JEE (Main) · Chemistry
Principles Related to Practical Chemistry: JEE Main PYQ, 2020–2025
Principles Related to Practical Chemistry (Chemistry) has been asked 47 times in the 111 JEE (Main) shift papers from 2020–2025: about 0.4 questions a paper, 2% of all Chemistry questions. It is the 20th most-asked of 21 Chemistry chapters. The 2025 shifts had 5 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 2in 14 papers |
| 2021 | 13in 26 papers |
| 2022 | 8in 24 papers |
| 2023 | 4in 12 papers |
| 2024 | 15in 18 papers |
| 2025 | 5in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Titrimetric analysis21
- Qualitative salt analysis14
- Detection of elements and functional groups8
- Preparation of inorganic and organic compounds3
- Physical chemistry experiments1
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10 Principles Related to Practical Chemistry questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Principles Related to Practical Chemistry
Benzoic acid is prepared by oxidising toluene with hot alkaline potassium permanganate. After the MnO2 is filtered off, the clear filtrate is acidified with dilute hydrochloric acid. Why is the acid added?
Answer: (A)
In alkaline medium toluene is oxidised to potassium benzoate, which stays dissolved. Adding HCl protonates the carboxylate: C6H5COO⁻ + H⁺ gives C6H5COOH. Benzoic acid has low solubility in cold water, so it precipitates and is filtered and recrystallised.
Q2 · Chemistry · Principles Related to Practical Chemistry
Three isomeric alcohols of formula C₄H₁₀O, X, Y and Z, are shaken separately with anhydrous ZnCl₂ in concentrated HCl (Lucas reagent) at room temperature. X gives turbidity immediately, Y gives turbidity after about five minutes, and Z shows no turbidity at room temperature. X, Y and Z are respectively:
Answer: (A)
The Lucas test distinguishes the classes of alcohol by the rate at which the insoluble alkyl chloride forms, seen as turbidity. The reaction proceeds through a carbocation, so the order is 3° > 2° > 1°. 2-Methylpropan-2-ol (3°) gives turbidity immediately, butan-2-ol (2°) after about five minutes, and butan-1-ol (1°) not at room temperature.
Q3 · Chemistry · Principles Related to Practical Chemistry
A freshly prepared solution of a salt gives a deep blue precipitate with potassium hexacyanoferrate(III), K₃[Fe(CN)₆]. With potassium hexacyanoferrate(II), K₄[Fe(CN)₆], it gives a white precipitate that turns blue on standing in air. The solution gives no blood-red colour with KSCN until it is oxidised. The cation present is:
Answer: (D)
Principle: the colour depends on both the iron oxidation state in the cation and which cyanoferrate reagent is used. Fe³⁺ with [Fe(CN)₆]⁴⁻ gives Prussian blue at once and gives blood-red [Fe(SCN)]²⁺ with KSCN. Fe²⁺ with [Fe(CN)₆]³⁻ gives a deep blue precipitate. Fe²⁺ with [Fe(CN)₆]⁴⁻ gives a white precipitate that turns blue as air oxidises Fe²⁺ to Fe³⁺. The absence of a KSCN colour before oxidation also points to Fe²⁺. Zn²⁺ and Cu²⁺ fit neither the blue precipitate nor the white-to-blue change.
Q4 · Chemistry · Principles Related to Practical Chemistry
A solution containing Cu²⁺ ions is treated with a few drops of dilute NaOH, and then excess aqueous ammonia is added. What is the colour of the final solution?
Answer: (A)
Adding NaOH first precipitates pale blue Cu(OH)₂. Excess NH₃ then dissolves this precipitate by forming the soluble, deep blue tetraammine copper(II) complex, [Cu(NH₃)₄]²⁺. The colour change from the hydrated ion's pale blue to deep blue is the recognition feature. Remember that the deep blue colour comes from the ammine complex, not from Cu²⁺ alone.
Q5 · Chemistry · Principles Related to Practical Chemistry
On adding dilute HCl to a salt solution, a white precipitate forms. It dissolves completely on boiling with water. The hot solution gives a yellow precipitate with KI, which dissolves on heating and reappears on cooling as golden spangles. Which cation is present?
Answer: (A)
Group I cations (Pb²⁺, Ag⁺, Hg₂²⁺) are precipitated as white chlorides by dilute HCl. Lead chloride is more soluble in hot water than AgCl or Hg₂Cl₂, so the precipitate dissolves on boiling. The hot solution gives yellow PbI₂ with KI. PbI₂ dissolves on heating and crystallises as golden spangles on cooling. Ag⁺ and Hg₂²⁺ are separated by their behaviour with NH₄OH, and Cu²⁺ is not precipitated by HCl.
Q6 · Chemistry · Principles Related to Practical Chemistry
In a rate study of Na2S2O3 + 2HCl → 2NaCl + H2O + SO2 + S↓, the time for a fixed turbidity of sulfur to hide a cross marked under the flask is noted. Run 1: 40 mL of 0.10 M Na2S2O3 + 10 mL water + 5 mL of 2 M HCl gives t = 20 s. Run 2: 10 mL of 0.10 M Na2S2O3 + 40 mL water + 5 mL of 2 M HCl, at the same temperature. Assume the reaction is first order in thiosulfate and that the HCl concentration is effectively the same in both runs. For the same turbidity, the time is inversely proportional to the rate. What is t for Run 2?
Answer: (A)
Since the same amount of sulfur must be formed to hide the cross, t ∝ 1/rate. The total volume is 55 mL in both runs, so [S2O3²⁻] in Run 2 is 10/40 = 1/4 of Run 1. HCl is the same in both runs. For a first-order dependence, rate2 = rate1/4, so t2 = 4 × 20 s = 80 s.
Q7 · Chemistry · Principles Related to Practical Chemistry
A pure substance is heated and then cooled slowly, and its temperature is recorded against time. During the time when the substance is changing from liquid to solid at its freezing point, what happens to its temperature?
Answer: (A)
Principle: during a phase change at constant pressure, heat is exchanged as latent heat while the temperature stays constant. On cooling, the liquid releases latent heat as it freezes, so the temperature holds at the freezing point until solidification is complete. The cooling curve therefore shows a flat plateau.
Q8 · Chemistry · Principles Related to Practical Chemistry
In the laboratory preparation of potash alum, scrap aluminium is warmed with hot aqueous KOH until effervescence stops, and the filtered solution is treated dropwise with dilute H₂SO₄ with constant stirring. A white gelatinous precipitate appears first and then dissolves as more acid is added. What is the precipitate, and why does it dissolve?
Answer: (A)
Aluminium dissolves in hot KOH to give potassium tetrahydroxoaluminate, K[Al(OH)₄], with hydrogen evolved. Adding dilute H₂SO₄ first neutralises the OH⁻ and the aluminate: [Al(OH)₄]⁻ + H⁺ → Al(OH)₃↓ + H₂O. Al(OH)₃ is amphoteric, so more acid dissolves it: Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O. The solution then contains K⁺, Al³⁺ and SO₄²⁻ in the 1:1:2 ratio. Concentrating and cooling it gives potash alum crystals. Reasoning that the precipitate is a sulphate, or that excess OH⁻ redissolves it, ignores that the acid is the reagent being added.
Q9 · Chemistry · Principles Related to Practical Chemistry
In the laboratory preparation of potash alum, scrap aluminium is warmed with aqueous KOH and filtered. Dilute H2SO4 is then added dropwise to the filtrate. A white gelatinous precipitate forms first and then redissolves as more acid is added. The solution is concentrated and cooled, and octahedral crystals of KAl(SO4)2·12H2O separate. Which statement correctly identifies the precipitate and explains where the K2SO4 part of the alum comes from?
Answer: (A)
Aluminium dissolves in hot KOH: 2Al + 2KOH + 6H2O → 2K[Al(OH)4] + 3H2. The filtrate therefore holds K[Al(OH)4] and some unreacted KOH. Adding dilute H2SO4 first neutralises the free KOH to K2SO4 and then precipitates amphoteric Al(OH)3. More acid dissolves it: 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O. The solution now contains K2SO4 and Al2(SO4)3 in a 1:1 mole ratio of K to Al, which is what the alum KAl(SO4)2·12H2O needs. On concentrating and cooling, the less soluble double salt crystallises. K2SO4 does not have to be added.
Q10 · Chemistry · Principles Related to Practical Chemistry
In the detection of carbon in an organic compound, the compound is heated with copper(II) oxide and the gas evolved is passed through lime water. Which observation confirms the presence of carbon?
Answer: (A)
Carbon in an organic compound is oxidised by CuO to CO2. Passing CO2 through lime water gives insoluble CaCO3, which makes the solution milky. Turbidity is the confirmatory observation for carbon.
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All 47 real JEE Main questions from Principles Related to Practical Chemistry.
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About Principles Related to Practical Chemistry in JEE Main.
How many questions come from Principles Related to Practical Chemistry in JEE Main?
Principles Related to Practical Chemistry (Chemistry) has been asked 47 times in the 111 JEE (Main) shift papers from 2020–2025: about 0.4 questions a paper, 2% of all Chemistry questions. It is the 20th most-asked of 21 Chemistry chapters. The 2025 shifts had 5 from it in all.
Which topics of Principles Related to Practical Chemistry are asked most in JEE Main?
Titrimetric analysis (21), Qualitative salt analysis (14), Detection of elements and functional groups (8), Preparation of inorganic and organic compounds (3), counted over the papers from 2020–2025.
Is Principles Related to Practical Chemistry important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 20 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Principles Related to Practical Chemistry JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Principles Related to Practical Chemistry with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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