Expression of Concentration of Solutions

1.0Introduction

An accurate grasp of solution concentration is fundamental for solving JEE chemistry problems—from reaction stoichiometry to colligative properties. Mastering different concentration units (molarity, molality, mole fraction, etc.) helps you interpret lab data, calculate reagent quantities, and work efficiently in theoretical and experimental contexts. This guide demystifies each expression, shows how to interconvert them, and gives you practice at JEE‑level rigor.

2.0Concentration of Solution

The concentration of a solution represents the amount of solute present in a given quantity of solvent or solution.

• Dilute solution → small amount of solute.
• Concentrated solution → large amount of solute.

In Chemistry, different quantitative methods are used to express concentration, depending on the type of calculation.

3.0Mass Percentage

Mass percentage, also known as weight percentage, expresses the mass of the solute as a percentage of the total mass of the solution. It is a temperature-independent unit.

• Formula:

$\text{Mass Percentage}=\frac{\text{Mass of Solute}}{\text{Mass of Solution}}\times 100$

• Example: A 10% solution of glucose by mass means that 10 grams of glucose is present in 100 grams of the solution (10 g glucose + 90 g solvent).

Volume Percentage

Volume percentage is used when the solute and solvent are liquids. It represents the volume of the solute as a percentage of the total volume of the solution. This is a temperature-dependent unit.

• Formula:

$\text{Volume Percentage}=\frac{\text{Volume of Solute}}{\text{Volume of Solution}}\times 100$

• Example: A 20% solution of ethanol by volume means that 20 mL of ethanol is present in 100 mL of the solution.

Mass by Volume Percentage

This unit expresses the mass of the solute dissolved in 100 mL of the solution. It is commonly used in pharmaceuticals and medicine. This is a temperature-dependent unit.

• Formula:

$\text{Mass by Volume Percentage}=\left(\frac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}}\right)\times 100$

• Example: A 5% w/v solution means 5 grams of solute are present in 100 mL of the solution.

Parts Per Million (ppm) & Parts Per Billion (ppb)

These units are used to express the concentration of a very dilute solution, where the solute is present in trace amounts.

• Formula:

$\text{ppm}=\frac{\text{Mass/Volume of Component}}{\text{Total Mass/Volume of Solution}}\times 10^6$

$\text{ppb}=\frac{\text{Mass/Volume of Component}}{\text{Total Mass/Volume of Solution}}\times 10^9$

• Applications: Commonly used to measure the concentration of pollutants in water or the atmosphere.

Molarity (M)

Molarity is defined as the number of moles of solute dissolved per liter of solution. It is a temperature-dependent unit because volume changes with temperature.

• Formula:

$M=\frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}}$

• Units: mol/L or M.
• Fact: Molarity is the most common unit for concentration in laboratory settings for quantitative analysis.

Example:

How many moles of sulfuric acid (H₂SO₄ ) are in 63.5 mL of a 3.0 M solution?

Given:

• Volume of solution = 63.5 mL = 0.0635 L
• Molarity (M) = 3.0 mol/L

Formula:

Moles=M×V

Substitute:

Moles of H₂SO₄=3.0×0.0635

 =0.1905 mol

Answer: 0.190 mol is correct.

Molality (m)

Molality is defined as the number of moles of solute dissolved per kilogram of solvent. It is a temperature-independent unit because it is based on mass, which does not change with temperature.

• Formula:

$m=\frac{\text{Moles of Solute}}{\text{Mass of Solvent (kg)}}$

• Units: mol/kg or m.
• Fact: Molality is preferred over molarity for experiments involving temperature changes, such as freezing-point depression and boiling-point elevation, which are colligative properties.

Example:

100 g solution of urea in water has 40 g urea (molar mass = 60 g/mol). What is the molality of the urea solution?

Solution

Mass of urea and water = 100 g
Mass of urea = 40 g

Moles of urea=\text{Moles of urea} = \frac{40}{60} = 0.66

Mass of water = 60 g

\text{Moles of water} = \frac{60}{18} = 3.333

Weight of solvent (water) in kg

= \frac{60}{1000} = 0.06 \text{ kg}

Molality (m)

$m=\frac{\text{Moles of Solute}}{\text{Mass of Solvent (kg)}}=\frac{0.666\text{ mol}}{0.06\text{ kg}}=11.11\text{ m}$

Mole Fraction (x)

Mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component (solute or solvent) to the total number of moles of all components in the solution.

• Formula:

$X_{\text{solute}}=\frac{\text{moles of solute}}{\text{total moles of all components}}$

$\text{(}{X}_{solvent})=\frac{\text{Moles of Solvent}}{\text{Total Moles of All Components}}$

• Property: The sum of the mole fractions of all components in a solution is always equal to 1.
  

Formality (F)

Formality is similar to molarity but is specifically used for ionic compounds that exist as aggregates in the solid state. It is defined as the number of gram formula masses of the solute per liter of the solution.

• Formula:

$F=\frac{\text{Gram Formula Mass of Solute}}{\text{Volume of Solution (L)}}$

• Note: For most practical purposes, especially for non-ionic compounds, formality is numerically the same as molarity.

Normality (N)

Normality is a measure of concentration based on the concept of equivalent weight. It is defined as the number of gram equivalents of solute per liter of solution.

• Formula:

$N=\frac{\text{Gram Equivalents of Solute}}{\text{Volume of Solution (L)}}$

• Valence Factor: The valence factor (or n-factor) depends on the type of reaction:
• • For acids: It is the number of replaceable ions (basicity).
  • For bases: It is the number of replaceable ions m(acidity).
  • For salts: It is the total positive or negative charge on the ions.
  • For redox reactions, It is the number of electrons gained or lost per molecule.
• Application: Normality is particularly useful in titration and redox reactions.