Calculate molarity, use it to convert between moles and solution volume, prepare and dilute solutions, and carry out solution stoichiometry (MA STE supporting content, concentration and quantitative solution chemistry).

What this topic is asking

The qualitative words "dilute" and "concentrated" are not precise enough for calculation, so chemists measure concentration as molarity. A Massachusetts high school chemistry course expects you to calculate molarity, convert between moles and volume, prepare and dilute solutions, and use molarity in stoichiometry. This is the quantitative skill HS-PS1-7(MA) calls for, applied to reactions in solution.

Molarity

A 2.0 M solution contains 2.0 moles of solute in every liter. The volume must be in liters, so convert milliliters by dividing by 1000. Molarity gives two everyday conversions:

• Moles from a solution: $n = M \times V$.
• Volume needed for a number of moles: $V = \dfrac{n}{M}$.

To prepare a solution of known molarity, weigh out the required moles of solute (using molar mass), dissolve it, and add solvent up to the final volume in a volumetric flask. Note the volume is the final volume of solution, not the volume of solvent added: the dissolved solute takes up some space, so you fill to the mark after dissolving, not before.

Chemists prefer molarity over loose words like "dilute" because it ties concentration directly to the mole, the unit reactions are counted in. Other units exist (mass per volume, or percent by mass), but molarity is the one that plugs straight into a balanced equation, which is why it dominates quantitative solution chemistry. A bottle labelled 6.0 M tells you exactly how many moles are in any measured volume, with no further information needed.

Dilution

Since moles equal molarity times volume, and the moles do not change when you add water, the product $M \times V$ is the same before and after: $M_1V_1 = M_2V_2$. This lets you find any one of the four quantities given the other three, and it is how concentrated stock solutions are diluted to a working strength in the lab.

Solution stoichiometry

For a reaction in solution, molarity is simply the tool that converts a measured volume into moles. The full path follows the same mole bridge as Module 3:

1. Convert the known to moles. For a solution, multiply molarity by volume in liters.
2. Apply the mole ratio from the balanced equation.
3. Convert the moles of the unknown to the units asked, including back to a volume or concentration if needed.

This is exactly the method used in titrations, developed in neutralization and titration.

Try this

Q1. Find the molarity of a solution with 0.40 mol of solute in 2.0 L. [1]

• Cue. $M = \dfrac{0.40}{2.0} = 0.20$ M.

Q2. 50 mL of 4.0 M solution is diluted to 200 mL. Find the new concentration. [1]

• Cue. $M_2 = \dfrac{4.0 \times 50}{200} = 1.0$ M.