## JEE advanced important topics : Molarity, Molality and Normality

In this post I will talk about JEE advanced important topics : Molarity, Molality and Normality.
The concentration is the physical quantity that expresses the quantity of an element or a compound per unit volume. In the SI, the mol · m-3 units are used. Each substance has a solubility that is the maximum amount of solute that can be dissolved in a solution, and depends on conditions such as temperature, pressure, and other substances dissolved or suspended.
In chemistry, to quantitatively express the ratio between a solute and the solvent in a solution, different units are used: molarity, normality, molality, formality, percentage by weight, percentage in volume, mole fraction, parts per million, parts per billion, parts per trillion, etc. It can also be expressed qualitatively using terms such as diluted, for low concentrations, or concentrated, for highs.
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The molarity (M) is the number of moles of solute per liter of solution. For example, if 0.5 moles of solute are dissolved in 100 mL of solution, there is a concentration of that solute of 5.0 M (5.0 molar). To prepare a solution of this concentration, the solute is usually dissolved first in a smaller volume, for example 30 mL, and that solution is transferred to a graduated flask, and then filled with more solvent up to 100 mL.
M = n / V = moles of solute / liters of solution (mol / l ≡ molar)
It is the most common method of expressing concentration in chemistry, especially when working with chemical reactions and stoichiometric relationships. However, it has the disadvantage that the volume changes with temperature.

## The Molar fraction

It is a chemical unit to express the absolute concentration in a solution. It expresses to us the proportion in which the mole of solute is found with respect to the total moles of dissolution, which are calculated by adding the moles of solute (s) and solvent. To calculate the mole fraction of a homogeneous mixture, the following expression is used:
$x_{i}= \frac{n_{i}}{n_{t}}$
Where ni is the number of the solutes, and nt the total number of moles in the whole solution (both solutes and solvent).
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As the volume of a solution depends on temperature and the pressure; when they change, the volume changes with them. Thanks to the fact that the molar fraction is not a function of volume, it is independent of temperature and pressure. It should also be noted that in ideal gas the volume variation will be proportional for each one of the solutes, and therefore also for the solution. In this way there is a direct relationship between the molar fractions and the partial volumes.

### Sum of molar fractions

For example, in a binary mixture of 6 moles of ethanol and 4 moles of water, which gives a total of 10 moles, the molar fraction of ethanol is 6/10 = 0.6; while the mole fraction of water is 4/10 = 0.4. All molar fractions of a solution will always be less than 1, and the sum of these will result in 1.
Molality
The molality (m) is the number of moles of solute per kilogram of solvent. To prepare solutions of a certain molality in a solute, a volumetric flask is not used as in the case of molarity, but it can be done in a beaker and weighing with an analytical balance, previous weight of the empty glass to be able subtract the corresponding value.
m = moles of solute / kilograms of solvent (mol / kg ≡ molal)
The main advantage of this method of measurement with respect to molarity is that as the volume of a solution depends on the temperature and the pressure, when these change, the volume changes with them. Because molality is not a function of volume, it is independent of temperature and pressure, and can be measured more accurately.
It is less used than molarity.
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The normality (N) is the number of equivalents (n) of solute (st) per liter of solution (sc).
$N= \frac{n_{st}}{V_{st}}$

#### Acid-base normality

It is the normality of a solution when it is used for a reaction such as acid or base. This is why they are usually titrated using pH indicators.
In this case, the equivalents can be expressed as follows:
n = moles * H + for an acid, or n = moles * HO- for a base.
Where:
n: is the amount of equivalents.
moles: is the amount of moles.
H +: Is the amount of protons given by one mole of acid.
HO-: It is the quantity of hydroxyl yielded by one mole of the base.
For this, we can say the following:
N = M * H + for an acid, or N = M * HO- for a base.
Where:
N: it is the normality of the solution.
M: is the molarity of the solution.
H +: Is the amount of protons given by one mole of acid.
HO-: It is the quantity of hydroxyl yielded by one mole of the base.

#### Examples:

A 1 M solution of HCl yields 1 M of H +, therefore, equals 1 N.
A 1 M solution of Ca (OH) 2 yields 2 M of HO-, therefore, equals 2 N.
In this case, the equivalents can be expressed as follows:
n = moles * e-.
Where:
n: is the amount of equivalents.
moles: is the amount of moles.
e-: It is the quantity of electrons exchanged in the oxidation or reduction hemireaction.
For this, we can say the following:
N = M * e-.
Where:
N: it is the normality of the solution.
M: is the molarity of the solution.
e-: It is the quantity of electrons exchanged in the oxidation or reduction hemireaction.

#### Examples:

In the following case we see that the nitrate anion in acid medium (for example nitric acid, can act as an oxidant, where a 1 M solution, is 3 Nox.
4 H + + NO3-1 + 3 e- ↔ NO + 2 H2O
In the following case we see that the iodide anion, can act as a reducer, where a solution 1 M, is 1 Nrd.
2 I- – 2 e- ↔ I2
In the following case we see that the silver cation can act as an oxidant, where a 1 M solution is 1 Nox.
1 Ag + + 1 e- ↔ Ag0