Concentration — Molarity, Molality & Dilution

The one-sentence answer: concentration tells you how much solute is present relative to a chosen reference amount, and the form you pick depends on whether that reference is liters of solution (molarity) or kilograms of solvent (molality). Keep the denominator straight and most concentration problems fall apart easily.

Molarity vs. Molality at a Glance

Here $n$ is moles of solute, $V$ is the final solution volume in liters, and $m_{\mathrm{solvent}}$ is the solvent mass in kilograms. The two can be numerically close in dilute aqueous solutions, but they are defined differently and are not interchangeable.

Reading the Table: What Each Row Means

Molarity counts solute against the whole solution. A $0.50\ \mathrm{M}$ solution holds $0.50$ mole of solute per liter of finished solution. The phrase "of solution" matters: if you dissolve a solute and then fill the flask to $1.00\ \mathrm{L}$, that final volume is what you use. Because volume can drift with temperature, molarity can shift slightly when the temperature changes enough.

Molality counts solute against the solvent alone. Dissolve $0.50$ mole of solute in $1.00\ \mathrm{kg}$ of water and the molality is $0.50\ m$. Since mass does not change with temperature, molality stays put even when conditions vary, which is why it shows up in topics like colligative properties.

When to Use Which

• Use molarity when the problem hands you, or asks for, a solution volume. It dominates lab preparation, titrations, and solution stoichiometry because volumes are easy to measure with flasks and pipettes.
• Use molality when the problem hands you, or asks for, a solvent mass, or when temperature changes would make volume-based measurements unreliable.

Worked Example: One Sample, Two Numbers

Dissolve $0.300$ mole of glucose to make $600\ \mathrm{mL}$ of solution, using $0.500\ \mathrm{kg}$ of water as solvent.

Molarity first. Convert $600\ \mathrm{mL}$ to $0.600\ \mathrm{L}$, because molarity uses liters of solution:

Now molality, using the solvent mass:

The same sample carries two different concentration values because each definition uses a different denominator: $0.600\ \mathrm{L}$ of solution for molarity, $0.500\ \mathrm{kg}$ of solvent for molality.

The Dilution Relationship

Dilution adds solvent but keeps the solute amount fixed. If nothing reacts and nothing is lost, the moles before and after are equal, $n_1 = n_2$. Since $n = MV$ for molarity, this gives

For example, taking $100\ \mathrm{mL}$ of $1.50\ \mathrm{M}$ solution and diluting to $300\ \mathrm{mL}$:

The concentration drops because the same solute is now spread through a larger volume.

A Quick Decision Check

Before you finish any concentration problem, ask:

1. Did I use moles of solute?
2. Did I use liters of solution for molarity, or kilograms of solvent for molality?
3. If I used $M_1 V_1 = M_2 V_2$, did the solute amount really stay constant?

Common Confusions

Using Solution Mass for Molality

Molality uses kilograms of solvent, not kilograms of solution.

Using Solvent Volume for Molarity

Molarity uses the final volume of the whole solution.

Applying $M_1 V_1 = M_2 V_2$ in the Wrong Situation

That shortcut is only for diluting the same solute, where moles are conserved. It fails if a reaction changes the solute amount.

Treating Molarity and Molality as Interchangeable

They may be numerically close in some dilute cases, but the definitions differ. Do not swap them without a proper conversion.

Test Your Reading of the Definitions

Rework the example so the same $0.300$ mole of glucose is made up to $1.20\ \mathrm{L}$ instead of $600\ \mathrm{mL}$. Recalculate the molarity, then decide whether the molality changes under that new setup. The result tells you whether you truly know which denominator each definition uses.