Search the World of Chemistry

×

Molarity of Pure Substances

23rd Nov 2019 @ 2 min read

Analytical Chemistry

The molarity (aka molar concentration) of a pure substance is the ratio of the density of the pure substance to its molar mass. It is calculated by the following formula.

Here, c is the molarity, ρ is the density, and M is the molar mass of pure compound.

Derivation

Molarity is the number of moles of a substance (usually a solute) present in a litre of the solution.

Here, ci is the molarity of a solute, mi is moles of the solute, and V is the volume of the solution.

In a pure substance, there is no solute or solvent, only one chemical. Thus, ci = c and ni = n.

Volume is mass divided by density; .

Here, is molar mass (mass divided by the number of moles).

Note: The term the molarity of a pure substance is inappropriate because molarity is a property of solutions. A solution is a homogeneous mixture of two or substances. And pure substances are not solutions. Molar density is the appropriate term to use here, not molarity or molar concentration. The ratio of density of a substance to the molar mass of the substance gives its molar density.

In short, molar density and molarity are the same for pure substances ignorantly.

Examples

The molar concentration (or molar density correctly) of some common chemicals is calculated below.

Example 1: Water

ρH2O = 1000 g L−1 and MH2O = 18.0 g mol−1.

Example 2: Ethanol

ρC2H6O = 789 g L−1 and MC2H6O = 46.1 g mol−1.

Example 3: Methanol

ρCH4O = 792 g L−1 and MCH4O = 32.0 g mol−1.

Example 4: Benzene

ρC6H6 = 876 g L−1 and MC6H6 = 78.1 g mol−1.

Example 5: Acetic acid

ρC2H4O2 = 1050 g L−1 and MC2H4O2 = 60.1 g mol−1.

Example 6: Glycerol

ρC3H8O3 = 1260 g L−1 and MC3H8O3 = 92.1 g mol−1.

Associated articles

If you appreciate our work, consider supporting us on ❤️ patreon.
Molarity

Copy Article Cite