Solved NCERT Questions For Class 12 Chemistry Chapter 5 In Hindi - Free PDF
FAQs on NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry in Hindi - 2025-26
1. Where can I find accurate and step-by-step NCERT Solutions for Class 12 Chemistry Chapter 5, Surface Chemistry?
You can find reliable and comprehensive NCERT Solutions for Class 12 Chemistry Chapter 5 on Vedantu. These solutions are prepared by subject matter experts and provide a detailed, step-by-step explanation for all in-text and exercise questions, strictly following the CBSE methodology to help students understand the correct way to answer in exams.
2. Is Chapter 5, Surface Chemistry, part of the CBSE Class 12 Chemistry syllabus for the 2025-26 exams?
No, as per the rationalised syllabus issued by NCERT for the academic year 2025-26, Chapter 5, Surface Chemistry, has been removed from the CBSE Class 12 Chemistry curriculum. However, these NCERT solutions remain a valuable resource for students preparing for competitive exams like JEE and NEET, or those studying under boards that still include this chapter, as the concepts are fundamental to chemistry.
3. How should one approach solving numerical problems based on the Freundlich adsorption isotherm from the NCERT textbook?
To solve numericals on the Freundlich adsorption isotherm, follow these steps as per the NCERT guidelines:
1. Use the logarithmic form of the equation: log(x/m) = log K + (1/n) log P.
2. Identify the given variables, such as the extent of adsorption (x/m) and pressure (P).
3. If a graph is required, plot log(x/m) on the y-axis against log P on the x-axis.
4. The slope of the line will give you the value of 1/n, and the y-intercept will give you the value of log K. This method ensures you derive the constants correctly.
4. What is the correct way to present the difference between physisorption and chemisorption in an exam according to the NCERT solutions pattern?
For maximum marks, the NCERT solutions format suggests presenting the differences in a clear, tabular form. Key points of comparison should include:
- Nature of Forces: Weak van der Waals forces in physisorption versus strong chemical bonds in chemisorption.
- Specificity: Physisorption is not specific, while chemisorption is highly specific.
- Reversibility: Physisorption is reversible, whereas chemisorption is irreversible.
- Enthalpy of Adsorption: Low enthalpy (20-40 kJ/mol) for physisorption and high enthalpy (80-240 kJ/mol) for chemisorption.
- Effect of Temperature: Physisorption decreases with an increase in temperature, while chemisorption first increases and then decreases.
5. How do the NCERT solutions for Chapter 5 explain the role of coagulation in purifying drinking water?
The NCERT solutions explain that raw water often contains suspended colloidal impurities, such as clay, which carry a negative charge. To purify the water, a coagulating agent like alum is added. The Al³⁺ ions from alum neutralise the negative charge on the colloidal particles. This causes the particles to clump together (coagulate) and settle down, allowing for their removal and resulting in clean water.
6. Why is adsorption different from absorption? How do NCERT solutions clarify this fundamental concept?
The NCERT solutions clarify this by defining their locations. Adsorption is a surface phenomenon where molecules of a substance (adsorbate) accumulate only on the surface of a solid or liquid (adsorbent). In contrast, absorption is a bulk phenomenon, where the substance is uniformly distributed throughout the body of the solid or liquid. For example, water vapour is adsorbed by silica gel but absorbed by anhydrous calcium chloride.
7. What are the key steps involved in the mechanism of enzyme catalysis as detailed in the NCERT Class 12 solutions?
The NCERT solutions explain the enzyme catalysis mechanism, often called the lock-and-key model, in two main steps:
Step 1: Formation of the Enzyme-Substrate Complex. The substrate (the 'key') binds to the specific active site on the enzyme (the 'lock'), forming an activated enzyme-substrate complex.
Step 2: Decomposition to Products. The enzyme catalyses the chemical reaction within the complex, which then decomposes to release the product(s) and regenerates the original enzyme, ready to catalyse another reaction.
8. How do the NCERT solutions explain the cleansing action of soaps and detergents using the concept of micelle formation?
The solutions explain that soap molecules have a hydrophilic (water-loving) head and a hydrophobic (oil-loving) tail. Above a certain concentration (Critical Micelle Concentration), these molecules form spherical aggregates called micelles. During cleaning, the hydrophobic tails trap oily dirt at the centre of the micelle, while the hydrophilic heads remain on the surface, interacting with water. This allows the oil and grease to be washed away with water.

















