"Molarity" redirects here. It is not to be confused with
Molality.
In chemistry, the molar concentration, $c\_i$ is defined as the amount of a constituent $n\_i$ (usually measured in moles – hence the name) divided by the volume of the mixture $V$:^{[1]}
 $c\_i\; =\; \backslash frac\; \{n\_i\}\{V\}$
It is also called molarity, amountofsubstance concentration, amount concentration, substance concentration, or simply concentration. The volume $V$ in the definition $c\_i\; =\; n\_i/V$ refers to the volume of the solution, not the volume of the solvent. One litre of a solution usually contains either slightly more or slightly less than 1 litre of solvent because the process of dissolution causes volume of liquid to increase or decrease.
The reciprocal quantity represents the dilution (volume) which can appear in Ostwald's law of dilution.
Notation
In addition to the notation $c\_i$ there is a notation using brackets and the formula of a compound like [A]. This notation is encountered especially in equilibrium constants and reaction quotients.
Units
The SI unit is mol/m^{3}. However, more commonly the unit mol/L is used.
A solution of concentration 1 mol/L is also denoted as "1 molar" (1 M).
 1 mol/L = 1 mol/dm^{3} = 1 mol dm^{−3} = 1 M = 1000 mol/m^{3}.
An SI prefix is often used to denote concentrations. Commonly used units are listed in the table hereafter:
Name

Abbreviation

Concentration

Concentration (SI unit)

millimolar

mM

10^{−3} mol/dm^{3}

10^{0} mol/m^{3}

micromolar

μM

10^{−6} mol/dm^{3}

10^{−3} mol/m^{3}

nanomolar

nM

10^{−9} mol/dm^{3}

10^{−6} mol/m^{3}

picomolar

pM

10^{−12} mol/dm^{3}

10^{−9} mol/m^{3}

femtomolar

fM

10^{−15} mol/dm^{3}

10^{−12} mol/m^{3}

attomolar

aM

10^{−18} mol/dm^{3}

10^{−15} mol/m^{3}

zeptomolar

zM

10^{−21} mol/dm^{3}

10^{−18} mol/m^{3}

yoctomolar

yM^{[2]}

10^{−24} mol/dm^{3} (1 particle per 1.6 L)

10^{−21} mol/m^{3}

Related quantities
Number concentration
The conversion to number concentration $C\_i$ is given by:
 $C\_i\; =\; c\_i\; \backslash cdot\; N\_\{\backslash rm\; A\}$
where $N\_\{\backslash rm\; A\}$ is the Avogadro constant, approximately 6.022×10^{23} mol^{−1}.
Mass concentration
The conversion to mass concentration $\backslash rho\_i$ is given by:
 $\backslash rho\_i\; =\; c\_i\; \backslash cdot\; M\_i$
where $M\_i$ is the molar mass of constituent $i$.
Mole fraction
Main article:
Mole fraction
The conversion to mole fraction $x\_i$ is given by:
 $x\_i\; =\; c\_i\; \backslash cdot\; \backslash frac\{M\}\{\backslash rho\}\; =\; c\_i\; \backslash cdot\; \backslash frac\{\backslash sum\_i\; x\_i\; M\_i\}\{\backslash rho\}$
 $x\_i=\; c\_i\; \backslash cdot\; \backslash frac\{\backslash sum\; x\_j\; M\_j\}\{\backslash rho\; \; c\_i\; M\_i\}$
where $M$ is the average molar mass of the solution, $\backslash rho$ is the density of the solution and j is the index of other solutes.
A simpler relation can be obtained by considering the total molar concentration namely the sum of molar concentrations of all the components of the mixture.
 $x\_i\; =\; \backslash frac\{c\_i\}\{c\}\; =\; \backslash frac\{c\_i\}\{\backslash sum\; c\_i\}$
Mass fraction
The conversion to mass fraction $w\_i$ is given by:
 $w\_i\; =\; c\_i\; \backslash cdot\; \backslash frac\{M\_i\}\{\backslash rho\}$
Molality
The conversion to molality (for binary mixtures) is:
 $b\_2\; =\; \backslash frac\; \backslash ,$
where the solute is assigned the subscript 2.
For solutions with more than one solute, the conversion is:
 $b\_i\; =\; \backslash frac\; \backslash ,$
Properties
Sum of molar concentrations  normalizing relation
The sum of molar concentrations gives the total molar concentration, namely the density of the mixture divided by the molar mass of the mixture or by another name the reciprocal of the molar volume of the mixture. In an ionic solution ionic strength is proportional to the sum of molar concentration of salts.
Sum of products molar concentrationspartial molar volumes
The sum of products between these quantities equals one.
 $\backslash sum\_i\; c\_i\; \backslash cdot\; \backslash bar\{V\_i\}\; =\; 1$
Dependence on volume
Molar concentration depends on the variation of the volume of the solution due mainly to thermal expansion. On small intervals of temperature the dependence is :
 $c\_i\; =\; \backslash frac\; \}$
where $c\_\{i,T\_0\}$ is the molar concentration at a reference temperature, $\backslash alpha$ is the thermal expansion coefficient of the mixture.
Spatial variation and diffusion
Molar and mass concentration have different values in space where diffusion happens.
Examples
Example 1: Consider 11.6 g of NaCl dissolved in 100 g of water. The final mass concentration $\backslash rho$(NaCl) will be:
 $\backslash rho$(NaCl) = 11.6 g / (11.6 g + 100 g) = 0.104 g/g = 10.4 %
The density of such a solution is 1.07 g/mL, thus its volume will be:
 $V$ = (11.6 g + 100 g) / (1.07 g/mL) = 104.3 mL
The molar concentration of NaCl in the solution is therefore:
 $c$(NaCl) = (11.6 g / 58 g/mol) / 104.3 mL = 0.00192 mol/mL = 1.92 mol/L
Here, 58 g/mol is the molar mass of NaCl.
Example 2: Another typical task in chemistry is the preparation of 100 mL (= 0.1 L) of a 2 mol/L solution of NaCl in water. The mass of salt needed is:
 $m$(NaCl) = 2 mol/L * 0.1 L * 58 g/mol = 11.6 g
To create the solution, 11.6 g NaCl are placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL.
Example 3: The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02=0.055 mol/g). Therefore, the molar concentration of water is:
 $c$(H_{2}O) = 1000 g/L / (18.02 g/mol) = 55.5 mol/L
Likewise, the concentration of solid hydrogen (molar mass = 2.02 g/mol) is:
 $c$(H_{2}) = 88 g/L / (2.02 g/mol) = 43.7 mol/L
The concentration of pure osmium tetroxide (molar mass = 254.23 g/mol) is:
 $c$(OsO_{4}) = 5.1 kg/L / (254.23 g/mol) = 20.1 mol/L.
Example 4: Proteins in bacteria, such as E. coli, usually occur at about 60 copies, and the volume of a bacterium is about $10^\{15\}$ L. Thus, the number concentration $C$ is:
 $C$ = 60 / (10^{−15} L)= 6×10^{16} L^{−1}
The molar concentration is:
 $c\; =\; C\; /\; N\_A$ = 6×10^{16} L^{−1} / (6×10^{23} mol^{−1}) = 10^{−7} mol/L = 100 nmol/L
If the concentration refers to original chemical formula in solution, the molar concentration is sometimes called formal concentration. For example, if a sodium carbonate solution has a formal concentration of $c$(Na_{2}CO_{3}) = 1 mol/L, the molar concentrations are $c$(Na^{+}) = 2 mol/L and $c$(CO_{3}^{2}) = 1 mol/L because the salt dissociates into these ions.
References
External links
 Molar Solution Concentration Calculator
 Experiment to determine the molar concentration of vinegar by titration
This article was sourced from Creative Commons AttributionShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, EGovernment Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a nonprofit organization.