Class 12 Physics Detailed Notes (West Bengal Board - WBCHSE)

Class 12 Chemistry
Chapter: Solutions
Lecture Notes & Formula Sheet
Part 1 – Introduction, Types of Solutions & Concentration Terms

Author: Mahadeb Bar




Learning Objectives
After studying this chapter, students will be able to:

 Define a solution and identify its components.
 Differentiate between solute and solvent.
 Classify different types of solutions.
 Calculate concentration using various methods.
 Solve basic numerical problems related to concentration.




1. Introduction to Solutions
A solution is a homogeneous mixture of two or more substances. The composition and
properties are uniform throughout the mixture.

Components of a Solution

 Solute: The substance present in a smaller amount and dissolved in the solvent.
 Solvent: The substance present in a larger amount that dissolves the solute.

Examples:

 Salt in water → Salt = Solute, Water = Solvent
 Sugar in water → Sugar = Solute, Water = Solvent




2. Types of Solutions
Solutions can exist in different physical states.

,Solute Solvent Example
Gas Gas Air
Gas Liquid Carbon dioxide in soft drinks
Liquid Liquid Ethanol in water
Solid Liquid Salt in water
Solid Solid Brass (Copper + Zinc)



3. Characteristics of a True Solution
 Homogeneous mixture
 Particle size less than 1 nm
 Transparent appearance
 Stable composition
 Cannot be separated by ordinary filtration




4. Concentration of Solutions
The concentration of a solution tells us how much solute is present in a given amount of
solution or solvent.

(a) Mass Percentage

Mass Percentage=Mass of SoluteMass of Solution×100\text{Mass
Percentage}=\frac{\text{Mass of Solute}}{\text{Mass of
Solution}}\times100Mass Percentage=Mass of SolutionMass of Solute×100

(b) Volume Percentage

Volume Percentage=Volume of SoluteVolume of Solution×100\text{Volume
Percentage}=\frac{\text{Volume of Solute}}{\text{Volume of
Solution}}\times100Volume Percentage=Volume of SolutionVolume of Solute×100

(c) Mass by Volume Percentage

Mass by Volume Percentage=Mass of Solute (g)Volume of Solution (mL)×100\text{Mass by
Volume Percentage}=\frac{\text{Mass of Solute (g)}}{\text{Volume of Solution
(mL)}}\times100Mass by Volume Percentage=Volume of Solution (mL)Mass of Solute (g)
×100

(d) Parts Per Million (ppm)

ppm=Mass of SoluteMass of Solution×106\text{ppm}=\frac{\text{Mass of
Solute}}{\text{Mass of Solution}}\times10^6ppm=Mass of SolutionMass of Solute×106

, Used for very dilute solutions.




5. Mole Fraction
The mole fraction of a component is the ratio of the number of moles of that component to
the total number of moles in the solution.

X=Moles of ComponentTotal MolesX=\frac{\text{Moles of Component}}{\text{Total
Moles}}X=Total MolesMoles of Component

The sum of mole fractions of all components equals 1.




6. Molarity (M)
Molarity is the number of moles of solute present in one litre of solution.

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

Unit: mol L⁻¹




7. Molality (m)
Molality is the number of moles of solute present in one kilogram of solvent.

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

Molality does not change with temperature.




Solved Example
Question: Calculate the molarity of a solution containing 2 moles of NaCl dissolved to make
1 litre of solution.

Solution:

M=21=2 mol L−1M=\frac{2}{1}=2\; \text{mol L}^{-1}M=12=2mol L−1

Answer: 2 M