Molarity is a measure of the concentration of a solute dissolved per liter (dm3) of solution. It is the most used unit of concentration. It is also called the molar concentration of the solution. It is represented by the capital letter ‘M’ and its unit is mol/dm3. For instance, if you have 2 moles of salt (solute) dissolved in 1000 milliliters of water (solvent), the molarity of the solution would be 2 moles/dm3.

Molarity is widely used because it is related to the mass of the solute and moles of the solute, and the total volume of the solution. It depends upon temperature and pressure because changes in temperature and pressure affect the volume of the solution.

Molarity is used to express the concentration of a wide variety of substances, including acids, bases, and other chemicals.

Outline

## The formula of molarity

The most commonly used formula for molarity is

Molarity= mass/molar mass x 1/volume of solutions in dm3

### Units

Its unit is mol/dm3 or mol dm-3. We can also write it as mol L-1.

## How to calculate the molarity?

Molarity is calculated by dividing moles of solute by the liter of solution. There are different formulas for molarity calculations. Using these formulas we can calculate the molarities of liquid, stock solutions, and standardized solutions.

Examples

Calculating the molarity of 1.186g of BaCl2.2H2O dissolved in 300 mL of solution.

Given date:

Mass of solute = 1.186g

Volume of solution = 300mL or 0.3L

The molar mass of BaCl2.2H2O = 244.3 g/mol

To find:

Molarity =?

Solution:

Molarity= mass/molar mass x 1/volume of solutions in dm3

= 1.186/244.3 x 1/0.3

=0.016 mol/dm3

Hence, the molarity of BaCl2.2H2O is 0.016 M

### Example 2

How to prepare a 5.0 L solution of 0.1M NaOH?

Given data:

Molarity = 0.1 M

Volume of solution = 5 dm3

Molar mass of NaOH = 40 g/mol

To find:

grams of solute =?

Solution:

By rearranging the equation (i)

Mass = molarity x molar mass x volume of solution

= 0.1 x 40 x 5

=20 gram

So, by adding 20 grams of NaOH in 5L water we can prepare 5.0L of 0.1M NaOH.

### Formula for molarity of a stock solution

We can calculate the molarity of the stock solution by using the percentage purity (%P) and density of that solution.

Formula:

Molarity= (percentage purity x density x 10) ÷ (molar mass) – (ii)

Example

How to Calculate the molarity of 37% of the HCl solution.

Given data:

Percentage purity of HCl = 37%

The density of HCl = 1.18 g/mL

Molecular Mass = 36.5 g/mol

To find

Molarity =?

Solution:

By using equation (ii)

Molarity = (37 x 1.18 x 10) ÷ (36.5)

Molarity= 11.9 mol/dm3

### The formula used to find out the unknown molarity of the solution using titration

M1V1= M2V2

where,

• M1 is the molarity of titrant
• V1 is the volume of titrant used
• M2 is the molarity of the analyte
• V2 is the volume of the analyte solution

This formula is commonly used for titrimetric calculations and also for the standardization of solutions. Using this formula, we can find the unknown molarity of known solutions. During titration, we have two solutions. One of the solutions is with known volume but unknown molarity whereas, the second solution is of known molarity and unknown volume.

### Example

Standardization of 30 mL of HCl with 1.0M solution of sodium hydroxide (NaOH)

The average volume used is 15 mL this is calculated after performing an acid-base titration

Given data:

M1 = 1.0M

V1= 15 cm3

V2 = 30 cm3

To find:

M2 = ?

Solution:

By putting values in the formula we can calculate the molarity of HCl,

Similarly, we can use this formula to make a solution of the required molarity. For example, we have to prepare 500 mL 2.0 M HCl.

Given data:

Molarity of solution to prepare (M1) = 2.0 M

The volume of the solution to prepare (V1) = 500 cm3

Molarity of (37%) stock solution (M2) = 11.96 g/ mol

To find:

The required volume of stock solution (V2) =?

Formula:

Putting values in formula

2.0 x 500 = 11.96 x V2

So we need 83.6 mL of 37% HCl to make 500 mL of 2.0 M of HCl.

## Factors affecting molarity

Temperature:

Molarity depends upon the temperature because it deals with the volume of the solution. It is inversely proportional to the temperature.

As the temperature increases the molarity decreases. This is due to an increase in the volume of the solution whereas the amount of solute remains constant. In fact, the volume of the solution is inversely proportional to the molarity of the solution. So, upon heating, the volume increase to decrease the molarity of a solution.

Pressure:

Molarity is also affected by changes in pressure. Molarity is directly proportional to the pressure.

As the pressure increases it decreases the volume of the solution but the amount of solute remains constant. That will ultimately increase the molarity of the solution.

Related resources

## Concepts berg

How to calculate molar mass?

Molar mass is calculated by adding the individual masses of all the atoms of molecules. For example,

The molar mass of H2O is calculated as:

Mass of H =1.007 g/mol

Mass of O = 16 g/mol

H+H+O

1+1+16= 18

So the molar mass of H2O is 18 g/mol.

What is the molar mass of CO2?

The molar mass of CO2 is 44 g/mol.

How can we prepare one molar solution?

We can prepare one molar solution by slowly dissolving the formula weight (in grams) of solute in a solvent and making up a volume of one liter.

How to make one molar solution of NaOH?

Dissolve 40g of NaOH in water and make a volume of upto 1000ml (1L).

Differentiate between molarity and molality.

Molarity is the measure of moles of solute per liter of solution whereas molality is the measure of moles of solute per kilogram of solvent.

Define normality.

Normality is the number of grams equivalent of solute per dm3 of solution. It is also known as equivalent concentration. It is represented by the capital letter N.

Does a change in temperature affect the molarity of the solution?

Yes, the molarity of the solution decreases when the temperature of the solution.

What is meant by one molar solution?

One molar solution is the one mole of solute dissolved in one liter of solution.

References

• Basic Chemistry Concepts and Exercises by John Kenkel.
• The second edition of Chemistry by Julia Burdge.
• The tenth edition of Chemistry by Zumdahl and DeCoste