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Molarity Calculator: Concentration in mol/dm³

Molarity is concentration in moles per cubic decimetre: c = n / V, where n is the number of moles of solute and V is the volume of solution in dm³. From a mass, find the moles first: c = m / (M × V).

Molarity formula: c = n / V

From moles
c = nV
c = nVn = c·VV = nc
From mass
c = mM·V
c = mM·Vm = c·M·VV = mc·M
From % by mass
c = w·ρ·1000M
c = w·ρ·1000Mw = c·Mρ·1000
c
concentration, mol/dm³
n
number of moles, mol
V
volume of solution, dm³
m
mass of solute, g
M
molar mass, g/mol
w
percentage by mass
ρ
density of solution, g/cm³

The legend lists the base units: g, mol, dm³ and g/mol. You can enter mg, kg, cm³ and other units; the calculator converts them and shows the conversion as a step. Relative atomic masses are rounded as in the GCSE periodic table, for example Cl = 35.5 and Cu = 63.5.

H₂SO₄M(H₂SO₄) = 98 g/mol

2.33 mol/dm³
g/mol
cconcentration
2.33 mol/dm³
concentration: 2.33 mol/dm³
A solution of H₂SO₄ with w = 0.2 and a density of 1.14 g/cm³ (M = 98 g/mol) has a concentration of 2.33 mol/dm³.

Step-by-step solution

  1. Molar massM(H₂SO₄) = 2·1 + 32 + 4·16 = 98 g/mol
  2. Formulac = w·ρ·1000M
  3. Converting unitsw = 20 % = 0.2
  4. Substitutec = 0.2 · 1.14 g/cm³ · 100098 g/mol
  5. Answerc ≈ 2.33 mol/dm³

Molarity examples

  1. Example 1

    0.25 mol of sodium chloride is dissolved in water to make 0.5 dm³ of solution. Calculate the concentration in mol/dm³.

    Given:n = 0.25 molV = 0.5 dm³
    1. Formulac = nV
    2. Substitutec = 0.25 mol0.5 dm³
    3. Answerc = 0.5 mol/dm³

  2. Example 2

    A technician needs to make 250 cm³ of sodium hydroxide solution, NaOH, with a concentration of 2 mol/dm³. Calculate the mass of sodium hydroxide needed.

    Given:NaOHc = 2 mol/dm³V = 250 cm³
    1. Molar massM(NaOH) = 23 + 16 + 1 = 40 g/mol
    2. Formulam = c·M·V
    3. Converting unitsV = 250 cm³ = 0.25 dm³
    4. Substitutem = 2 mol/dm³ · 40 g/mol · 0.25 dm³
    5. Answerm = 20 g

  3. Example 3

    A student dissolves 10.6 g of anhydrous sodium carbonate, Na₂CO₃, in water and makes the solution up to 500 cm³ in a volumetric flask. Calculate the concentration of the solution in mol/dm³.

    Given:Na₂CO₃m = 10.6 gV = 500 cm³
    1. Molar massM(Na₂CO₃) = 2·23 + 12 + 3·16 = 106 g/mol
    2. Formulac = mM·V
    3. Converting unitsV = 500 cm³ = 0.5 dm³
    4. Substitutec = 10.6 g106 g/mol · 0.5 dm³
    5. Answerc ≈ 0.2 mol/dm³

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What is molarity?

Molarity, or concentration in mol/dm³, tells you how many moles of solute are dissolved in each cubic decimetre of solution. The formula is concentration = moles ÷ volume, c = n / V. A 1 mol/dm³ solution of sodium hydroxide (Mr = 40) contains 40 g of NaOH in every 1 dm³. The volume is that of the whole solution, not of the water added.

Working in moles is what makes concentrations useful for reactions. In a titration, you measure the volume of one solution that exactly reacts with a known volume of another, then use the mole ratio from the balanced equation to find the unknown concentration. That is why mol/dm³ appears in the GCSE Higher tier and throughout A-level chemistry, while g/dm³ is mainly used to describe how a solution was made.

Volumes are often given in cm³, so divide by 1000 to get dm³: 25.0 cm³ is 0.0250 dm³. If you know the mass, convert it to moles first: 4 g of NaOH is 0.1 mol, and made up to 0.25 dm³ it gives 0.4 mol/dm³. To convert from g/dm³, divide by Mr. From a percentage by mass and a density, c = w × ρ × 1000 ÷ M, with w as a decimal and ρ in g/cm³.

Common mistakes

  • Leaving the volume in cm³: 25 cm³ must become 0.025 dm³ before you divide.
  • Putting the mass into c = n / V. Divide the mass by Mr first to get moles.
  • Using the volume of water added instead of the final volume of the solution.
  • Confusing mol/dm³ with g/dm³. Multiply mol/dm³ by Mr to get g/dm³.
  • Using a percentage as a whole number in c = w × ρ × 1000 ÷ M: 20% must go in as 0.20.

More chemistry formulas

ConcentrationMolesRelative formula massPeriodic tableAll chemistry equations

Frequently asked questions

Concentration in mol/dm³ = moles of solute ÷ volume of solution in dm³, or c = n / V.

Divide the mass by Mr to get moles, then divide by the volume in dm³. For example, 11.7 g of NaCl (Mr = 58.5) is 0.2 mol; in 0.5 dm³ that is 0.4 mol/dm³.

2 M is shorthand for 2 mol/dm³: two moles of solute in every cubic decimetre of solution. GCSE papers write the unit in full.

Multiply by Mr. Hydrochloric acid at 0.5 mol/dm³ (Mr = 36.5) is 0.5 × 36.5 = 18.25 g/dm³.

Molarity is one way of giving concentration, in mol/dm³. Concentration can also be given in g/dm³ or as a percentage by mass.