Molarity Solver

Molarity ($M$, also written $c$) is the amount of dissolved solute per litre of solution:

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

• $n$ — moles of solute (mol).
• $V$ — volume of the final solution in litres, not the volume of solvent used.
• $M$ — molarity, in mol/L. The symbol M is read "molar": 0.200 M means 0.200 mol per litre.

When the solute is weighed out, moles come from the molar mass $\mathcal{M}$ (g/mol):

$n = \frac{m}{\mathcal{M}} \qquad\Longrightarrow\qquad M = \frac{m}{\mathcal{M}\,V}$

What molarity assumes. It is a volume-based concentration, so it is tied to the temperature at which the volume was measured — solutions expand when warmed, which lowers the molarity slightly without any solute leaving. For temperature-independent work, molality (mol solute per kg solvent) is used instead.

Formal versus actual concentration. A label of 0.10 M acetic acid states how much acid was dissolved, not how much exists as free ions. For a strong electrolyte such as NaCl the dissolved species are fully separated, so 0.10 M NaCl really is 0.10 M in $\mathrm{Na^+}$; for a weak electrolyte only a small fraction ionises.

The three rearrangements

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

From grams to molarity

1. Find the molar mass of the solute by adding the atomic masses in its formula.
2. Convert mass to moles: $n = m/\mathcal{M}$.
3. Convert the volume to litres — 250.0 mL is 0.2500 L. This is where most errors enter.
4. Divide: $M = n/V$.

From molarity back to moles

Multiply: $n = MV$. This is the standard bridge from a measured volume of solution to an amount that can be used in a reaction ratio, which is why almost every titration calculation starts here.

Dilution

Adding solvent changes the volume but not the number of moles of solute, so $n$ is conserved:

$M_1V_1 = M_2V_2$

Because only a ratio of volumes appears, $V_1$ and $V_2$ may both be in mL.

Significant figures

The answer takes the fewest significant figures among the measurements. Molar masses from a periodic table are normally quoted to more figures than the balance reading, so the mass usually sets the precision.

• Dividing by the volume of solvent. Molarity uses the total volume of solution. Dissolving solute in 1.00 L of water does not give exactly 1.00 L of solution.
• Leaving the volume in millilitres. 35.0 mL is 0.0350 L; using 35.0 makes the answer 1000 times too small.
• Forgetting the formula's subscripts in the molar mass. $\mathrm{Ca(NO_3)_2}$ contains two nitrogens and six oxygens, not one and three.
• Ignoring ion stoichiometry. 0.10 M $\mathrm{Na_2SO_4}$ is 0.20 M in $\mathrm{Na^+}$; the label refers to the formula unit.
• Confusing molarity with molality. Molarity divides by litres of solution, molality by kilograms of solvent. They are close in dilute aqueous solutions and diverge in concentrated ones.
• Using $M_1V_1 = M_2V_2$ for a reaction. The dilution equation assumes the moles of solute are unchanged; if the solute is consumed by a reaction, use the balanced mole ratio instead.