MOLARITY: CONCENTRATION, UNITS, CALCULATION, EXERCISES - CHEMISTRY - 2025

The molarity is the concentration of a solution expressed in moles of solute per liter of solution. It is abbreviated as M and expresses a relationship between the mass of the solute and the volume of the solution (m / v); although in traditional form this relation is expressed as weight to volume.

A mole is the number of atoms or molecules contained in the atomic or molecular weight; It is expressed in grams / mole. One mole equals 6.02 · 10 ²³ atoms or molecules, known as Avogadro's number.

Molarity formula. Source: Gabriel Bolívar.

There are other ways to express the relationship between the mass of a solute and the volume, including: the percentage relationship between the mass of the solute and the volume of the solution, and normality. The latter is expressed as the number of equivalents of a solute per liter of solution.

Meanwhile, molality, which is defined as the number of moles per kilogram of solvent, usually water, expresses a relationship between the mass of the solute and the mass of the solvent (m / m).

The form of expression of the concentration of a solution in parts per million (ppm), expresses a relationship between one part of the solute and one million parts of the solution, being generally used to express a mass-mass relationship (m / m). But you can express a mass-volume relationship (m / v).

The molar concentration, in addition to being expressed in moles per liter, can be expressed as millimoles / liter (millimolar solution); micromoles / liter (micromolar solution); etc.

Molarity problems can be solved using an analytical method and by using the “rule of three”. The choice of a method will depend on the skill you have in the use of one of the methods.

Molar concentration

It is used to express the concentration of a solute in a given volume of solution.

M = n / V (1)

Where M is equal to molarity, n is the number of moles, and V is the volume of the solution. So, the molarity of a solution is the number of moles of solute per volume of solution expressed in liters.

On the other hand the number of moles is

n = m / PM (2)

Where m is the mass of the dissolved solute and PM its molar mass.

Substituting (2) in (1):

M = (m / PM) / V

Units

The unit in the International System for molar concentration is mol / m ³. This corresponds to a millimolar solution, because one m ³ equals 1,000 liters. In the branches of chemistry and biology, molar concentration is usually expressed as moles / L. This unit is expressed with the M (capital letter).

A solution of one mole / L is equivalent to a solution M; a solution of 10 ^-3 moles / L, equals 1mM (millimolar); and a solution of 10 ^-6 moles / L, is equivalent to 1 µM (micromolar).

How to calculate molarity?

It is convenient to use the expressions indicated above, because this ensures that the result of applying the formula will be in moles / liters, which is the definition of molar concentration.

Then, to calculate the molarity of a solution, it is necessary to express the concentration of the solute in g / L. Then find the molecular weight of the solute (g / mol) and find the ratio between the concentration and the molecular weight. The result obtained is the molarity expressed in moles / liter.

Go from molarity to molality

A necessary piece of information to go from molarity to molality is to know the density of the solution. This allows the mass of the solution to be known, an essential requirement for calculating molality.

First step

First you have to go from the molar concentration to grams / liters. To do this, simply multiply the molarity of the solution by the molecular weight of the solute.

Grams / liter of solute = molarity (moles / liter) · molecular weight of solute (grams / mole).

This allows to obtain the mass of the solute in 1 L of solution.

Second step

Then it is necessary to calculate the mass of the solution. For this, the density of the same is used. Density is usually expressed in grams / cubic centimeter or milliliter.

Mass of solution (g) = volume of solution (mL) density (g / mL)

Third step

Obtaining the mass of the solvent. Since the mass of the solution is equal to the mass of the solute plus the mass of the solvent, to obtain the mass of the latter, it is enough to subtract the mass of the solute (step 1) from the mass of the solution (step 2).

Fourth step

Finally, the solute mass (g) of the solution must be transferred to the solute mass that corresponds to 1,000 g or 1 kg of solvent. To do this, it would be enough to perform a simple rule of three or another equivalent mathematical operation.

Fifth step

Divide g of solute / 1000 g of solvent by the molecular weight of the solute (g / mol) to obtain the molality of the solution.

Numerical example

A 3 M glucose solution (molecular weight 180 g / mol) has a density of 1.15 g / mL. Calculate the molality of this solution.

We first calculate the grams of glucose dissolved in a liter of solution:

g / L = 3 moles / L 180 g / mol

= 540 g / L

Next, we calculate the mass of the solution from its density:

g of solution (mass) = 1,000 mL 1.15 g / mL

= 1,150 g

The mass of solvent is given by the difference:

Mass of solvent = mass of solution - mass of solute

= 1,150 g - 540 g

= 610 g

But 610 g is not 1000 g of solvent as stipulated by the definition of molality. Therefore, it must be calculated how many grams of glucose are dissolved in 1000 g of solvent:

Mass of solute = 540 g solute (1,000 g solvent / 610 g solvent)

= 885.25 g

And finally the molality is calculated by returning the grams to moles:

Molality = (885.25 g of solute / 180 g / mol)

= 4.92 mol solute / kg of solvent

= 4.92 m

Problems solved

Problem 1

How much cupric sulfate will it take to prepare 500 mL of a 0.4 M solution? Express the result in grams. Molecular weight of cupric sulfate (CuSO ₄): 160 g / mol.

We first determine the moles that must be dissolved in such a solution:

M = n / V

n = M V

n = (0.4 mol / L) 0.5 L

= 0.2 moles

Knowing then the number of moles of cupric sulfate its mass can be obtained

n = m / PM

m = nmolecular weight

m = 0.2 mol 160 g / mol

= 32 g of CuSO ₄

That is, 32 grams of this salt must be dissolved in 500 mL of solvent.

Problem 2

What volume of a solution is needed so that when 0.4 mol of solute dissolves, it has a concentration of 0.25 M?

The volume of the solution is obtained from the concept of molarity

M = n / V

V = n / M

V = 0.4 moles / (0.25 moles / L)

= 1.6 L

It means that the solution must have a volume of 1.6 liters to obtain such a concentration of 0.25 M.

Problem 3

A mass of 8 g sodium hydroxide (NaOH) is dissolved in 60 g of a solution that has a density of 1.6 g / mL. What will be the molarity of the solution? Molecular weight of sodium hydroxide: 40 g / mol.

The moles of NaOH must first be calculated:

n = m / PM

= 8 g sodium hydroxide / (40 g / mol)

= 0.2 moles

Now we proceed to calculate the volume of the solution:

m = V d

v = 60 g / (1.6 g / mL)

v = 37.5 mL

To obtain the molarity, the volume of the solution in liters must be placed:

V = 37.5 ml 10 ^-3 L / mL

= 0.0375 L

M = 0.2 moles / 0.0375 L

5.33 moles / L

5.33 M

Problem 4

Calculate the molarity of a hydrochloric acid (HCl) solution with a density of 1.25 g / mL and a 35% concentration, expressed mass / mass. Molecular weight of hydrochloric acid: 36.5 g / mol.

Determine the mass of 35% hydrochloric acid

m = V d

m = 1,000 mL 1.25 g / mL

= 1,250 g

But not everything is HCl, there is also water:

mass HCl = 1,250 g (35/100)

= 437.5 g

Which is the same as saying that in a liter of 35% HCl solution there are 437.5 grams of HCl.

Then, the moles of HCl are calculated, to immediately determine the molarity:

n = m / PM

n = (437.5 g / L) / (36.5 g / mol)

= 11.98 moles / L

Molarity = 11.98 M

Problem 5

Calculate the molarity of a solution containing 5 g of NaCl in 2 L of solution. Molecular weight of sodium chloride: 58.5 g / mol.

You can obtain the moles / L of NaCl in a single step:

molarity = (5 g NaCl / 2 L solution) x (1 mol NaCl / 58.5 g NaCl)

= 0.043 mol / L

Molarity = 0.043 M

Another procedure could be:

g / L NaCl = 5 g / 2 L

= 2.5 g / L

moles / L = (2.5 g / L) / (58.5 g / mol)

= 0.043 mol / L

= 0.043 M

References

1. Rodríguez, M. (2007). Chemistry. Salesian Editorial Foundation
2. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
3. Wikipedia. (2019). Molarity Recovered from: es.wikipedia.org
4. Atarés Huerta. (sf). Molarity and molality.. Recovered from: riunet.upv.es
5. Softchools. (2019). Molarity formula. Recovered from: softschools.com