Molarity measures the concentration of a solution. It is defined as the quantity of a substance (solute) dissolved in a given volume of a liquid (solution). Represented by the symbol M, it is widely used in physical chemistry, science, and everyday life. ^[1-4]

Formula

The molarity of a solution is calculated using the following formula: ^[1-5]

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

The number of moles of the solute is given by:

Number of moles (n) = (Mass of solute)/(Molar mass of solute)

The molar mass is expressed in grams per mole (g/mol).

Units

The SI unit of molarity is moles per liter (mol/L). Therefore, molarity measures the number of moles of solute in one liter of solution. 

Example: If 2 moles of solute are dissolved in 0.5 liters of solution, the molarity is:

M = 2 mol/0.5 L = 4 mol/L 

A solution with a concentration of 1 mol/L is called 1 molar solution or 1 M.

Dilution is a related concept often used to adjust solution concentrations.

Dilution of a Solution

Dilution involves adding more solvent to reduce the solution’s concentration. The number of moles of the solute remains constant, but the total volume increases, decreasing the solution’s concentration. Dilution is commonly used in laboratories to prepare solutions of desired concentrations from more concentrated stock solutions. ^[1-5]

The following formula gives the relationship between the initial and final concentrations and volumes during dilution:

C₁V₁ = C₂V₂

Where:

– C₁​: initial concentration of the solution

– V₁​: initial volume of the solution

– C₂​: final concentration of the solution

– V₂​: final volume of the solution

This equation can be derived from the fact that the number of moles of the solute, given by n = C x V, remains the same before and after dilution.

Applications ^[1-5]

• Laboratory: Finding the unknown concentration of a solution by reacting it with a solution of known molarity
• Medical: Preparation of saline solutions and IV fluids with accurate concentrations
• Environmental Monitoring: Measuring pollutants, such as nitrates or heavy metals, in water
• Food Industry: Determining the acidity of vinegar or the sugar concentration in beverages
• Pharmaceuticals: Ensuring the right dosage of drugs in liquid medicines

Example Problems with Solutions

Problem 1: Calculate the molarity of a solution containing 2 moles of NaCl in 1 liter of water.

Solution

Use the molarity formula:

Molarity (M) = (Moles of solute)/(liters of solution)

=> M = 2 mol/1 L = 2 mol/L

Problem 2: How many grams of KCl are needed to make 500 mL of a 1 M solution? (Molar mass of KCl = 74.55 g/mol)

Solution

1. Convert the volume to liters:

500 mL = 0.500 L

2. Use the molarity formula to find the number of moles of KCl:

Molarity (M) = (Moles of KCl)/(Volume of solution)

=>  Moles of KCl = M x Volume of solution

=> Moles of KCl = 1 mol/L x 0.500 L = 0.500 mol

3. Convert moles of KCl to grams of KCl:

Mass of KCl = Moles of KCl x Molar mass of KCl = 0.500 mol x 74.55 g/mol = 37.28 g

Problem 3: A solution has a molarity of 0.5 M. How many moles of solute are in 250 mL of the solution?

Solution

1. 250 mL = 0.250 L

2. Molarity (M) = (Moles of solute)/Volume of solution)

 =>  Moles of solute = M x Volume of solution

=> Moles of solute​ = 0.5 mol/L x 0.250 L = 0.125 mol

Problem 4: What is the molarity of a solution made by dissolving 10 grams of NaOH in 250 mL of water? (Molar mass of NaOH = 40 g/mol)

Solution

1. Moles of NaOH = (Mass of NaOH)/(Molar mass of NaOH) = 10 g/40 g/mol = 0.25 mol

2. 250 mL = 0.250 L

3. Molarity (M) = (Moles of NaOH)/(Volume of solution)

=> M = 0.25 mol/0.250 L = 1 mol/L

Problem 5: You have 50 mL of a 2.0 M sodium chloride (NaCl) solution, and you want to dilute it to a concentration of 0.5 M. What will be the final volume of the solution after dilution? How much liquid must be added to achieve this concentration?

Solution

Given:

Initial concentration: C₁ = 2.0 M

Initial volume: V₁ = 50 mL

Final concentration: C₂ = 0.5 M

Use the dilution formula and substitute the given values:

C₁V₁ = C₂V₂

=> 2.0 M x 50 mL = 0.5 M x V₂

​=> V₂ = (2.0 M x 50 mL)/0.5 M = 200 mL

The final volume of the solution after dilution is 200 mL. To prepare this solution, add 200 – 50 = 150 mL of water to the original 50 mL of 2.0 M NaCl solution. Thus, adding water achieves the desired concentration.