The **molarity**also called **molar concentration**is a way of measuring the concentration of a solution that is defined as the **number of moles of solute per liter of solution**understanding dissolution (or solution) as a homogeneous mixture of substances.

The substance that is in greater proportion is called solvent or solvent. The other components of the solution are called solutes. If there are several solutes in a solution, the molar concentration for each of them can be calculated.

The **formula** general to calculate the molarity would be the following:

**molarity = mol/L (moles per liter)**

And to calculate the number of moles:

**moles = mass (g)/molar mass (g)**

Thus, the calculation of the molarity of a solution will always go through these **step**yes:

- Find the number of
**moles of solute** - Find the volume of the solution in
**liters**(eye, the volume of the solution, not the volume of solvent) **Split**the number of moles divided by the number of liters

To calculate the number of liters, the corresponding conversion is applied according to the scale used. For example, 100 ml is 0.1 L (1 L is 1000 ml, so 100 ml = 100/1000 = 0.1 L).

To calculate the number of moles from the mass, we first convert the mass to grams (if we don't already have it in grams) and then divide the **grams divided by the molar mass**.

The molar mass of an element can be found in most periodic tables as **atomic weight or mass**.

If the solute is not an element but a compound or a molecule, then **add the atomic weight** of each element that forms the molecule to obtain the **total molar mass of the molecule**.

## What is molarity or molar concentration?

One mole of a pure substance, whether it is an element or a compound, contains one **number of entities equal to Avogadro's number**either of atoms or molecules, depending on the composition of the substance:

**1 mole = 6.022 140 857(74) ×10 ^{23} entities (atoms or molecules)**

The **atomic mass** corresponds to the **number of grams in 1 mole**and using this relationship we can easily go from mass to number of moles.

For example, if the atomic mass of oxygen is 15.9994 u, it means that in 15.9994 g there are 1 mole of oxygen atoms, that is, a number of atoms equal to Avogadro's number.

If we have **molecular oxygen, O _{two}**the usual form of oxygen in the atmosphere, the molar mass will be the

**mass of oxygen atom times two**that is, 31.9988 g.

This means that 1 mole of atomic oxygen (O) and 1 mole of molecular oxygen (O_{two}) contain the same number of entities (Avogadro's number), but 1 mol of atomic oxygen has a mass of 15.9994 g and 1 mol of molecular oxygen (O_{two}) has a mass of 31.9988 g.

Using moles instead of mass allows solutions of substances to be prepared in terms of number of atoms or number of molecules, which in turn allows **calculate stoichiometric ratios**which are the ratios in which chemical reactions occur.

But counting the number of atoms or molecules is much more difficult than obtaining their weight or mass. That is why it is so important to calculate the molarity or molar concentration (mass/volume) of a solution from the mass/volume concentration.

## Calculate molarity: examples

**Example: Obtain the molar concentration of 25 g of common salt (ClNa) in 1 L of solution**

The first thing we have to do is calculate the number of moles that correspond to 25 g of ClNa, and for this we must first calculate the molar mass of ClNa. We consult a periodic table, take the mass or atomic weight of each element and add it:

**atomic mass of chlorine**: 35g**sodium atomic mass**: 23g**molar mass of sodium chloride**: 35 g + 23 g = 58 g (this means that 1 mole of ClNa has a mass of 58 g)

Now that we know the molar mass, we divide the mass we have (25 g) by the molar mass and get the number of moles:

**number of moles in 25 g of ClNa**: 25 gx (1 mol/58 g) = 25/58 = 0.43 mol

**Result**: if we dissolve **25g NaCl** in sufficient quantity of water to complete **1 L of solution**the **molar concentration** of salt in this solution is **0.43mol/L**.

**Example: We dissolve 23.7 g of potassium permanganate (KMnO _{4}) in a quantity of water sufficient to prepare 750 ml of solution. Calculate the molar concentration of KMnO_{4}.**

The objective is to obtain the number of moles and the volume in liters and then divide both. So we have to convert the grams of permanganate to number of moles, and the milliliters of solution to liters.

The volume in liters is the easy part, just divide 750 by 1000 to go from milliliters to liters:

**750ml = 0.75L**

To calculate the number of moles we have to consult a periodic table in which the atomic weight or mass is indicated, and write down the atomic mass of the elements that make up potassium permanganate: manganese, oxygen and potassium.

Then we multiply the atomic mass of each by the number of atoms they have in the molecule (KMnO_{4}):

**K**: atomic mass of 39.1 g, for 1 atom = 39.1 g**min**: 54.9 g x 1 = 54.9 g**EITHER**: 16.0 g x 4 = 64.0 g_{4}

Finally, we add all the masses of each element separately to obtain the total molar mass of the potassium permanganate molecule:

- 1 mole of KMnO4 = 39.1 g + 54.9 g + 64.0 g = 158 g

In the statement of the problem we are given 23.7 g of KMnO_{4}if in 158 g there is 1 mole, following a rule of three, in 23.7 there will be 0.15 moles:

- moles of KMnO
_{4}= 23.7 gx (1 mole /158 g) = 23.7/158 = 0.15 moles

Thus, we have arrived at that our dissolution has **0.15 moles of KMnO4 per 0.75 L**. If we divide, we will obtain the number of moles that would be in 1 L, and thus we obtain the molarity or molar concentration:

**molarity = 0.15 mol / 0.75 L = 0.2 mol/L**

With this data, **How many molecules of permanganate are there in the solution?** Well, there are 0.2 times Avogadro's number, that is, 0.2 × 6.022 140 857(74) ×1023.

We have seen two examples. In the first we started with mass in grams (25 g) and volume in liters (1 L). In the second we started with mass in grams (23.7 g) and volume in milliliters (750 ml).

But in any of the cases, to calculate the molarity it is always necessary to arrive at **number of moles** and to **volume of solution to liters**, and once you have these data, just divide to obtain the amount in mol/L. Having this clear, the calculation of molarity is quite simple.

**Note**: the unit of molar concentration in the International System is **mol/m ^{3}**but the use of

**mol/L**or its equivalent

**mol/dm**in literature it has traditionally been much higher. It is also common for mol/L to be expressed as

^{3}**M**. For example, a concentration of 0.43 M (0.43 molar) is equal to 0.43 mol/L.