Molar Concentration - Examples

Examples

Example 1: Consider 11.6 g of NaCl dissolved in 100 g of water. The final mass concentration (NaCl) will be:

(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:

= (11.6 g + 100 g) / (1.07 g/mL) = 104.3 mL

The molar concentration of NaCl in the solution is therefore:

(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:

(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:

(H₂O) = 1000 g/L / (18.02 g/mol) = 55.5 mol/L

Likewise, the concentration of solid hydrogen (molar mass = 2.02 g/mol) is:

(H₂) = 88 g/L / (2.02 g/mol) = 43.7 mol/L

The concentration of pure osmium tetroxide (molar mass = 254.23 g/mol) is:

(OsO₄) = 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 L. Thus, the number concentration is:

= 60 / (10−15 L)= 6×1016 L−1

The molar concentration is:

= 6×1016 L−1 / (6×1023 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 (Na₂CO₃) = 1 mol/L, the molar concentrations are (Na+) = 2 mol/L and (CO₃2-) = 1 mol/L because the salt dissociates into these ions.