What is the adsorption and colloids JEE 2017 question, and which options are correct?
The adsorption and colloids JEE 2017 question has exactly two correct options: A and B. This Chemistry question from JEE Advanced 2017, Paper 2 is multiple-correct, so more than one option can be correct. Judge each statement independently rather than searching for one “best” answer.
The question bank labels it medium difficulty, with difficulty tag 2 and a 90-second target. These are question-bank metadata, not an official exam time limit.
In plain language, the four statements claim:
Verify every essential clause. In particular, check the classification word in C and the independence claim in D.
Why are A and B correct?
Adsorption releases energy and restricts molecular motion, making A correct. For B, the supplied critical temperatures indicate that ethane is more readily liquefied and is therefore expected to undergo stronger physical adsorption. Check energy and disorder separately from the comparison between gases.
A, step 1: Follow the molecules. Molecules leave the bulk gas or solution phase and accumulate at the adsorbent surface. Adsorption is a surface process.
A, step 2: Check the energy change. Formation of adsorbate–adsorbent interactions releases energy. Adsorption is generally exothermic in the treatment used here:
A, step 3: Check freedom of motion. Surface-bound molecules have less freedom of motion, so randomness decreases:
Both parts hold: A is correct. Now compare the gases in B.
- B, step 1: Physical adsorption is stronger for a gas that is more readily liquefied.
- B, step 2: A higher critical temperature indicates stronger intermolecular attractions and easier liquefaction.
- B, step 3: Use the question’s supplied comparison:
The difference establishes the ordering, not an adsorption ratio. These are the question’s supplied values, not independently checked reference-table values.
Following the official method, ethane is expected to adsorb more strongly on equal amounts of activated charcoal at the same temperature because its supplied critical temperature is higher. Therefore, B is correct.
Why are C and D incorrect?
C gives the wrong colloid name; D falsely claims that particle size has no effect on Brownian motion. Each contains a correct clause, but one false essential clause makes the whole option incorrect. Check the phase pair in C and both dependencies in D.
C, step 1: Identify the phases. Tiny liquid water droplets form the dispersed phase. Air is the gaseous dispersion medium.
C, step 2: Match the phases to the name. Liquid dispersed in gas forms a liquid aerosol. An emulsion requires liquid dispersed in liquid.
C, step 3: Judge the complete statement. The phase description is correct, but the emulsion label is false. Therefore, C is incorrect.
- D, step 1: Brownian motion depends on particle size as well as the medium’s viscosity.
- D, step 2: Under otherwise comparable conditions, smaller particles show more vigorous Brownian motion. Greater viscosity reduces that motion.
- D, step 3: The correct viscosity clause does not rescue the false claim of size independence. Therefore, D is incorrect.
A true, B true, C false, D false. Correct options: A and B.
Why is checking only the phase description in C not enough?
A correct composition does not guarantee a correct classification. Selecting C after identifying water droplets and air leaves the word “emulsion” untested. The faulty reasoning is:
“Cloud contains liquid droplets in gas; that description is right, so C must be right.”
This checks the composition but skips the classification word “emulsion”. Replace that shortcut with a two-part test:
- Identify the dispersed phase and the dispersion medium.
- Verify the colloid name against that exact pair.
For a cloud, the pair is liquid in gas, so the name is liquid aerosol. An emulsion requires liquid in liquid.
The rule for multiple-correct questions is strict: every essential clause must be true before you select an option. One false label is enough to reject the statement.
How can I practise the same checks with three related questions?
Use these original related practice questions, not additional verified PYQs, to check critical-temperature comparisons, colloid classification and Brownian motion. Answer each before reading its reasoning, and keep the stated comparison conditions fixed.
Question 1: Which gas should adsorb more strongly?
Gases X and Y have critical temperatures satisfying:
Under comparable physical-adsorption conditions on the same adsorbent, X is expected to adsorb more strongly. Its higher critical temperature indicates easier liquefaction, which supports stronger physical adsorption. This ordering does not give a numerical ratio of amounts adsorbed.
Question 2: How should fog and milk be classified?
Fog is a liquid aerosol; milk is an emulsion. Identify both phases to justify each name:
- Fog: Liquid water droplets are the dispersed phase; gas is the dispersion medium. Liquid in gas makes it a liquid aerosol.
- Milk: Liquid fat is the dispersed phase; an aqueous liquid is the dispersion medium. Liquid in liquid makes it an emulsion.
Both have liquid dispersed phases. Their different dispersion media give them different colloid names.
Question 3: How do particle size and viscosity affect Brownian motion?
At fixed temperature, decreasing particle size at fixed viscosity makes motion more vigorous. Smaller particles show stronger Brownian motion under otherwise comparable conditions.
Separately, increasing viscosity at fixed particle size makes motion less vigorous. A more viscous medium offers greater resistance to movement.
Before checking an answer, write the reasoning chain. For a false statement, underline the exact clause that makes it false.
Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).
If that step was the hard part, work through Units and Measurements JEE 2018: Wire and Vernier Solution.
Frequently asked questions
Which options are correct in the adsorption and colloids JEE 2017 question?
The correct options are A and B in this JEE Advanced 2017 Paper 2 multiple-correct question. C incorrectly classifies a cloud as an emulsion, while D incorrectly claims that Brownian motion is independent of particle size.
Why do enthalpy and entropy decrease during adsorption?
In the treatment used in this question, adsorbate–adsorbent interactions release energy, so the enthalpy change is negative. Molecules become restricted at the surface and lose freedom of motion, so the entropy change is also negative.
Why does ethane adsorb more strongly than nitrogen on activated charcoal?
The question supplies a higher critical temperature for ethane than for nitrogen, indicating easier liquefaction. Under comparable physical-adsorption conditions, easier liquefaction supports stronger adsorption. The critical-temperature comparison establishes an ordering, not a numerical adsorption ratio.
Is a cloud an aerosol or an emulsion?
A cloud containing liquid water droplets dispersed in air is a liquid aerosol. An emulsion requires a liquid dispersed in another liquid. The liquid-in-gas description is correct in option C, but its emulsion label makes the option incorrect.
How do particle size and viscosity affect Brownian motion?
At fixed temperature and viscosity, smaller particles show more vigorous Brownian motion. At fixed temperature and particle size, increasing the medium’s viscosity reduces the motion. Brownian motion therefore depends on both particle size and viscosity.