Molarity — Formula, Dilution & Practice Problems

Molarity is concentration measured in moles of solute per liter of solution. One formula carries the whole topic:

where $M$ is molarity, $n$ is moles of solute, and $V$ is the final solution volume in liters. The single detail worth burning into memory is the denominator: liters of solution, not liters of solvent. When a problem says "make the solution up to $250\ \mathrm{mL}$," that final volume is the number you divide by.

Why The Formula Looks The Way It Does

Concentration means "amount per unit volume," so dividing moles of solute by liters of solution is just that definition written out. A $1.0\ \mathrm{M}$ solution holds $1.0$ mole of solute per $1.0$ liter of solution — it does not mean you poured the solute into exactly $1.0$ liter of water. After mixing, the solute itself takes up some room, so only the final solution volume is meaningful. That is also why molarity is convenient in the lab: final volume is something you measure directly with volumetric flasks, pipettes, and burettes.

Calculating Molarity From Moles Or Grams

The sequence is the same 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:

Worked Example: Molarity From Grams And Volume

Dissolve $5.84\ \mathrm{g}$ of NaCl and bring the final volume to $500\ \mathrm{mL}$.

Convert grams to moles, using $M_\mathrm{m} \approx 58.44\ \mathrm{g/mol}$:

Convert the volume, then divide:

That full path — grams → moles → liters → molarity — is the pattern behind most molarity questions.

The Dilution Shortcut, And Why It Works

When you dilute a solution you add solvent but keep the same amount of the same solute. The moles before and after are therefore equal, and since $n = MV$, equal moles means

This holds only when the same solute is being diluted and none is lost or consumed in a reaction. Example: take $25.0\ \mathrm{mL}$ of $1.20\ \mathrm{M}$ NaCl and dilute to $100.0\ \mathrm{mL}$:

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

Now You Try

Solve these two without peeking at the worked example, then check below:

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

Answers: (1) $0.250\ \mathrm{M}$. (2) $0.20\ \mathrm{M}$.

The Calculation Traps

• Milliliters used as liters. Feeding $250$ into $M = n/V$ instead of $0.250$ throws the answer off by $1000$.
• Solvent volume instead of solution volume. "Dilute to $1.00\ \mathrm{L}$" means use $1.00\ \mathrm{L}$ as the final volume.
• Skipping grams-to-moles. Mass does not go straight into the 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 reaction changes the amount of solute, go back to moles and the balanced equation.

Molarity shows up in solution preparation, titrations, dilution work, and solution stoichiometry — especially when a problem is built around measured volumes. Because it depends on volume, a large enough temperature change can shift it, though introductory problems usually ignore that unless the question flags it. For concentration logic applied inside a reaction, the natural next stop is Titration Calculations.