# Methods of Expressing Concentration of Solution for IIT JEE

## Methods of Expressing Concentration of Solution - Definition and Formula

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 is defined as the amount of solute (in gram) present in 100 grams of the solution. This can be calculated by using following formula:

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 can be expressed as the ratio of mass of solute and mass 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).

2. Percent by Volume:

This term is defined as the solute volume present in 100 mL of solution. This can be expressed by following formula:
Percent of solute by volume = (Volume of solute / Volume of solution) X 100

3. Percent mass by volume:

This term is defined as the solute mass present in 100 mL of solution. This can be expressed by following formula:
Percent of solute of mass by volume = (Mass of solute / Volume of solution) X 100

4. Strength orSolution Concentration (Concentration in Gram per liter):

This term can be defined as the amount of the solute mass (in gram) present in one liter of the solution. So, the unit of this term is gram per liter. It can be expressed by the following formula:
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):

This method is particularly important when the solute is present in trace quantities. It is easy to express concentration of solution in that case. Parts per million can be defined as the amount of the solute in gram present in 106 gram of the solution.
ppm = (Mass of solute / Mass of solution) X 106
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 106 units of air volume. 20 ppm of sulfur dioxide (SO2) in air means that 10 mL of SO2 is present in 106 mL of air volume.

6. Mole Fraction:

Mole fraction method can be used in the case when the solution is composed of two or more components. It is defined as the ratio of total moles of one component to the total number of moles of the given solution. (Note- The total number of moles of the solution includes the moles of all the components present in the solution.)

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 of the solution can be defined as the number of solute moles available or present in total 1 kg of the solvent. The symbol of molality is ‘m’.

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):

Molar concentration i.e. also known as molarity can be defined as the total number of solute moles per liter or say per dm3 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:

The term ‘normality’ can be defined as the number of gram equivalents of solute present per liter of the solution. Normality is denoted by ‘N’.
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:

Formality of the solution can be defined as the formula mass (in gram) present per liter of solution. When the formula mass is equal to the molecular mass, formality is equal to the molarity. Formality is also dependent on temperature, just like normality and molarity of the solution. Formality is particularly useful when the solution contains ionic compounds in which molecule does not exist. (Formole is known as the mole of ionic compounds and formality is the molarity of ionic compounds.)