Molarity Calculator c = n/V: Moles, Volume and Concentration

About this calculator

The molarity calculator solves $c = n/V$ : given the amount of solute $n$ (mol) and the solution volume $V$ (L), it finds the amount concentration $c$ (mol/L); any two of the three quantities determine the third—how many mol/L is 58.44 g of salt made up to 1 L, how much solute is weighed out to prepare 500 mL of a 0.1 mol/L solution, and what volume is needed to turn 0.2 mol of solute into a 2 mol/L solution.

Chemistry has several senses of “concentration”: molarity, mass fraction, mass concentration, molality and ppm. This page is the anchor of the chemistry cluster: what a mole is, why molar mass is numerically equal to relative molecular mass, how the concentration measures convert, and why “1 mol dissolved in 1 L of water” is not 1 mol/L are all set out here. The two paired pages cover only what is unique to them: turning mass into amount on the Moles from Mass Calculator ( $n = m/M$ ), and diluting a concentrated solution to a target concentration on the Dilution Calculator ( $C_1V_1 = C_2V_2$ ).

What it does not do: it does not compute $n$ from a mass (use the moles calculator first); it does not convert mL ↔ L (enter 250 mL as 0.25); it does not compute mass fraction, density or activity; and it does not handle volume changes on dissolution.

Formula and variables

c = \frac{n}{V}

Inverse forms:

n = c\,V,\qquad V = \frac{n}{c}

Symbol	Meaning	Calculator field	Unit	Range and notes
$n$	amount of solute	Moles n (mol)	mol	$n \ge 0$ ; default 0.5. From mass: $n = m/M$
$V$	total volume of the solution (not the solvent volume)	Volume V (L)	L	$V > 0$ ; default 1. Enter 250 mL as 0.25
$c$	amount concentration (molarity)	Molarity c (mol/L)	mol/L	output; an input when solving in reverse. 1 mol/L is often written 1 M

Dimensions: mol/L is also written mol/dm³, identically. The formal IUPAC name is “amount concentration”; school texts call it molarity, laboratory speech says “molar”, and the symbol M (read as molar) is extremely common even though it is not an SI unit symbol: 0.1 M = 0.1 mol/L.

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Molarity” (/#/mathtools/formula?calculator=calculator.chem-molarity). On open, “Moles n” = 0.5 and “Volume V” = 1 are already filled in, and “Molarity c” carries a “calculate” marker and shows 0.5—a genuine result.
Find the concentrationchange $n$ or $V$ and the concentration updates immediately. The volume unit is L: enter 250 mL as 0.25 and 50 mL as 0.05.
Solve for the amount of substance(how much to weigh out): clear “Moles n”, fill in the target concentration $c$ and the volume $V$ , and the n field switches to “calculate” and shows $cV$ ; multiply by the molar mass for the mass in grams (or use the moles calculator to solve for mass).
Solve for volumeclear “Volume V” and enter $n$ and $c$ ; the volume field shows $n/c$ in L. With $c = 0$ it displays ∞.
Starting from a massfirst compute $n$ from $m$ and $M$ in the moles calculator, then copy the displayed value into the n field here.
Read the resultvalues keep at most 8 decimal places with trailing zeros removed.
Copy the formulathe copy button at the right of the panel title puts the LaTeX c = \frac{n}{V} on the clipboard.

Worked examples

All four examples were recomputed by the engine's chem-molarity compute routine; the display convention is at most 8 decimal places with trailing zeros removed, in units mol, L and mol/L.

Example 1: four beakers

Default inputs $n = 0.5$ , $V = 1$ : $c = 0.5$ . Interface shows: molarity c 0.5.

Change the volume to 0.5 (same solute, half the volume): the interface shows 1. Change $n$ to 1 and $V$ back to 1 (58.44 g of salt made up to 1 L): the interface shows 1. The second and third beakers have the same concentration even though the total amount of solute differs by a factor of two—concentration is “how much per litre of solution”, not “how much in total”. Fourth beaker: 0.1 mol of solute made up to 250 mL, $n = 0.1$ , $V = 0.25$ : $c = 0.4$ . Interface shows 0.4.

Figure 1: the four beakers of Example 1. Concentration is the density of the dots, not their total number

Example 2: how much to weigh out for 500 mL of 0.1 mol/L saline

Solving for the amount of substance. Clear $n$ and enter $c = 0.1$ , $V = 0.5$ : $n = 0.1 \times 0.5 = 0.05$ . Interface shows: moles n 0.05. The molar mass of NaCl is 58.44 g/mol, so weigh out $0.05 \times 58.44 = 2.922$ g.

The order of operations is the convention this page stresses: weigh 2.922 g of solid into a 500 mL volumetric flask, add water to dissolve it, then add water to the graduation mark—the total solution volume is 500 mL. If you first measure 500 mL of water and pour the salt into it, the resulting solution volume is slightly more than 500 mL and the concentration slightly below 0.1.

Solving for volume: 0.2 mol of solute to be made into a 2 mol/L solution. Clear $V$ and enter $n = 0.2$ , $c = 2$ : $V = 0.1$ . Interface shows: volume V 0.1—100 mL.

Example 3: glucose, saline and two meanings of “concentration”

Glucose: 9 g of glucose ( $M = 180.156$ ) made up to 100 mL. First use the moles calculator to obtain $n = 0.0499567$ , then enter $n = 0.0499567$ , $V = 0.1$ : interface shows: molarity c 0.499567—about 0.5 mol/L.

Saline: 0.9% saline means 0.9 g of NaCl per 100 mL, i.e. 9 g/L—a mass concentration (g/L), not a molarity. Convert: $n = 9 / 58.44 = 0.15400411$ mol; enter $n = 0.15400411$ , $V = 1$ : the interface shows 0.15400411—0.154 mol/L. Both are labelled “9 g”, yet the glucose solution is 0.5 mol/L and the saline 0.154 mol/L, because one NaCl “unit” weighs about three times as much as one glucose molecule. Osmotic pressure depends on the number of particles (mol), not on grams, which is why medicine compares the two liquids in moles.

Seawater: the salt content of seawater is roughly equivalent to a 0.6 mol/L NaCl solution. $n = 0.6$ , $V = 1$ : the interface shows 0.6—nearly four times saline, the quantitative reason drinking seawater dehydrates you.

Example 4: one digit out of place, a thousandfold error

Enter the 250 mL of beaker 4 in Example 1 directly as $V = 250$ : $n = 0.1$ , $V = 250$ : $c = 0.0004$ . Interface shows: molarity c 0.0004—1000 times smaller than the correct 0.4. The engine does not know you were thinking of mL. The note in the bottom right of Figure 1 is a reminder of exactly this.

Conversely, if solving for volume returns 0.1, that is 0.1 L = 100 mL, not 0.1 mL.

The inputs of Example 1 can be entered directly into the panel to reproduce it; change the volume to 0.5, 0.25 and 0.1 and watch the concentration grow in inverse proportion to 1, 2 and 5 mol/L; change the amount to 0.0499567 and the volume to 0.1 to obtain the glucose solution of Example 3.

Principle and derivation

The mole: a quantity you can count

The amount of substance $n$ is one of the seven SI base quantities, with the unit mole (mol). The 9th edition of the SI, in force from 20 May 2019, defines it as: 1 mol contains exactly 6.022 140 76 × 10²³ elementary entities, the numerical value of the Avogadro constant $N_A$ . The old definition, from 1971 to 2019, was “the number of atoms in 0.012 kg of carbon-12”, with $N_A$ measured; the new one makes it exact, while the amount of substance in 12 g of carbon-12 becomes approximately 1 mol rather than exactly 1 mol (a deviation of order $10^{-9}$ ).

Why use the mole instead of the gram? Because chemical reactions proceed by number: one Na pairs with one Cl, two H₂ with one O₂. Molecules of vastly different mass can only be combined by number in an equation. The mole is a counting unit that scales up numbers by $6.022 \times 10^{23}$ , just as “a dozen” is 12.

Why molar mass equals relative molecular mass

The scale of relative atomic mass is “one twelfth of the mass of a carbon-12 atom = 1”; the old definition also fixed the amount of substance in 12 g of carbon-12 to exactly 1 mol. Together, the two stipulations guarantee that the mass of 1 mol of any substance in grams is numerically equal to its relative molecular mass: H₂O has relative molecular mass 18.015 and molar mass 18.015 g/mol. After 2019 this equality is exact only to $10^{-9}$ , entirely negligible for everyday calculations. Standard atomic weights are maintained by the IUPAC CIAAW committee: Na 22.990, Cl 35.45, C 12.011, H 1.008, O 15.999—the step from mass to moles is on the moles calculator.

The five concentration measures and how they convert

Name	Symbol	Definition	Unit	When it is used
Amount concentration (molarity)	$c$	$n_\text{solute} / V_\text{solution}$	mol/L	Reaction stoichiometry, titration, osmotic pressure—wherever things are compared by “number”
Mass fraction	$w$	$m_\text{solute} / m_\text{solution}$	dimensionless, usually %	Commercial reagent labels (37% HCl, 98% H₂SO₄), food, alloys
Mass concentration	$\rho_B$	$m_\text{solute} / V_\text{solution}$	g/L, mg/mL	Medicine (0.9% saline = 9 g/L), water quality
Molality	$b$	$n_\text{solute} / m_\text{solvent}$	mol/kg	Boiling-point elevation, freezing-point depression; independent of temperature
ppm / ppb	—	mass ratio $10^{-6}$ / $10^{-9}$	—	Trace analysis: in dilute aqueous solutions 1 ppm ≈ 1 mg/L

The bridge between them is density $\rho$ (g/mL) and molar mass $M$ (g/mol):

c = \frac{1000\,\rho\,w}{M}

Take commercial concentrated hydrochloric acid: $w = 37\%$ , $\rho = 1.18$ , $M_\text{HCl} = 36.46$ : $c = 1000 \times 1.18 \times 0.37 / 36.46 = 11.97 \approx 12\ \text{mol/L}$ —precisely the stock concentration that the dilution calculator uses as its default. Without density there is no conversion: mass fraction says “how many grams per 100 g of solution”, molarity says “how many moles per 1 L of solution”, and between them stands “how much does 1 L of solution weigh”.

Mass concentration and molarity differ only by $M$ : $c = \rho_B / M$ , which is how the saline in Example 3 converts. Molality and molarity are numerically close in dilute aqueous solutions (1 kg of water ≈ 1 L) and clearly different in concentrated ones (see the note after the examples).

Volumes are not additive and change with temperature

$V$ is the total volume of the solution, which is not the solvent volume plus the solute volume: 50 mL of ethanol plus 50 mL of water gives about 96 mL; salt dissolving in water also changes the volume. The standard preparation is therefore “dissolve first, then make up to the mark”, bringing the total volume exactly to the graduation of a volumetric flask. Flasks are calibrated at 20 °C, and deviations from that temperature introduce a systematic reading error; where temperature varies widely (boiling-point determination, low-temperature reactions), use $b$ or $w$ , which do not change with temperature.

The relationship to inverse proportion and dilution

With $n$ fixed, $c \propto 1/V$ —an inverse proportion: double the volume and the concentration halves (beakers 1 and 2 of Example 1). Writing that as “ $n$ is unchanged by dilution” gives $c_1 V_1 = c_2 V_2$ —the entire dilution calculator is this corollary plus volume units and preparation procedure.

Assumptions

$V$ is the solution volume: the total volume after making up to the mark, not the solvent volume.
A uniform solution: the solute is fully dissolved and evenly mixed.
Fixed temperature: volumes refer to a 20 °C basis; thermal expansion is not considered.
$n$ counts the “units” of the stated substance: 1 mol of NaCl solution contains 1 mol of Na⁺ and 1 mol of Cl⁻; ion concentrations are calculated separately.
SI units: mol, L, mol/L; no conversion is performed.
Interface conventions: when solving for volume, $c = 0$ displays ∞; when solving for $n$ , an unfilled $V$ participates as 0, giving 0.

Scope and limitations

Can calculate: any one of $n$ , $V$ , $c$ from the other two; any positive numbers.
Approximate only: a deviation from 20 °C shifts volumes by of order 0.1%; in concentrated solutions $c$ and $b$ are not interchangeable.
Cannot calculate: $n$ from a mass (use the moles calculator); direct conversion of mass fraction, mass concentration or ppm (needs density and molar mass; see the table under Principles); ion concentrations (multiply by the number of ions in the formula); activity (in concentrated solutions the “effective concentration” is below $c$ ).
Does not: convert mL ↔ L; judge saturation (solubility has a limit, and the formula will not tell you that the salt no longer dissolves).
Numerics: 8 decimal places, trailing zeros removed.

Common pitfalls

Entering mL directly as V: 250 mL entered as 250 gives a concentration 1000 times too small (Example 4). Enter 0.25.
Using solvent volume instead of solution volume: “1 mol in 1 L of water” is not 1 mol/L (see the note).
Mistaking mass concentration for molarity: 0.9% saline is 9 g/L, a molarity of 0.154 (Example 3).
Treating a mass fraction as mol/L: 37% is not 37 mol/L; density is needed (table under Principles).
Entering grams directly as n: $n$ is in mol; 9 g of glucose is 0.05 mol, so divide by the molar mass first.
Forgetting the number of ions: in 0.1 mol/L CaCl₂ the Cl⁻ concentration is 0.2 mol/L.
Confusing M and m: M is mol/L (molarity); m or b is mol/kg (molality).

Typical use cases

Teaching

The definition of the mole and its 2019 redefinition; molar mass and relative molecular mass; distinguishing and converting the five concentration measures; the make-up-to-volume procedure.

Laboratory solution preparation

Solving for the amount of substance to weigh out from a target concentration and volume (then multiplying by the molar mass); working out what volume can be prepared from the solute on hand; preparing working solutions from concentrated stock together with the dilution calculator.

Medicine and physiology

Converting g/L and mol/L for saline and glucose infusions; comparing osmotic pressure in moles; converting blood measures (mmol/L and mg/dL) requires the molar mass.

Environment and water quality

Converting mg/L (≈ ppm) and mol/L; unifying the units of pollutant limits.

FAQ

Is 1 M the same as 1 mol/L?

Yes. M, read as molar, is the customary symbol for mol/L, not an SI unit symbol; formal documents write mol/L or mol/dm³.

Can I enter the volume in mL?

No—the field unit is L. Enter 250 mL as 0.25. A solved volume is also in L.

Why is “1 L of water plus 1 mol of salt” not 1 mol/L?

The solute occupies volume; the total solution volume is about 1.02 L and the concentration about 0.98. The standard procedure is to dissolve first and then add water to the 1 L mark.

How do I convert a mass fraction to molarity?

$c = 1000\rho w / M$ , and you need the density. 37% hydrochloric acid (density 1.18) is about 12 mol/L.

How do I get the amount of substance from a mass of grams?

$n = m/M$ , using the moles calculator; then enter $n$ on this page.

Molarity or molality—which is better?

Use molarity for reaction stoichiometry and titration (measuring liquid by volume is convenient); use molality where temperature varies or for boiling and freezing points (it does not change with temperature). In dilute aqueous solutions the two are numerically close.

The 2019 redefinition changed the mole—do calculations change?

No. $N_A$ went from a measured value to the exact 6.022 140 76 × 10²³, and the equality of molar mass with relative molecular mass is no longer exact only at the $10^{-9}$ level, which has no effect on everyday calculations.

References and further reading

The chem-molarity definition in the CalcX engine source src/data/formulasScience.ts (compute(known, solveFor) solves for c / n / V, with defaults n = 0.5, V = 1); every figure on this page was recomputed by that engine.
BIPM, The International System of Units (SI), 9th edition （访问日期：2026-09-09）—the 2019 edition; the current definition of the mole and the exact value of $N_A$ .
IUPAC, Compendium of Chemical Terminology (Gold Book): amount concentration （访问日期：2026-09-09）—the formal definition and symbol of amount concentration.
IUPAC CIAAW, Standard Atomic Weights （访问日期：2026-09-09）—the source of the standard atomic weights of Na, Cl, C, H, O and others.
Wikipedia, Molar concentration （访问日期：2026-09-09）—conversion formulas to mass fraction and molality.
Wikipedia, Mole (unit) （访问日期：2026-09-09）—the history of becoming a base unit in 1971 and the 2019 redefinition.
Wikipedia, Saline (medicine) （访问日期：2026-09-09）—the composition of 0.9% saline and its 154 mmol/L.

Privacy

All inputs and calculations run inside your browser and are never uploaded to a server. The text on this page is static content; it neither contains nor records any user input.

Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

Related calculators

Open Molarity Calculator c = n/V: Moles, Volume and Concentration in CalcX