What is Solutions Class 12: Complete Guide for Chemistry Students

Definition and Basics of Solutions in Class 12 Chemistry

A solution is a homogeneous mixture formed when one substance (solute) dissolves in another (solvent). In Class 12 NCERT Chemistry, solutions are studied to understand how substances interact at the molecular level.

• Solute: The substance dissolved (e.g., salt)
• Solvent: The substance that dissolves the solute (e.g., water)

Solutions can be in various states: solid, liquid, or gas. However, liquid solutions like saltwater are the most commonly studied.

Example: When 10 g of salt dissolves in 100 g of water, a salt solution is formed.

Understanding solutions helps explain many natural and industrial processes.

Types of Solutions and Their Classification

Solutions can be classified based on different criteria:

1. Physical State:

• Gas in gas (air)
• Gas in liquid (carbonated water)
• Liquid in liquid (alcohol in water)
• Solid in liquid (salt in water)
• Solid in solid (alloys like brass)

2. Nature of Solvent:

• Aqueous solutions (water as solvent)
• Non-aqueous solutions (alcohol, benzene)

3. Concentration:

• Dilute (small amount of solute)
• Concentrated (large amount of solute)

4. Saturation:

• Unsaturated (more solute can dissolve)
• Saturated (maximum solute dissolved)
• Supersaturated (more solute than saturation)

This classification helps in understanding solution behaviour and properties.

Concentration Terms: Molarity, Molality, and Mole Fraction

Concentration measures how much solute is present in a solution. Key units include:

Example: Calculate molarity of 2 moles of NaCl in 0.5 L of solution:

$$M = \frac{2}{0.5} = 4\, mol/L$$

These units are crucial for solving problems related to solution properties.

Colligative Properties and Their Importance

Colligative properties depend on the number of solute particles, not their identity. Important colligative properties include:

• Vapour Pressure Lowering: Presence of solute lowers solvent vapour pressure
• Boiling Point Elevation: Solution boils at a higher temperature
• Freezing Point Depression: Solution freezes at a lower temperature
• Osmotic Pressure: Pressure required to stop solvent flow through a semipermeable membrane

These properties are explained by Raoult's Law:

$$P_{solution} = x_{solvent} \times P^0_{solvent}$$

where $P_{solution}$ is vapour pressure of solution, $x_{solvent}$ is mole fraction of solvent, and $P^0_{solvent}$ is vapour pressure of pure solvent.

Colligative properties have applications in everyday life and industries, such as antifreeze in cars and preservation of food.

Solubility and Factors Affecting It

Solubility is the maximum amount of solute that can dissolve in a solvent at a given temperature and pressure.

• Expressed in grams of solute per 100 g of solvent
• Varies with temperature and pressure

Factors affecting solubility:

• Temperature: Solubility of solids usually increases with temperature; gases decrease
• Pressure: Mainly affects gas solubility (Henry’s Law)

Henry’s Law:

$$C = kP$$

where $C$ is gas solubility, $k$ is Henry’s constant, and $P$ is gas pressure.

Understanding solubility is important for chemical reactions, pharmaceuticals, and environmental science.

Worked Example: Calculating Molality and Freezing Point Depression

Problem: Calculate the freezing point depression when 10 g of NaCl is dissolved in 500 g of water. (Given: $K_f$ for water = 1.86 K kg/mol, molar mass of NaCl = 58.5 g/mol)

Solution:

1. Calculate moles of NaCl:

$$n = \frac{10}{58.5} = 0.171\, mol$$

2. Calculate molality:

$$m = \frac{0.171}{0.5} = 0.342\, mol/kg$$

3. Since NaCl dissociates into 2 ions, van’t Hoff factor $i = 2$

4. Calculate freezing point depression:

$$\Delta T_f = i \times K_f \times m = 2 \times 1.86 \times 0.342 = 1.27 K$$

5. New freezing point:

$$0 - 1.27 = -1.27 ^\circ C$$

The solution freezes at -1.27 °C.

This example shows how to apply formulas from the Solutions chapter.