Objectives of Colligative Properties in Class 12 Chemistry NCERT

Understanding the Objectives of Colligative Properties

Colligative properties are essential in Class 12 Chemistry as they help explain how the presence of solute particles affects physical properties of solutions. The primary objective is to study properties that depend solely on the number of dissolved particles, not their chemical nature. This leads to understanding:

• How vapour pressure of a solvent decreases when a non-volatile solute is added
• Why boiling point rises and freezing point lowers in solutions
• The concept of osmotic pressure and its role in biological and chemical systems

These objectives enable students to apply theoretical knowledge to practical problems such as determining molar masses of unknown solutes using colligative property measurements.

Relative Lowering of Vapour Pressure: Objective and Explanation

One key objective is to understand how adding a solute lowers the vapour pressure of a solvent. According to Raoult's law, the vapour pressure of the solvent in a solution decreases proportionally to the mole fraction of the solute.

Formula:

$$ \frac{P^0 - P}{P^0} = x_2 $$

Where:

• $P^0$ = vapour pressure of pure solvent
• $P$ = vapour pressure of solution
• $x_2$ = mole fraction of solute

This relative lowering of vapour pressure explains why solutions behave differently from pure solvents and is foundational for understanding boiling point elevation and freezing point depression.

Boiling Point Elevation and Freezing Point Depression Objectives

Another objective is to understand how solutes affect phase change temperatures:

• Boiling Point Elevation: Adding solute lowers vapour pressure, so the solution boils at a higher temperature.
• Freezing Point Depression: The solution freezes at a lower temperature due to decreased vapour pressure.

Both effects are proportional to the molality ($m$) of the solution:

$$ \Delta T_b = K_b \times m $$ $$ \Delta T_f = K_f \times m $$

Where $K_b$ and $K_f$ are the ebullioscopic and cryoscopic constants respectively, specific to each solvent.

Example: Calculate freezing point depression when 1 mol of solute is dissolved in 1 kg of water ($K_f = 1.86$ °C kg/mol):

$$ \Delta T_f = 1.86 \times 1 = 1.86 ^\circ C $$

So, freezing point lowers by 1.86 °C.

Osmotic Pressure: Objective and Importance in Solutions

Osmotic pressure is the pressure required to stop solvent flow through a semipermeable membrane from pure solvent to solution. The objective here is to understand how osmotic pressure relates to solute concentration and temperature.

Formula:

$$ \Pi = CRT $$

Where:

• $\Pi$ = osmotic pressure
• $C$ = molarity of solution
• $R$ = gas constant
• $T$ = temperature in Kelvin

Osmotic pressure is crucial in biological systems (e.g., cell membranes) and industrial processes. Understanding this objective helps students link chemistry concepts to real-world applications.

Role of Van't Hoff Factor in Colligative Properties

The Van't Hoff factor ($i$) accounts for solute dissociation or association in solutions, affecting colligative properties. The objective is to understand that:

• Electrolytes dissociate into ions, increasing the number of particles and thus colligative effects.
• Nonelectrolytes do not dissociate, so $i = 1$.

Modified formulas:

$$ \Delta T_b = i K_b m $$ $$ \Delta T_f = i K_f m $$ $$ \Pi = i C R T $$

Example: For NaCl dissociating into Na$^+$ and Cl$^-$, $i \approx 2$.

This adjustment is important for accurate molar mass calculations and understanding abnormal molar masses.

Comparison of Concentration Units Relevant to Colligative Properties

Understanding concentration units is vital for solving colligative property problems. Here's a comparison of molarity and molality:

Molality is preferred because it remains constant with temperature changes, ensuring accurate colligative property measurements.

Worked Example: Determining Molar Mass Using Freezing Point Depression

Problem: A solution is prepared by dissolving 10 g of an unknown solute in 100 g of water. The freezing point of the solution is found to be -1.86 °C. Calculate the molar mass of the solute. ($K_f$ for water = 1.86 °C kg/mol)

Solution:

1. Calculate freezing point depression:

$$ \Delta T_f = 0 - (-1.86) = 1.86 ^\circ C $$

2. Calculate molality:

$$ m = \frac{\Delta T_f}{K_f} = \frac{1.86}{1.86} = 1 \text{ mol/kg} $$

3. Calculate moles of solute:

$$ \text{moles} = m \times \text{kg solvent} = 1 \times 0.1 = 0.1 \text{ mol} $$

4. Calculate molar mass:

$$ M = \frac{\text{mass}}{\text{moles}} = \frac{10}{0.1} = 100 \text{ g/mol} $$

Answer: Molar mass of solute is 100 g/mol.