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How do we describe the composition of a solution quantitatively using molarity?

Topic 3.7 Solutions and Mixtures: define solute, solvent and solution, and calculate and use molarity to relate moles, volume and concentration, including dilutions.

A focused answer to AP Chemistry Topic 3.7, covering solute and solvent, the molarity concentration formula, preparing solutions, and dilution calculations with the M1V1 equals M2V2 relationship, with full worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Solute, solvent and solution
Molarity
Dilution
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What this topic is asking

The College Board (Topic 3.7) wants you to describe a solution as a homogeneous mixture of a solute dissolved in a solvent, and to quantify its composition using molarity. You must convert confidently between moles of solute, volume of solution and concentration, prepare solutions, and carry out dilution calculations. Molarity is the working concentration unit for the rest of the course, from titrations to equilibrium.

Solute, solvent and solution

Because a solution is uniform, a sample taken from anywhere in it has the same concentration. This is what lets us describe the whole solution with a single number, the molarity.

Molarity

To prepare a solution of known molarity, you weigh out the calculated mass of solute, dissolve it, and then add solvent up to the final volume mark (not before). For an ionic compound, the concentration of each ion is the molarity multiplied by the number of those ions in the formula: a $0.10$ M solution of $\text{CaCl}_2$ is $0.10$ M in $\text{Ca}^{2+}$ and $0.20$ M in $\text{Cl}^-$ , because each formula unit releases two chloride ions.

Dilution

When you dilute a solution by adding more solvent, you do not add or remove any solute, so the moles of solute are conserved. Since $n = MV$ , that means

M_1 V_1 = M_2 V_2

where the subscripts label the concentrated (1) and diluted (2) states. The concentration falls in the same proportion as the volume rises: doubling the volume halves the concentration. This conservation-of-moles idea is the key to every dilution problem.

A practical point: when you prepare a diluted solution, $V_1$ is the volume of the concentrated stock you measure out, and $V_2$ is the total final volume after adding solvent, not the volume of solvent added. Mixing up these two volumes is the most common dilution error. Because the relationship depends only on moles, it works for any pair of concentration and volume units, provided you use the same units on both sides; you do not even need to convert milliliters to liters, since the volume units cancel.

Preparing a solution and diluting it

A chemist dissolves $0.0200$ mol of potassium nitrate ( $\text{KNO}_3$ ) in water to make $100.0$ mL of solution, then dilutes $25.0$ mL of it to $250.0$ mL. Find the molarity at each stage.

step 1 Find the original molarity

$M_1 = \dfrac{n}{V} = \dfrac{0.0200\ \text{mol}}{0.1000\ \text{L}} = 0.200$ M.

step 2 Set up the dilution

Moles of solute are conserved on dilution, so $M_1 V_1 = M_2 V_2$ with $V_1 = 25.0$ mL and $V_2 = 250.0$ mL.

step 3 Solve for the new molarity

$M_2 = \dfrac{M_1 V_1}{V_2} = \dfrac{(0.200)(25.0)}{250.0} = 0.0200$ M.

step 4 Check the reasoning

The volume increased tenfold ( $25.0$ to $250.0$ mL), so the concentration fell tenfold ( $0.200$ to $0.0200$ M), which is consistent with moles of solute being conserved.

Try this

Q1. Calculate the molarity of a solution containing $0.500$ mol of glucose in $2.00$ L of solution. [1 point]

Cue. $M = \dfrac{0.500}{2.00} = 0.250$ M.

Q2. Calculate the concentration of sulfate ions in a $0.150$ M solution of sodium sulfate, $\text{Na}_2\text{SO}_4$ . [1 point]

Cue. Each formula unit gives one sulfate ion, so $[\text{SO}_4^{2-}] = 0.150$ M (and $[\text{Na}^+] = 0.300$ M).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A student prepares a solution by dissolving

5.85

g of sodium chloride (

\text{NaCl}

) in enough water to make

250.0

mL of solution. (a) Calculate the molarity of the solution. (b) Calculate the concentration of chloride ions. (c) The student dilutes

50.0

mL of this solution to a final volume of

200.0

mL. Calculate the new molarity. (d) Justify whether the number of moles of

\text{NaCl}

changes on dilution.

Show worked answer →

A 4-point quantitative FRQ on molarity and dilution ( $M(\text{NaCl}) = 58.44$ g/mol).

(a) Molarity (1 point): $n = \dfrac{5.85}{58.44} = 0.100$ mol; $M = \dfrac{0.100}{0.2500} = 0.400$ M.
(b) Chloride (1 point): each $\text{NaCl}$ gives one $\text{Cl}^-$ , so $[\text{Cl}^-] = 0.400$ M.
(c) Dilution (1 point): $M_2 = \dfrac{M_1 V_1}{V_2} = \dfrac{(0.400)(50.0)}{200.0} = 0.100$ M.
(d) Justify (1 point): the moles of $\text{NaCl}$ do not change on dilution; adding solvent increases the volume but not the amount of solute, which is why concentration falls.

Markers reward a correct molarity, recognizing the 1:1 chloride ratio, correct use of $M_1V_1 = M_2V_2$ , and the reasoning that dilution conserves moles of solute.

AP 2021 (style)1 marksSection I (multiple choice). What volume of

2.00

M stock solution is needed to prepare

500.

mL of

0.250

M solution? (A)

31.3

mL (B)

62.5

mL (C)

125

mL (D)

250

mL. Justify your reasoning.

Show worked answer →

A 1-point quantitative MCQ. The answer is (B).

Using $M_1 V_1 = M_2 V_2$ : $V_1 = \dfrac{M_2 V_2}{M_1} = \dfrac{(0.250)(500.)}{2.00} = 62.5$ mL of the stock solution, which is then diluted to $500.$ mL. The other options come from inverting the ratio or arithmetic slips.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)