Expressing Concentration of Solutions

When you dissolve sugar in water or mix acid into a flask for a reaction, one of the first things you need to know is: how much solute is actually present in the solution? Answering that question precisely is what concentration is all about. There are four standard ways to express concentration, and each one captures the same basic idea from a different angle. Choosing between them depends on what you are trying to do: compare compositions by mass, work with moles, measure out volumes in the lab, or set up calculations that stay valid even when the temperature changes.

Mass Per Cent: Expressing Concentration by Weight

The simplest way to describe how concentrated a solution is: take the mass of the solute, divide it by the total mass of the solution, and multiply by 100. That gives you the mass per cent (also called weight per cent or w/w %).

$\text{Mass per cent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

Notice that the denominator is the mass of the entire solution (solute + solvent combined), not just the solvent.

Worked Example: Problem 1.6

Question: A solution is prepared by adding 2 g of substance A to 18 g of water. What is the mass per cent of the solute?

Step 1:Find the total mass of the solution.

$\text{Mass of solution} = \text{Mass of A} + \text{Mass of water} = 2 \text{ g} + 18 \text{ g} = 20 \text{ g}$

Step 2:Apply the formula.

$\text{Mass per cent of A} = \frac{2 \text{ g}}{20 \text{ g}} \times 100 = 10\%$

Answer: The solution is 10% by mass in substance A.

A common slip is dividing by the mass of solvent (18 g) instead of the total solution mass (20 g). Always remember: the denominator includes both solute and solvent.

Mole Fraction: Thinking in Terms of Particle Counts

Sometimes you need to express concentration not by mass but by the relative number of particles. That is where mole fraction comes in. It tells you what fraction of the total moles in the solution belongs to a particular component.

Suppose substance A dissolves in substance B, and you have $n_A$ moles of A and $n_B$ moles of B. The mole fractions are:

$x_A = \frac{n_A}{n_A + n_B}$

$x_B = \frac{n_B}{n_A + n_B}$

There is a neat built-in check here. If you add the two mole fractions together:

$x_A + x_B = \frac{n_A}{n_A + n_B} + \frac{n_B}{n_A + n_B} = \frac{n_A + n_B}{n_A + n_B} = 1$

The mole fractions of all components in any solution must always add up to exactly 1. If they do not, something has gone wrong in the calculation.

Mole fraction is a dimensionless (no units) quantity, and it works regardless of the number of components. For a three-component system, $x_A + x_B + x_C = 1$, and so on.

Molarity: The Lab’s Favourite Concentration Unit

Of all the concentration measures, molarity (symbol M) is the one you will encounter most often in a chemistry lab. It tells you how many moles of solute are present in one litre of solution.

$\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}$

A “1 M” solution of $NaOH$, for instance, contains 1 mole of $NaOH$ dissolved in enough water to make exactly 1 litre of solution.

Worked Example: Problem 1.7

Question: Calculate the molarity of $NaOH$ in a solution prepared by dissolving 4 g of $NaOH$ in enough water to form 250 mL of solution.

Step 1:Convert mass to moles.

The molar mass of $NaOH$ is:

$23 + 16 + 1 = 40 \text{ g/mol}$

$\text{Moles of } NaOH = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}$

Step 2:Convert volume to litres.

$250 \text{ mL} = \frac{250}{1000} = 0.250 \text{ L}$

Step 3:Calculate molarity.

$M = \frac{0.1 \text{ mol}}{0.250 \text{ L}} = 0.4 \text{ M}$

Answer: The solution is 0.4 M in $NaOH$.

An Important Warning About Temperature

Molarity depends on the volume of the solution, and volume changes with temperature (liquids expand when heated, contract when cooled). This means the molarity of the same solution will be slightly different at different temperatures, even though you have not added or removed anything. Keep this in mind when precision matters.

Preparing a Dilute Solution from a Concentrated One

In a real lab, you rarely prepare a solution from scratch every time. Instead, you keep a stock solution (a solution of known higher concentration) and dilute it to get whatever lower concentration you need.

Here is the key idea behind dilution: you are only adding solvent, not solute. The number of moles of solute stays exactly the same before and after dilution. What changes is the volume, and therefore the concentration.

Since moles = molarity $\times$ volume, and moles stay the same:

$M_1 \times V_1 = M_2 \times V_2$

where $M_1$ and $V_1$ are the molarity and volume of the concentrated (stock) solution, and $M_2$ and $V_2$ are the molarity and volume of the diluted solution you want.

Worked Example: Diluting 1 M NaOH to 0.2 M

Question: You have a 1 M $NaOH$ stock solution and you need to prepare 1 litre of 0.2 M $NaOH$. How much of the stock solution should you use?

Approach 1: Thinking in moles.

For 1 litre of 0.2 M solution, you need 0.2 mol of $NaOH$. The stock solution has 1 mol per litre (that is what 1 M means), so 0.2 mol is contained in:

$\frac{1000 \text{ mL}}{1 \text{ mol}} \times 0.2 \text{ mol} = 200 \text{ mL}$

Take 200 mL of the stock and add enough water to bring the total volume to 1000 mL.

Approach 2: Using the dilution formula.

$M_1 \times V_1 = M_2 \times V_2$

$1.0 \text{ M} \times V_1 = 0.2 \text{ M} \times 1000 \text{ mL}$

$V_1 = \frac{0.2 \times 1000}{1.0} = 200 \text{ mL}$

Both approaches give the same result: measure out 200 mL of the 1 M stock solution, then add water until the final volume reaches 1 litre.

Notice what happened to the solute: there were 0.2 mol of $NaOH$ in the 200 mL you took, and there are still exactly 0.2 mol in the final 1000 mL. The moles did not change; only the concentration dropped because the volume increased.

Molality: A Temperature-Proof Concentration Unit

The fourth method, molality (symbol m), gets around molarity’s temperature problem entirely. Instead of dividing by the volume of solution, you divide by the mass of the solvent in kilograms.

$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Why does this solve the temperature issue? Because mass does not change with temperature. A kilogram of water is a kilogram of water whether it is at 5 degrees or 50 degrees Celsius. Volume, on the other hand, expands and contracts. By anchoring the concentration to mass rather than volume, molality remains constant regardless of temperature.

Worked Example: Problem 1.8: Converting Molarity to Molality

Question: The density of a 3 M $NaCl$ solution is 1.25 g/mL. Calculate the molality of the solution.

This problem is a great example of how molarity and molality are connected, and why you need the density to convert between them.

Step 1:Find the mass of $NaCl$ in 1 litre of solution.

A 3 M solution has 3 moles of $NaCl$ per litre. The molar mass of $NaCl$ is:

$23.0 + 35.5 = 58.5 \text{ g/mol}$

$\text{Mass of } NaCl = 3 \times 58.5 = 175.5 \text{ g}$

Step 2:Find the total mass of 1 litre of solution using the density.

$\text{Mass of 1 L solution} = 1000 \text{ mL} \times 1.25 \text{ g/mL} = 1250 \text{ g}$

Step 3:Find the mass of water (solvent) by subtraction.

$\text{Mass of water} = 1250 - 175.5 = 1074.5 \text{ g} = 1.0745 \text{ kg}$

Step 4:Calculate molality.

$m = \frac{3 \text{ mol}}{1.0745 \text{ kg}} = 2.79 \text{ m}$

Answer: The molality of the solution is 2.79 m.

Notice that although the molarity was 3 M, the molality turned out to be 2.79 m. These two values differ because molarity divides by the volume of the whole solution, while molality divides by only the mass of the solvent. They would only be numerically equal in very dilute solutions where the solvent makes up nearly all of the solution’s mass and volume.

Choosing the Right Unit: A Quick Comparison

Property	Mass per cent	Mole fraction	Molarity (M)	Molality (m)
What it measures	Mass of solute per 100 g of solution	Fraction of total moles belonging to one component	Moles of solute per litre of solution	Moles of solute per kg of solvent
Units	%	Dimensionless (no units)	mol/L (or M)	mol/kg (or m)
Depends on temperature?	No (mass-based)	No (mole-based)	Yes (volume changes with temperature)	No (mass-based)
Most useful when	Comparing compositions by weight, food labelling, industrial formulations	Thermodynamic calculations, vapour pressure problems	Everyday lab work, volumetric analysis, titrations	Precise physical chemistry, colligative property calculations

The table highlights a practical rule of thumb: if your work involves measuring volumes (pipetting, titrating), use molarity. If your results must be independent of temperature, use molality.