Concentration Terms and Interconversions

CHEConcentration & Solubility

This subtopic covers the standard ways to express concentration — molarity, molality, mole fraction, mass percentage, ppm — and converting between them using density. Knowing which terms are temperature-dependent is a frequent conceptual question.

Molarity and molality

M = \dfrac{n_{solute}}{V_{solution}\,(\text{L})} \qquad m = \dfrac{n_{solute}}{w_{solvent}\,(\text{kg})}

n

moles of solute,

V

litres of solution,

w

kg of solvent

Mole fraction

x_A = \dfrac{n_A}{n_A + n_B}, \qquad x_A + x_B = 1

binary solution of components

A

and

B

Mass percent to molarity

M = \dfrac{10 \times w \times d}{M_r}

w

mass percent (% w/w),

d

density in g/mL,

M_r

solute molar mass

Molarity to molality

m = \dfrac{1000\,M}{1000\,d - M\,M_r}

d

density in g/mL,

M_r

solute molar mass

Parts per million

\text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^6

used for very dilute solutions

Molarity $M = \dfrac{\text{moles of solute}}{\text{volume of solution in L}}$ ; molality $m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$ .
Temperature-independent terms: molality, mole fraction and mass percentage (mass-based); molarity is temperature-dependent because solution volume expands on heating.
Mole fraction $x_A = \dfrac{n_A}{n_A + n_B}$ , and $x_A + x_B = 1$ for a binary solution.
Mass percentage to molarity: from $w$ % w/w and density $d$ (g/mL), $M = \dfrac{10 \times w \times d}{M_r}$ .
Molarity to molality: $m = \dfrac{1000\,M}{1000\,d - M\,M_r}$ , where $d$ is in g/mL and $M_r$ is solute molar mass.
ppm $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$ — used for very dilute solutions like dissolved gases in water.
Mass percentage (w/w) is mass of solute per $100\,\text{g}$ of solution; volume percentage (v/v) and mass/volume percentage (w/V) are distinct and must not be confused.
Mass of solution = mass of solute + mass of solvent, and mass of solution = density $\times$ volume — the bridge that lets density convert mass-based terms into volume-based ones.
Molality and mole fraction are interconvertible without density: for a binary aqueous solution, $m = \dfrac{x_{solute}}{x_{solvent}} \times \dfrac{1000}{M_{solvent}}$ .
Choosing a basis simplifies conversions: $100\,\text{g}$ of solution for w/w data, or $1\,\text{L}$ of solution (mass $= 1000d$ ) for molarity data.
Only molarity needs temperature correction because it alone is built on solution volume; all mass- and mole-based terms are unaffected by heating.
Trap: take a convenient basis (e.g. $100\,\text{g}$ of solution for w/w problems or $1\,\text{L}$ for molarity) to avoid juggling unknown amounts.

Where the marks go

Using mass of solution in the molality formula instead of mass of solvent — molality is per kg of solvent, not solution.
Forgetting to subtract solute mass: in molarity-to-molality, the solvent mass is $1000d - M\,M_r$ , not $1000d$ .
Mixing up % w/w, % v/v and % w/V; only mass percentage is directly temperature-independent and basis-friendly.
Calling molality temperature-dependent — it is molarity that changes with temperature because of volume expansion.
Slipping a factor of 10 or $10^6$ when converting density units (g/mL vs kg/L) or computing ppm.

How the board asks it

Conversionmass percent to molarity using density
A commercial sample of concentrated $HCl$ is $36\%$ by mass and has a density of $1.18\,\text{g/mL}$ . Calculate its molarity. (Molar mass of $HCl = 36.5\,\text{g mol}^{-1}$ .)
Numericalmolarity and molality definitions
Calculate the molarity and molality of a solution prepared by dissolving $4.0\,\text{g}$ of $NaOH$ in water to make $250\,\text{mL}$ of solution, given the density of the solution is $1.04\,\text{g/mL}$ . (Molar mass of $NaOH = 40\,\text{g mol}^{-1}$ .)
Conversionmolarity to molality interconversion
An aqueous solution of glucose is $2\,\text{M}$ and has a density of $1.12\,\text{g/mL}$ . Calculate the molality of the solution. (Molar mass of glucose $= 180\,\text{g mol}^{-1}$ .)
Give reasonstemperature-dependence of concentration terms
Give reasons: The molarity of a solution decreases when its temperature is raised, whereas its molality remains unchanged.
Numericalmole fraction of a binary solution
Calculate the mole fraction of ethanol ( $C_2H_5OH$ ) and of water in a solution containing $46\,\text{g}$ of ethanol dissolved in $90\,\text{g}$ of water.
Numericalppm for dilute solutions
Define parts per million (ppm), and calculate the concentration in ppm of dissolved oxygen if $5.8\times10^{-3}\,\text{g}$ of $O_2$ is present in $1\,\text{kg}$ of water.

Practise this topic

NextSolubility and Henry's Law

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.