Chemistry Class 12 NCERT Solutions Chapter 2

Introduction

Chemistry, as a subject, is crucial in understanding the fundamental principles that govern the physical world around us. For students of Class 12, mastering the subject is essential not only for academic success but also for laying a solid foundation for future scientific studies. Chapter 2 of the NCERT Class 12 Chemistry textbook deals with "Solutions," a topic that is pivotal for both theoretical understanding and practical applications.

In this article, we will delve into the key concepts and important questions from Chapter 2 of the NCERT Class 12 Chemistry textbook. We aim to provide a comprehensive guide to help students prepare effectively for their exams by covering essential topics, solving important questions, and offering insights into common problem areas.

Overview of Chapter 2: Solutions

Chapter 2, "Solutions," focuses on the concept of solutions, their types, properties, and methods of expressing concentration. The chapter introduces students to various forms of solutions, including:

Types of Solutions: Solutions can be solid, liquid, or gaseous. This chapter emphasizes the properties of these solutions and how they differ from each other.
Concentration Terms: The chapter covers different ways to express the concentration of a solution, such as molarity, molality, and percentage composition.
Colligative Properties: These are properties that depend on the number of solute particles in a solution, not on the nature of the solute. Important colligative properties discussed include boiling point elevation and freezing point depression.
Raoult's Law: This law describes the relationship between the vapor pressure of a solvent and the concentration of the solute in a solution.
Practical Applications: The chapter also explores the practical applications of solutions in everyday life and industrial processes.

Important Questions and Solutions

1. Define a solution. Discuss the types of solutions and provide examples.

Answer:
A solution is a homogeneous mixture of two or more substances. The substance present in the largest amount is called the solvent, while the other substances present in smaller amounts are called solutes.

Types of Solutions:

Solid Solutions: These are solutions where the solute and solvent are both in solid form. Example: Alloys like brass (a solution of zinc in copper).
Liquid Solutions: In these solutions, the solvent is liquid, and the solute can be a solid, liquid, or gas. Example: Saltwater (salt in water), alcohol in water.
Gaseous Solutions: Both the solute and solvent are gases. Example: Air (a solution of oxygen in nitrogen).

2. Explain the different ways to express the concentration of a solution.

Answer:
Concentration is a measure of the amount of solute present in a given quantity of solution. Different ways to express concentration include:

Molarity (M): It is defined as the number of moles of solute per liter of solution.
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$
Molality (m): It is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$
Percent Composition: This includes:
- Mass Percent:
  $\text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
- Volume Percent:
  $\text{Volume \%} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100$
Normality (N): It is the number of equivalents of solute per liter of solution.
$N = \frac{\text{Number of equivalents of solute}}{\text{Volume of solution in liters}}$

3. What are colligative properties? Explain how they are affected by the nature of the solute.

Answer:
Colligative properties are those properties of solutions that depend on the number of solute particles in a given quantity of solvent and not on the nature of the solute. The four primary colligative properties are:

Boiling Point Elevation: The boiling point of a solution is higher than that of the pure solvent. This elevation depends on the number of solute particles in the solution.
$\Delta T_b = i \cdot K_b \cdot m$ where $\Delta T_b$ is the boiling point elevation, $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution.
Freezing Point Depression: The freezing point of a solution is lower than that of the pure solvent.
$\Delta T_f = i \cdot K_f \cdot m$ where $\Delta T_f$ is the freezing point depression, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.
Osmotic Pressure: The pressure required to stop the osmotic flow of solvent through a semipermeable membrane.
$\Pi = i \cdot C \cdot R \cdot T$ where $\Pi$ is the osmotic pressure, $C$ is the concentration of the solution, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
Vapor Pressure Lowering: The vapor pressure of a solution is lower than that of the pure solvent. This decrease in vapor pressure is proportional to the concentration of the solute.
$\Delta P = P_0 - P$ where $\Delta P$ is the decrease in vapor pressure, $P_0$ is the vapor pressure of the pure solvent, and $P$ is the vapor pressure of the solution.

4. Derive Raoult's Law for a solution.

Answer:
Raoult's Law states that the partial vapor pressure of each volatile component in a solution is proportional to its mole fraction in the solution. For a solution of a volatile solute and solvent, Raoult's Law is expressed as:

$P_A = X_A \cdot P_A^0$ $P_B = X_B \cdot P_B^0$

where:

$P_A$ and $P_B$ are the partial vapor pressures of components A and B,
$P_A^0$ and $P_B^0$ are the vapor pressures of the pure components A and B,
$X_A$ and $X_B$ are the mole fractions of components A and B.

For an ideal solution, the total vapor pressure is given by:

$P_{\text{total}} = P_A + P_B$

5. Discuss the factors affecting the solubility of a solute in a solvent.

Answer:
The solubility of a solute in a solvent depends on several factors:

Nature of Solute and Solvent: The "like dissolves like" principle suggests that polar solutes dissolve in polar solvents, and non-polar solutes dissolve in non-polar solvents. For example, salt (a polar solute) dissolves in water (a polar solvent).
Temperature: For most solid solutes, solubility increases with an increase in temperature. For gases, solubility decreases with an increase in temperature.
Pressure: Pressure has a significant effect on the solubility of gases in liquids. According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
Presence of Other Substances: The presence of other substances can affect solubility through various interactions. For instance, the common ion effect can decrease the solubility of a salt when one of its ions is already present in the solution.

6. Solve a numerical problem involving the calculation of boiling point elevation.

Question: A solution is prepared by dissolving 0.5 moles of a non-electrolyte solute in 1 liter of water. The ebullioscopic constant for water is 0.512 °C kg/mol. Calculate the boiling point elevation.

Answer:
To find the boiling point elevation, use the formula:

$\Delta T_b = i \cdot K_b \cdot m$

where $i$ is the van't Hoff factor (1 for non-electrolytes), $K_b$ is the ebullioscopic constant, and $m$ is the molality. Here, molality $m$ is calculated as:

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5}{1} = 0.5 \text{ mol/kg}$

So:

$\Delta T_b = 1 \cdot 0.512 \cdot 0.5 = 0.256 \text{ °C}$

The boiling point of the solution will be elevated by 0.256 °C above the boiling point of pure water.

7. Explain the concept of osmotic pressure and its application.

Answer:
Osmotic pressure is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane, which allows the solvent but not the solute to pass through. It is given by the formula:

$\Pi = C \cdot R \cdot T$

where $\Pi$ is the osmotic pressure, $C$ is the molar concentration of the solution, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin.

Applications:

Medical: Osmotic pressure is used in medical treatments like dialysis, where it helps in removing waste products from the blood.
Food Preservation: Osmotic pressure helps in preserving food by preventing microbial growth through the use of salt or sugar solutions.

Conclusion

Chapter 2 of the NCERT Class 12 Chemistry textbook, focusing on "Solutions," covers fundamental concepts that are crucial for a deeper understanding of chemistry. By mastering the types of solutions, methods of expressing concentration, colligative properties, and Raoult's Law, students can develop a strong grasp of how solutions work and how they can be manipulated in practical scenarios.

This comprehensive guide aims to provide clarity on important questions from the chapter and help students prepare effectively for their exams. Regular practice and understanding these core concepts will enhance students' ability to solve problems related to solutions and excel in their chemistry studies.

Chemistry Class 12 NCERT Solutions Chapter 2 – Important Questions