Molarity — Formula, Dilution & Practice Problems

Molarity is the concentration of a solution in moles of solute per liter of solution. To calculate it, use $M = n/V$, convert the final volume to liters, and convert grams to moles first if needed.

The core formula is

$$M = \frac{n}{V}$$

where $M$ is molarity, $n$ is moles of solute, and $V$ is the final solution volume in liters.

If you only remember one detail, remember the denominator: molarity uses liters of solution, not liters of solvent. If a problem says "make the solution up to $250\ \mathrm{mL}$," that final solution volume is the number you need.

What the molarity formula means

A $1.0\ \mathrm{M}$ solution contains $1.0$ mole of solute per $1.0$ liter of solution. It does not mean you added the solute to exactly $1.0$ liter of water. After mixing, the final solution volume is what matters.

This is why molarity is useful in lab work: final solution volume is something you can measure directly with flasks, pipettes, and burettes.

How to calculate molarity from moles or grams

Use the same sequence every time:

1. Find the amount of solute in moles.
2. Convert the final solution volume to liters.
3. Divide moles by liters of solution.

If the problem gives mass instead of moles, convert first:

$$n = \frac{\text{mass}}{\text{molar mass}}$$

For example, if you know the solute mass in grams and its molar mass in $\mathrm{g/mol}$, this step gives you the number of moles needed for the molarity formula.

Worked example: molarity from grams and volume

Suppose $5.84\ \mathrm{g}$ of NaCl is dissolved and the final solution volume is $500\ \mathrm{mL}$. Find the molarity.

First convert grams to moles. Using a molar mass of about $58.44\ \mathrm{g/mol}$ for NaCl,

$$n = \frac{5.84}{58.44} \approx 0.100\ \mathrm{mol}$$

Now convert the volume:

$$500\ \mathrm{mL} = 0.500\ \mathrm{L}$$

Then apply the molarity formula:

$$M = \frac{0.100}{0.500} = 0.200\ \mathrm{mol/L}$$

So the solution is

$$0.200\ \mathrm{M}$$

That full path is the pattern behind many molarity calculations: grams -> moles -> liters -> molarity.

When the dilution formula works

When you dilute a solution, you add solvent but keep the amount of the same solute unchanged. Under that condition, the moles before and after dilution are equal, which gives

$$M_1 V_1 = M_2 V_2$$

This equation works only if the same solute is being diluted and no solute is lost or consumed in a reaction.

Quick example

If you take $25.0\ \mathrm{mL}$ of a $1.20\ \mathrm{M}$ NaCl solution and dilute it to $100.0\ \mathrm{mL}$, then

$$M_2 = \frac{M_1 V_1}{V_2} = \frac{(1.20)(25.0)}{100.0} = 0.300\ \mathrm{M}$$

The concentration drops because the same amount of solute is spread through a larger final volume.

Common mistakes in molarity calculations

Using milliliters as if they were liters

If you put $250\ \mathrm{mL}$ into $M = n/V$ as $250$ instead of $0.250$, your answer will be off by a factor of $1000$.

Using solvent volume instead of solution volume

Molarity is based on the final volume of the whole solution. If the problem says "dilute to $1.00\ \mathrm{L}$," use $1.00\ \mathrm{L}$.

Skipping the grams-to-moles conversion

Mass does not go directly into the molarity formula. You need moles first.

Using $M_1 V_1 = M_2 V_2$ in the wrong problem

That shortcut is for dilution only. If a chemical reaction changes the amount of solute, use moles and the balanced equation instead.

Where molarity is used in chemistry

Molarity shows up in solution preparation, titrations, dilution work, and solution stoichiometry. It is especially useful when the problem is built around measured volumes.

Because molarity depends on volume, it can change if temperature changes enough to change the solution volume noticeably. In most introductory problems, that effect is ignored unless the question points to it.

Two quick molarity practice problems

Try these without looking back at the worked example:

1. What is the molarity of a solution made by dissolving $0.250\ \mathrm{mol}$ of glucose and making the final volume $1.00\ \mathrm{L}$?
2. What is the new concentration if $50.0\ \mathrm{mL}$ of a $0.80\ \mathrm{M}$ solution is diluted to $200.0\ \mathrm{mL}$?

Answers:

1. $0.250\ \mathrm{M}$
2. $0.20\ \mathrm{M}$

Try a similar problem

Try your own version by changing just one number in the worked example, such as the solute mass or the final volume, and solve it again from the start. If you want a nearby case that uses the same concentration logic inside a reaction, continue with Titration Calculations.