There are numerous methods by using which the concentration of a solution can be expressed. In this section, following methods are explained:

1. Percent by mass or mass percentage

2. Percent by volume

3. Percent mass by volume

4. Strength or concentration

5. Parts per million

6. Mole fraction

7. Molality

8. Molar concentration (molarity)

9. Normality

10. Formality

**1.**** ****Percent by mass or mass percentage**

Mass percentage or percent by mass = (Mass of solute / Mass of solution) X 100

Here, mass of solution can be expressed by the following formula:

Mass of solution = Mass of solute + Mass of solvent

The mass of solution can also be expressed as:

Mass of solution = Volume of solution X Density of solution

**Mass Fraction:**

Mass fraction = Mass of solute / Mass of solution

Hence, the mass percentage of solute can be calculated by multiplying mass fraction with 100.

20% solution of glucose means that 20 gram of glucose (solute) is present in 100 grams of the solution, i.e., 20 gram of glucose has been dissolved in 80 gram of water (solvent).

**2. Percent by Volume:**

Percent of solute by volume = (Volume of solute / Volume of solution) X 100

**3. Percent mass by volume:**

Percent of solute of mass by volume = (Mass of solute / Volume of solution) X 100

**4. ****Strength o****rS****olution ****Concentration ****(Concentration in Gram per lite****r****):**

Concentration of solution = (Solute mass in gram / Solution volume in liters)

If the unit of volume is in mL, then the overall formula should be multiplied by 1000.

Concentration of solution = (Solute mass in gram / Solution volume in mL) X 1000

The strength of the solution can be defined as the concentration of solution in gram per liter. Suppose, 10 g of glucose (solute) is present in 1 L of solution, then the strength or concentration of the solution is 10 g/L.

**5. PPM (Parts per million)****:**^{6} gram of the solution.

ppm = (Mass of solute / Mass of solution) X 10^{6}

The example in which the solute is present in trace quantity is the atmospheric pollution in cities. This concentration refers to the pollutant volume in 10^{6} units of air volume. 20 ppm of sulfur dioxide (SO_{2}) in air means that 10 mL of SO_{2} is present in 10^{6} mL of air volume.

**6. Mole Fraction:**

Suppose there are three components present in the solution, namely component A, B and C. The moles of component A is a, the moles of component B is b, and the moles of component C is c. Hence, the total moles of solution is (a + b + c).

According to the definition of mole fraction, the mole fraction of component A, B, and C can be calculated by following way.

Mole fraction of A =*f *_{A} = a / (a + b + c)

Mole fraction of B =*f *_{B} = b / (a + b + c)

Mole fraction of C =*f *_{C} = c / (a + b + c)

Fact on Mole fraction: The addition of mole fraction of all components is always equal to 1, i.e.,

*f** *_{A}+ *f *_{B} + *f *_{C} = 1

In case of binary solution (i.e. the condition when the solution is made up of only one solute and remaining is solvent.),

Total mole fraction = Mole fraction of solute + Mole fraction of solvent = 1

Consider the case of a solution which is made up of solute A and solvent B. The moles of solute A is a and the moles of solvent B is b. Then the mole fraction is:

Mole fraction of solute A =*X *_{A}= a / (a + b)

Mole fraction of solute B =*X *_{B}= b / (a + b)

Hence, the total mole fraction is:

Total Mole Fraction =*X *_{A} + *X *_{B} = 1

Note- The mole fraction of solution is independent of temperature of the solution.

**7. Molality:**

Molality can be represented by following formula:

Molality (m) = [Number of solute moles / Number of kilograms (kg) of the solvent present]

Suppose, a gram of solute of molecular mass b be present in c gram of the solvent, then the molality can be calculated as:

Molality (m) = [a / (b X c)] X 1000

Molality (m) = [A / c] X 1000

Where, A is total number of moles of solute i.e. a / b.

Please note following points on molality.

i. Molality involves the mass of liquids during the calculation rather than involving volumes of liquids. Hence, this method is most convenient method of expressing concentration of solution.

ii. This method is, unlike mole fraction, independent of the variation in temperature.

iii. There is a relation between the molality and solubility which can be mentioned as follow:

Molality = (Solubility X 10) / Molecular mass of solute

Solubility can be expressed as:

Solubility = (Mass of solute in gram / Mass of solvent in gram) X 100

**8. Molar Concentration (Molarity):**^{3} of the solution.

According to the definition, it can be calculated by using the following formula:

Molarity (M) = Number of moles of solute / Number of liters of solution

OR

Molarity (M) = (Number of moles of solute / Number of mL of solution) X 1000

OR

Number of moles of solute = Molarity (M) X Number of liters of solution

Consider a solution containing a gram of solute of molecular mass b and it is dissolved in V liter of solution.

Then, the molarity of the solution is

Molarity (M) = a / (b X V)

OR

Molarity of solution X b = a / V = Strength of the solution

If the volume component is taken in mL unit, then following modification is needed:

Molarity (M) of the solution = [a / (b X V)] X 1000

The formula can also be written as:

Molarity (M) = (1000 X Weight of solute in gram) / (Molecular mass of solute X Volume of solvent in mL)

**9. Normality:**

According to definition, it can be expressed by the following formula:

Normality (N) = Number of gram equivalents of solute / Number of liters of the solution

OR

Number of gram equivalents of solute = Normality X Number of liters of the solution

Let A gram of the solute of equivalent mass B be present in V liter of the final solution, then the normality can be calculated by:

Normality (N) = (A / B) / V

OR

Normality (N) = A / (B X V)

OR

Normality (N) X Equivalent mass = A / V = Strength of the solution in g/L

The solutions are expressed as 1 N, 2 N, etc.

**10. Formality:**

Mass percentage or percent by mass = (Mass of solute / Mass of solution) X 100

Here, mass of solution can be expressed by the following formula:

Mass of solution = Mass of solute + Mass of solvent

The mass of solution can also be expressed as:

Mass of solution = Volume of solution X Density of solution

Mass fraction = Mass of solute / Mass of solution

Hence, the mass percentage of solute can be calculated by multiplying mass fraction with 100.

20% solution of glucose means that 20 gram of glucose (solute) is present in 100 grams of the solution, i.e., 20 gram of glucose has been dissolved in 80 gram of water (solvent).

Percent of solute by volume = (Volume of solute / Volume of solution) X 100

Percent of solute of mass by volume = (Mass of solute / Volume of solution) X 100

Concentration of solution = (Solute mass in gram / Solution volume in liters)

If the unit of volume is in mL, then the overall formula should be multiplied by 1000.

Concentration of solution = (Solute mass in gram / Solution volume in mL) X 1000

The strength of the solution can be defined as the concentration of solution in gram per liter. Suppose, 10 g of glucose (solute) is present in 1 L of solution, then the strength or concentration of the solution is 10 g/L.

ppm = (Mass of solute / Mass of solution) X 10

The example in which the solute is present in trace quantity is the atmospheric pollution in cities. This concentration refers to the pollutant volume in 10

Suppose there are three components present in the solution, namely component A, B and C. The moles of component A is a, the moles of component B is b, and the moles of component C is c. Hence, the total moles of solution is (a + b + c).

According to the definition of mole fraction, the mole fraction of component A, B, and C can be calculated by following way.

Mole fraction of A =

Mole fraction of B =

Mole fraction of C =

Fact on Mole fraction: The addition of mole fraction of all components is always equal to 1, i.e.,

In case of binary solution (i.e. the condition when the solution is made up of only one solute and remaining is solvent.),

Total mole fraction = Mole fraction of solute + Mole fraction of solvent = 1

Consider the case of a solution which is made up of solute A and solvent B. The moles of solute A is a and the moles of solvent B is b. Then the mole fraction is:

Mole fraction of solute A =

Mole fraction of solute B =

Hence, the total mole fraction is:

Total Mole Fraction =

Note- The mole fraction of solution is independent of temperature of the solution.

Molality can be represented by following formula:

Molality (m) = [Number of solute moles / Number of kilograms (kg) of the solvent present]

Suppose, a gram of solute of molecular mass b be present in c gram of the solvent, then the molality can be calculated as:

Molality (m) = [a / (b X c)] X 1000

Molality (m) = [A / c] X 1000

Where, A is total number of moles of solute i.e. a / b.

Please note following points on molality.

Molality = (Solubility X 10) / Molecular mass of solute

Solubility can be expressed as:

Solubility = (Mass of solute in gram / Mass of solvent in gram) X 100

According to the definition, it can be calculated by using the following formula:

Molarity (M) = Number of moles of solute / Number of liters of solution

OR

Molarity (M) = (Number of moles of solute / Number of mL of solution) X 1000

OR

Number of moles of solute = Molarity (M) X Number of liters of solution

Consider a solution containing a gram of solute of molecular mass b and it is dissolved in V liter of solution.

Then, the molarity of the solution is

Molarity (M) = a / (b X V)

OR

Molarity of solution X b = a / V = Strength of the solution

If the volume component is taken in mL unit, then following modification is needed:

Molarity (M) of the solution = [a / (b X V)] X 1000

The formula can also be written as:

Molarity (M) = (1000 X Weight of solute in gram) / (Molecular mass of solute X Volume of solvent in mL)

According to definition, it can be expressed by the following formula:

Normality (N) = Number of gram equivalents of solute / Number of liters of the solution

OR

Number of gram equivalents of solute = Normality X Number of liters of the solution

Let A gram of the solute of equivalent mass B be present in V liter of the final solution, then the normality can be calculated by:

Normality (N) = (A / B) / V

OR

Normality (N) = A / (B X V)

OR

Normality (N) X Equivalent mass = A / V = Strength of the solution in g/L

The solutions are expressed as 1 N, 2 N, etc.