IMAT Chemistry · Solutions and Concentration

Molarity, Mass Concentration and Other Units

Concentration says how much solute is in a given amount of solution; the units differ in whether they count moles or grams, and per volume or per mass.

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Why this matters

Converting between mol/L and g/L was asked in 2011, 2016 and 2019, and the 2024 ministry paper asked for the moles of one ion in a volume of a salt solution.

Concept 1 of 4: Molarity: moles of solute per litre of solution

Molarity tells you how many moles of solute are in each litre of solution. Multiply by the volume you take and you know how many moles you have. If the solute splits into ions, each ion has its own concentration, set by how many of it come from one formula unit.

Definition

Molar concentration (molarity) $c$ : moles of solute per litre of solution, unit mol/L, written M. So 0.5 M means 0.5 mol/L.
Volumes must be in litres: $1\ \text{mL} = 1\ \text{cm}^3 = 10^{-3}\ \text{L}$ , and $1\ \text{dm}^3 = 1\ \text{L}$ .
For a strong electrolyte, the concentration of each ion is $c$ times the number of that ion in the formula: 0.2 M $\mathrm{CaCl_2}$ is 0.2 M in $\mathrm{Ca^{2+}}$ and 0.4 M in $\mathrm{Cl^-}$ .
Taking a sample from a solution does not change its concentration, only the moles in your sample.

Molarity

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

$c$ concentration, in mol/L
$n$ moles of solute, in mol
$V$ volume of solution, in L

Worked example

11.1 g of calcium chloride,

\mathrm{CaCl_2}

(

M = 111\ \text{g/mol}

), is dissolved in water and made up to 500 mL. Find the concentration of the solution and of the chloride ions.

Practice this conceptself-check · 4 quick reps

Millilitres must become litres

Molarity is per litre. Multiplying 0.30 mol/L by 200 (mL) instead of 0.200 (L) gives an answer 1000 times too large. Convert the volume first.

Concept 2 of 4: Mass concentration in g/L and converting it to mol/L

Grams per litre and moles per litre describe the same solution in two currencies. The molar mass is the exchange rate: one mole weighs

M

grams, so a concentration in mol/L times

M

gives grams per litre.

Definition

Mass concentration: grams of solute per litre of solution, unit g/L.
From mol/L to g/L: multiply by the molar mass. From g/L to mol/L: divide by it.
Mass of solute in a volume: $m = c \times V \times M$ .

Mass concentration

\rho_{\text{solute}} = c \times M

$\rho_{\text{solute}}$ mass concentration, in g/L
$c$ molar concentration, in mol/L
$M$ molar mass of the solute, in g/mol

Worked example

A sulfuric acid solution is 0.25 mol/L. What is its concentration in g/L, and what mass of acid is in 200 mL of it? (

M(\mathrm{H_2SO_4}) = 98\ \text{g/mol}

)

Practice this conceptself-check · 3 quick reps

g/L and mol/L differ by the molar mass

A 0.1 mol/L solution of a substance with

M = 40

is 4 g/L, not 0.1 g/L. Check which unit the question asks for; options often give the right number in the wrong unit.

Concept 3 of 4: Percentage concentrations and parts per million

Medicine and everyday labels use percentages instead of moles: a saline drip is 0.9%, a disinfectant 3%. Each percentage says how many grams (or mL) of solute are in 100 of something, so it is quick to turn into a mass. For very small amounts, parts per million is easier to read.

Definition

% w/w (mass/mass): grams of solute per 100 g of solution.
% w/v (mass/volume): grams of solute per 100 mL of solution. So 1% w/v = 10 g/L.
% v/v (volume/volume): mL of solute per 100 mL of solution, used for liquids such as ethanol.
ppm (parts per million): grams of solute per $10^6$ g of solution, which is mg per kg. For dilute water solutions, 1 ppm is about 1 mg/L.
The mass of the solution is solute plus solvent.

Percentage by mass and by volume

\%\,w/w = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100 \qquad \%\,w/v = \frac{m_{\text{solute}}\,(\text{g})}{V_{\text{solution}}\,(\text{mL})} \times 100

$m_{\text{solution}}$ mass of solute plus solvent
$V_{\text{solution}}$ volume of the solution, in mL

Worked example

Physiological saline is 0.9% w/v sodium chloride. What mass of NaCl is in a 500 mL bag, and what is its molarity? (

M = 58.5\ \text{g/mol}

)

Practice this conceptself-check · 4 quick reps

Divide by the mass of solution, not of solvent

In % w/w the bottom of the fraction is solute plus solvent. 20 g of sugar in 80 g of water is 20%, not 25%. The same rule applies to ppm.

Concept 4 of 4: Molality, mole fraction and converting with density

Molarity uses the volume of the solution, and volumes change slightly with temperature. Molality uses the mass of the solvent instead, which never changes, so it is the unit used for boiling and freezing points. Mole fraction simply asks what share of all the particles is the solute.

Definition

Molality $b$ : moles of solute per kilogram of solvent, unit mol/kg.
Mole fraction $x$ : moles of one component divided by the total moles of all components. It has no unit, and the mole fractions of all components add up to 1.
For dilute water solutions, molality and molarity are almost equal, because 1 L of solution contains nearly 1 kg of water.
To convert between per-volume and per-mass units you need the density of the solution: mass of 1 L of solution = density (g/mL) times 1000 mL.

Molality and mole fraction

b = \frac{n_{\text{solute}}}{m_{\text{solvent}}\,(\text{kg})} \qquad x_A = \frac{n_A}{n_A + n_B}

$b$ molality, in mol/kg
$m_{\text{solvent}}$ mass of solvent only, in kg
$x_A$ mole fraction of A

Worked example

(a) 6.0 g of urea (

M = 60\ \text{g/mol}

) is dissolved in 250 g of water. Find the molality. (b) 46 g of ethanol (

M = 46

) is mixed with 72 g of water (

M = 18

). Find the mole fraction of ethanol.

Practice this conceptself-check · 4 quick reps

Molality uses kilograms of solvent, molarity uses litres of solution

Molality divides by the mass of the solvent alone, in kilograms. Molarity divides by the volume of the whole solution, in litres. Using grams, or the mass of the solution, gives a wrong molality.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Molarity: moles of solute per litre of solution
Molarity
$c = \frac{n}{V} \qquad n = c\,V$
Mass concentration in g/L and converting it to mol/L
Mass concentration
$\rho_{\text{solute}} = c \times M$
Percentage concentrations and parts per million
Percentage by mass and by volume
$\%\,w/w = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100 \qquad \%\,w/v = \frac{m_{\text{solute}}\,(\text{g})}{V_{\text{solution}}\,(\text{mL})} \times 100$
Molality, mole fraction and converting with density
Molality and mole fraction
$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}\,(\text{kg})} \qquad x_A = \frac{n_A}{n_A + n_B}$

Watch out for (4)

Millilitres must become litres
→ Molarity: moles of solute per litre of solution
g/L and mol/L differ by the molar mass
→ Mass concentration in g/L and converting it to mol/L
Divide by the mass of solution, not of solvent
→ Percentage concentrations and parts per million
Molality uses kilograms of solvent, molarity uses litres of solution
→ Molality, mole fraction and converting with density

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