Adsorption + Colloids + Tyndall Effect
Surface Chemistry
42
JEE Qs
5%
Hard
75
min
Master the definitions, classifications, and applications, as most questions from this topic are direct and concept-based rather than numerical.
🧮 Key Formulas
✅ Key Points for JEE
- 1Distinguish clearly between physisorption (multilayer, low enthalpy, reversible, favored by low T and high P) and chemisorption (monolayer, high enthalpy, irreversible, favored by high T, needs activation energy).
- 2Understand the conditions for the Tyndall effect: diameter of dispersed particles not much smaller than wavelength of light, and refractive indices of dispersed phase and dispersion medium differ significantly.
- 3Categorize colloids based on physical state of phases, nature of interaction (lyophilic/lyophobic), and type of particles (multimolecular, macromolecular, associated). Lyophilic are more stable due to solvation.
- 4Recall the Hardy-Schulze rule: The coagulating power of an electrolyte is directly proportional to the fourth power of the valency of the effective ion (ion opposite in charge to the colloidal particles).
- 5Be familiar with common applications of adsorption (e.g., gas masks, chromatography, catalysts) and colloids (e.g., medicines, purification of water, photography).
⚠️ Common Mistakes
- ✕Confusing adsorption with absorption; adsorption is a surface phenomenon while absorption involves uniform distribution throughout the bulk.
- ✕Incorrectly applying the Hardy-Schulze rule by not identifying the effective ion or by miscalculating its valency.
- ✕Failing to distinguish between true solutions, colloids, and suspensions based on particle size and properties like Tyndall effect and filterability.
- ✕Mixing up the characteristics and factors affecting physisorption and chemisorption, especially regarding temperature and pressure dependence.
📝 Practice Questions
See allQ60.The purification method based on the following physical transformation is : (1) Distillation (2) Extraction (3) Sublimation (4) Crystallization
Q70.If C (diamond ) →C (graphite ) + XkJmol−1 C( diamond ) + O2( g) →CO2( g) + YkJmol−1 C (graphite) +O2( g) →CO2( g) + ZkJmol−1 at constant temperature. Then (1) X=-Y+Z (2) -X=Y+Z (3) X=Y+Z (4) X=Y-Z
Q54.Given below are two statements : Statement (I) : Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode. Statement (II) : The rate of corrosion is more in alkaline medium than in acidic medium. In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are true (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are false 2025 (22 Jan Shift 2) JEE Main Previous Year Paper
Q31.Given below are two statements: Statement I: The hot water flows faster than cold water Statement II: Soap water has higher surface tension as compared to fresh water. In the light above statements, choose the correct answer from the options given below (1) Statement I is true but Statement II is false (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are false (4) Both Statement I and Statement II are true
Q54. In the given TLC, the distance of spot A & B are 5 cm & 7 cm, from the bottom of TLC plate, respectively. Rf value of B is x × 10−1 times more than A . The value of x is ______.
Q36.The technique used for purification of steam volatile water immiscible substance is: (1) Fractional distillation (2) Fractional distillation under reduced pressure (3) Distillation (4) Steam distillation
NCERT Chapters
- Class 12 Chemistry Part I Ch 5: Surface Chemistry