slightly when a liquid is heated and contracts when cooled — so the same solution, unchanged in composition, reports a marginally different molarity at different temperatures. Molality divides by the mass of the solvent instead, and mass does not change with temperature. This is why molality is the safer choice whenever a calculation must hold across a temperature range, even though its full use with colligative properties comes only in Class 12.

When I make up a solution, is the volume I use for molarity the water I poured in, or the final mixture?

Always the final mixture — the total volume of solution after the solute has been dissolved and the flask made up to the mark. Dissolving a solid in a liquid does not simply add the two volumes; the final solution volume is what you actually read off the volumetric flask, and that is the number the molarity formula demands. Using the volume of water added before the solute went in is the single most common molarity mistake.

With four concentration terms to choose from, how do I know which one a problem wants?

Look at what units the problem gives you and what it asks for. If masses are given and a percentage is asked, use mass percentage. If moles of two or more components are given with no reference to volume, use mole fraction. If a volume of solution and an amount of solute are involved — the everyday laboratory situation with burettes and pipettes — use molarity. If the problem explicitly mentions mass of solvent in kilograms, or hints at a temperature-sensitive calculation, it wants molality's definition, even though its full colligative use is a Class 12 topic.

Concentration of Solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads