Class 12 Chemistry MCQ Test – Solutions (Chapter 1) | Full Chapter Test
Class 12 Chemistry – Chapter: Solutions (Detailed Notes)
These notes are designed for CBSE, GSEB, JEE and NEET students. It includes detailed theory, concepts, formulas and exam-oriented explanations.
1. What is a Solution?
A solution is a homogeneous mixture of two or more substances whose composition is uniform throughout the system. This means that any small sample taken from the solution will have the same composition and properties.
A solution consists of:
- Solvent: The component present in larger amount and determines the physical state.
- Solute: The component present in smaller amount and dissolved in solvent.
Example: In salt water, water is solvent and salt is solute.
Types of Solutions
Solutions can exist in different physical states such as gaseous, liquid and solid depending on the nature of solute and solvent.
- Gas in Gas → Air
- Gas in Liquid → Oxygen in water
- Liquid in Liquid → Alcohol in water
- Solid in Liquid → Sugar in water
- Solid in Solid → Alloys (Brass)
2. Methods of Expressing Concentration
Concentration tells how much solute is dissolved in a given amount of solvent or solution. It is very important in chemistry as it helps in quantitative analysis.
1. Mass Percentage
It is defined as the mass of solute present in 100 g of solution.
\( \text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \)
Used in industrial solutions and medicines.
2. Molarity (M)
Molarity is defined as the number of moles of solute dissolved in one litre of solution.
\( M = \frac{n}{V(\text{L})} \)
Important: Molarity depends on temperature because volume changes with temperature.
3. Molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent.
\( m = \frac{n}{W(\text{kg})} \)
Molality is temperature independent because mass does not change.
4. Mole Fraction
It is the ratio of moles of one component to total moles in solution.
\( x_A = \frac{n_A}{n_A + n_B} \)
Total mole fraction is always:
\( x_A + x_B = 1 \)
3. Solubility of Gases – Henry’s Law
The solubility of a gas in a liquid depends on pressure and temperature.
Henry’s Law: At constant temperature, the solubility of a gas is directly proportional to its partial pressure.
\( p = K_H \cdot x \)
Where:
- \( p \) = Partial pressure of gas
- \( x \) = Mole fraction of gas
- \( K_H \) = Henry’s constant
Higher value of \( K_H \) means lower solubility.
Applications:
- Scuba diving (avoiding bends)
- Carbonated drinks
4. Vapour Pressure & Raoult’s Law
Vapour pressure is the pressure exerted by vapour in equilibrium with its liquid.
Raoult’s Law
The partial vapour pressure of each component is proportional to its mole fraction.
\( p_1 = p_1^\circ x_1 \)
\( p_2 = p_2^\circ x_2 \)
Total Pressure:
\( P = p_1^\circ x_1 + p_2^\circ x_2 \)
5. Ideal and Non-Ideal Solutions
Ideal Solutions
- Obey Raoult’s law completely
- \( \Delta H = 0 \)
- \( \Delta V = 0 \)
- Forces between molecules are similar
Non-Ideal Solutions
- Do not obey Raoult’s law
- \( \Delta H \neq 0 \)
- Show deviation
Types of deviation:
- Positive: weaker forces → higher vapour pressure
- Negative: stronger forces → lower vapour pressure
6. Colligative Properties
Colligative properties depend only on number of particles, not nature of solute.
1. Relative Lowering of Vapour Pressure
\( \frac{p^\circ - p}{p^\circ} = x_2 \)
2. Elevation of Boiling Point
\( \Delta T_b = K_b m \)
3. Depression of Freezing Point
\( \Delta T_f = K_f m \)
4. Osmotic Pressure
\( \pi = CRT \)
Highly important for determining molar mass of biomolecules.
7. Van’t Hoff Factor
Sometimes solutes dissociate or associate, changing number of particles.
\( i = \frac{\text{Observed property}}{\text{Calculated property}} \)
- \( i = 1 \) → normal
- \( i > 1 \) → dissociation
- \( i < 1 \) → association
Modified Equations
\( \Delta T_b = i K_b m \)
\( \Delta T_f = i K_f m \)
\( \pi = i CRT \)
⬇️ Attempt the test given below after revising these notes
Class 12 Chemistry
Chapter 1 : Solutions | Full Chapter MCQ Test
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