JEE Main Previous Year Questions (2025-2026): Principles Related to Practical Chemistry

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Principles Related to Practical Chemistry  
 
(January 2026) 
 
 
Practical Organic Chemistry 
 
Q1: A student has been given 0.314 g of an organic compound and asked to estimate 
Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. 
The percentage of sulphur present in the compound is _________.  
(Given Molar mass in g mol
-1
 S: 32, BaSO
4
: 233) 
A: 63.15% 
B: 21.05% 
C: 48.24% 
D: 42.10% 
Answer: B 
Explanation:  
Mass of compound taken = 0.314 g 
Mass of BaSO 4 obtained = 0.4813 g 
In BaSO 4, sulphur present per mole = 32 g 
Molar mass of BaSO 4 = 233 g mol ?¹ 
So, mass of sulphur in 0.4813 g of BaSO 4 is: 
m(S) = 0.4813 × 32/233 
m(S) = 0.4813 × 32/233 = 0.06608 g (approx.) 
Percentage of sulphur in the compound: 
%S = (0.06608 / 0.314) × 100 ˜ 21.05% 
Correct option: B) 21.05% 
 
Q2: Method used for separation of mixture of products ( B and C ) obtained in the 
following reaction is 
 
A: simple distillation 
B: Sublimation 
C: fractional distillation 
D: steam distillation 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Principles Related to Practical Chemistry  
 
(January 2026) 
 
 
Practical Organic Chemistry 
 
Q1: A student has been given 0.314 g of an organic compound and asked to estimate 
Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. 
The percentage of sulphur present in the compound is _________.  
(Given Molar mass in g mol
-1
 S: 32, BaSO
4
: 233) 
A: 63.15% 
B: 21.05% 
C: 48.24% 
D: 42.10% 
Answer: B 
Explanation:  
Mass of compound taken = 0.314 g 
Mass of BaSO 4 obtained = 0.4813 g 
In BaSO 4, sulphur present per mole = 32 g 
Molar mass of BaSO 4 = 233 g mol ?¹ 
So, mass of sulphur in 0.4813 g of BaSO 4 is: 
m(S) = 0.4813 × 32/233 
m(S) = 0.4813 × 32/233 = 0.06608 g (approx.) 
Percentage of sulphur in the compound: 
%S = (0.06608 / 0.314) × 100 ˜ 21.05% 
Correct option: B) 21.05% 
 
Q2: Method used for separation of mixture of products ( B and C ) obtained in the 
following reaction is 
 
A: simple distillation 
B: Sublimation 
C: fractional distillation 
D: steam distillation 
Answer: C 
Explanation:  
 
B & C separate by Fractional Distillation method Due to their different boiling point. 
Q3: In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The 
percentage of chlorine in the organic compound is 
A: 87.65% 
B: 53.58% 
C: 37.57% 
D: 34.79% 
Answer: B 
Explanation: 
Mass of organic compound taken = 0.2425g  
Mass of AgCl obtained = 0.5253g 
In Carius method for estimation of halogens, the halogen present in the organic compound is 
converted into silver halide. Here chlorine is obtained as AgCl. 
In AgCl, the fraction of chlorine is: 
 
So, mass of chlorine in 0.5253g of AgCl is: 
 
Therefore, percentage of chlorine in the compound is: 
 
 
Q4: Match List - I with List - II. 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Principles Related to Practical Chemistry  
 
(January 2026) 
 
 
Practical Organic Chemistry 
 
Q1: A student has been given 0.314 g of an organic compound and asked to estimate 
Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. 
The percentage of sulphur present in the compound is _________.  
(Given Molar mass in g mol
-1
 S: 32, BaSO
4
: 233) 
A: 63.15% 
B: 21.05% 
C: 48.24% 
D: 42.10% 
Answer: B 
Explanation:  
Mass of compound taken = 0.314 g 
Mass of BaSO 4 obtained = 0.4813 g 
In BaSO 4, sulphur present per mole = 32 g 
Molar mass of BaSO 4 = 233 g mol ?¹ 
So, mass of sulphur in 0.4813 g of BaSO 4 is: 
m(S) = 0.4813 × 32/233 
m(S) = 0.4813 × 32/233 = 0.06608 g (approx.) 
Percentage of sulphur in the compound: 
%S = (0.06608 / 0.314) × 100 ˜ 21.05% 
Correct option: B) 21.05% 
 
Q2: Method used for separation of mixture of products ( B and C ) obtained in the 
following reaction is 
 
A: simple distillation 
B: Sublimation 
C: fractional distillation 
D: steam distillation 
Answer: C 
Explanation:  
 
B & C separate by Fractional Distillation method Due to their different boiling point. 
Q3: In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The 
percentage of chlorine in the organic compound is 
A: 87.65% 
B: 53.58% 
C: 37.57% 
D: 34.79% 
Answer: B 
Explanation: 
Mass of organic compound taken = 0.2425g  
Mass of AgCl obtained = 0.5253g 
In Carius method for estimation of halogens, the halogen present in the organic compound is 
converted into silver halide. Here chlorine is obtained as AgCl. 
In AgCl, the fraction of chlorine is: 
 
So, mass of chlorine in 0.5253g of AgCl is: 
 
Therefore, percentage of chlorine in the compound is: 
 
 
Q4: Match List - I with List - II. 
 
A: A-IV, B-I, C-II, D-III 
B: A-III, B-IV, C-I, D-II 
C: A-III, B-IV, C-II, D-I 
D: A-IV, B-III, C-II, D-I 
Answer: A 
Explanation: 
For each functional group, recall the standard qualitative test observation (NCERT): 
A. Unsaturation (Baeyer’s test) 
Baeyer’s reagent is cold, dilute alkaline KMnO4 (pink/purple). With C = C / C = C, the pink colour 
gets discharged. 
So, A ? IV. 
B. Alcoholic group (Ceric ammonium nitrate test) 
Alcohols give red colour with ceric ammonium nitrate solution. 
So, B ? I. 
C. Aldehyde group (Tollen’s reagent) 
Aldehydes reduce Tollen’s reagent to metallic silver, giving a silver mirror. 
So, C ? II. 
D. Phenolic group (FeCl3 test) 
Phenols give violet colour with neutral FeCl3. 
So, D ? III. 
Hence the correct matching is: 
A-IV, B-I, C-II, D-III 
Correct option: A 
 
Q5: When 1 g of compound (X)is subjected to Kjeldahl's method for estimation of nitrogen, 15 
mL 1 M H
2
SO
4
 was neutralized by ammonia evolved. The percentage of nitrogen in compound 
(X) Is: 
A: 0.21 
B: 21 
C: 42 
D: 0.42 
Answer: C 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Principles Related to Practical Chemistry  
 
(January 2026) 
 
 
Practical Organic Chemistry 
 
Q1: A student has been given 0.314 g of an organic compound and asked to estimate 
Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. 
The percentage of sulphur present in the compound is _________.  
(Given Molar mass in g mol
-1
 S: 32, BaSO
4
: 233) 
A: 63.15% 
B: 21.05% 
C: 48.24% 
D: 42.10% 
Answer: B 
Explanation:  
Mass of compound taken = 0.314 g 
Mass of BaSO 4 obtained = 0.4813 g 
In BaSO 4, sulphur present per mole = 32 g 
Molar mass of BaSO 4 = 233 g mol ?¹ 
So, mass of sulphur in 0.4813 g of BaSO 4 is: 
m(S) = 0.4813 × 32/233 
m(S) = 0.4813 × 32/233 = 0.06608 g (approx.) 
Percentage of sulphur in the compound: 
%S = (0.06608 / 0.314) × 100 ˜ 21.05% 
Correct option: B) 21.05% 
 
Q2: Method used for separation of mixture of products ( B and C ) obtained in the 
following reaction is 
 
A: simple distillation 
B: Sublimation 
C: fractional distillation 
D: steam distillation 
Answer: C 
Explanation:  
 
B & C separate by Fractional Distillation method Due to their different boiling point. 
Q3: In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The 
percentage of chlorine in the organic compound is 
A: 87.65% 
B: 53.58% 
C: 37.57% 
D: 34.79% 
Answer: B 
Explanation: 
Mass of organic compound taken = 0.2425g  
Mass of AgCl obtained = 0.5253g 
In Carius method for estimation of halogens, the halogen present in the organic compound is 
converted into silver halide. Here chlorine is obtained as AgCl. 
In AgCl, the fraction of chlorine is: 
 
So, mass of chlorine in 0.5253g of AgCl is: 
 
Therefore, percentage of chlorine in the compound is: 
 
 
Q4: Match List - I with List - II. 
 
A: A-IV, B-I, C-II, D-III 
B: A-III, B-IV, C-I, D-II 
C: A-III, B-IV, C-II, D-I 
D: A-IV, B-III, C-II, D-I 
Answer: A 
Explanation: 
For each functional group, recall the standard qualitative test observation (NCERT): 
A. Unsaturation (Baeyer’s test) 
Baeyer’s reagent is cold, dilute alkaline KMnO4 (pink/purple). With C = C / C = C, the pink colour 
gets discharged. 
So, A ? IV. 
B. Alcoholic group (Ceric ammonium nitrate test) 
Alcohols give red colour with ceric ammonium nitrate solution. 
So, B ? I. 
C. Aldehyde group (Tollen’s reagent) 
Aldehydes reduce Tollen’s reagent to metallic silver, giving a silver mirror. 
So, C ? II. 
D. Phenolic group (FeCl3 test) 
Phenols give violet colour with neutral FeCl3. 
So, D ? III. 
Hence the correct matching is: 
A-IV, B-I, C-II, D-III 
Correct option: A 
 
Q5: When 1 g of compound (X)is subjected to Kjeldahl's method for estimation of nitrogen, 15 
mL 1 M H
2
SO
4
 was neutralized by ammonia evolved. The percentage of nitrogen in compound 
(X) Is: 
A: 0.21 
B: 21 
C: 42 
D: 0.42 
Answer: C 
Explanation: 
In Kjeldahl’s method, the ammonia (NH3) evolved neutralizes the acid taken. So,  
equivalents of H2SO4 = equivalents of ammonia 
Given: 15 mL of 1 M H2SO4 is neutralized. 
H2SO4 is a dibasic acid, so 1 mole of H2SO4 gives 2 equivalents. Therefore, equivalents of H2SO4 
used are: 
Ammonia is monobasic (one NH3 neutralizes one equivalent of acid), so: 
= moles of ammonia × 1 
So, moles of ammonia = moles of nitrogen (N), because each mole of NH3 contains 1 mole of N. 
Now, mass of nitrogen: 
 
Percentage of nitrogen in 1 g compound: 
 
 
Q6: In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, ?nd 
percentage of sulphur (molar mass 32 g mol
-1
). Molar mass of barium sulphate is 233 g mol
-1
. 
A: 4.55% 
B: 16.48% 
C: 10.30% 
D: 21.97% 
Answer: D 
Explanation: 
Moles of BaSO
4
 formed are 
 
In BaSO
4
, there is 1 atom of S per formula unit, so 
 
Mass of sulphur in the compound: 
 
Percentage of sulphur in the organic compound (mass of compound = 0.75 g): 
 
So, the correct option is D) 21.97% 
 
Q7: In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N
2
) at 
300 K temperature and 715 mm Hg pressure. The percentage composition of ' N ' in the 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Principles Related to Practical Chemistry  
 
(January 2026) 
 
 
Practical Organic Chemistry 
 
Q1: A student has been given 0.314 g of an organic compound and asked to estimate 
Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. 
The percentage of sulphur present in the compound is _________.  
(Given Molar mass in g mol
-1
 S: 32, BaSO
4
: 233) 
A: 63.15% 
B: 21.05% 
C: 48.24% 
D: 42.10% 
Answer: B 
Explanation:  
Mass of compound taken = 0.314 g 
Mass of BaSO 4 obtained = 0.4813 g 
In BaSO 4, sulphur present per mole = 32 g 
Molar mass of BaSO 4 = 233 g mol ?¹ 
So, mass of sulphur in 0.4813 g of BaSO 4 is: 
m(S) = 0.4813 × 32/233 
m(S) = 0.4813 × 32/233 = 0.06608 g (approx.) 
Percentage of sulphur in the compound: 
%S = (0.06608 / 0.314) × 100 ˜ 21.05% 
Correct option: B) 21.05% 
 
Q2: Method used for separation of mixture of products ( B and C ) obtained in the 
following reaction is 
 
A: simple distillation 
B: Sublimation 
C: fractional distillation 
D: steam distillation 
Answer: C 
Explanation:  
 
B & C separate by Fractional Distillation method Due to their different boiling point. 
Q3: In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The 
percentage of chlorine in the organic compound is 
A: 87.65% 
B: 53.58% 
C: 37.57% 
D: 34.79% 
Answer: B 
Explanation: 
Mass of organic compound taken = 0.2425g  
Mass of AgCl obtained = 0.5253g 
In Carius method for estimation of halogens, the halogen present in the organic compound is 
converted into silver halide. Here chlorine is obtained as AgCl. 
In AgCl, the fraction of chlorine is: 
 
So, mass of chlorine in 0.5253g of AgCl is: 
 
Therefore, percentage of chlorine in the compound is: 
 
 
Q4: Match List - I with List - II. 
 
A: A-IV, B-I, C-II, D-III 
B: A-III, B-IV, C-I, D-II 
C: A-III, B-IV, C-II, D-I 
D: A-IV, B-III, C-II, D-I 
Answer: A 
Explanation: 
For each functional group, recall the standard qualitative test observation (NCERT): 
A. Unsaturation (Baeyer’s test) 
Baeyer’s reagent is cold, dilute alkaline KMnO4 (pink/purple). With C = C / C = C, the pink colour 
gets discharged. 
So, A ? IV. 
B. Alcoholic group (Ceric ammonium nitrate test) 
Alcohols give red colour with ceric ammonium nitrate solution. 
So, B ? I. 
C. Aldehyde group (Tollen’s reagent) 
Aldehydes reduce Tollen’s reagent to metallic silver, giving a silver mirror. 
So, C ? II. 
D. Phenolic group (FeCl3 test) 
Phenols give violet colour with neutral FeCl3. 
So, D ? III. 
Hence the correct matching is: 
A-IV, B-I, C-II, D-III 
Correct option: A 
 
Q5: When 1 g of compound (X)is subjected to Kjeldahl's method for estimation of nitrogen, 15 
mL 1 M H
2
SO
4
 was neutralized by ammonia evolved. The percentage of nitrogen in compound 
(X) Is: 
A: 0.21 
B: 21 
C: 42 
D: 0.42 
Answer: C 
Explanation: 
In Kjeldahl’s method, the ammonia (NH3) evolved neutralizes the acid taken. So,  
equivalents of H2SO4 = equivalents of ammonia 
Given: 15 mL of 1 M H2SO4 is neutralized. 
H2SO4 is a dibasic acid, so 1 mole of H2SO4 gives 2 equivalents. Therefore, equivalents of H2SO4 
used are: 
Ammonia is monobasic (one NH3 neutralizes one equivalent of acid), so: 
= moles of ammonia × 1 
So, moles of ammonia = moles of nitrogen (N), because each mole of NH3 contains 1 mole of N. 
Now, mass of nitrogen: 
 
Percentage of nitrogen in 1 g compound: 
 
 
Q6: In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, ?nd 
percentage of sulphur (molar mass 32 g mol
-1
). Molar mass of barium sulphate is 233 g mol
-1
. 
A: 4.55% 
B: 16.48% 
C: 10.30% 
D: 21.97% 
Answer: D 
Explanation: 
Moles of BaSO
4
 formed are 
 
In BaSO
4
, there is 1 atom of S per formula unit, so 
 
Mass of sulphur in the compound: 
 
Percentage of sulphur in the organic compound (mass of compound = 0.75 g): 
 
So, the correct option is D) 21.97% 
 
Q7: In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N
2
) at 
300 K temperature and 715 mm Hg pressure. The percentage composition of ' N ' in the 
organic compound is _____________ % (Nearest integer)  
(Aqueous tension at 300 K = 15mmHg 
Answer: 15 
Explanation: 
In Dumas method, the nitrogen gas collected over water contains water vapour also. So ?rst we 
?nd the pressure of dry nitrogen gas by subtracting aqueous tension. 
 
 
Q8: Sodium fusion extract of an organic compound (Y) with CHCl3 and chlorine water gives 
violet color to the CHCl3 layer. 0.15 g of (Y) gave 0.12 g of the silver halide precipitate in Carius 
method. Percentage of halogen in the compound (Y) is _____. (Nearest integer) 
(Given : molar mass gmol?¹ C : 12, H : 1, Cl : 35.5, Br : 80, I : 127 ) 
Answer: 43 
Explanation: 
Violet colour in the CHCl3 layer on adding chlorine water indicates iodide ion in the sodium 
fusion extract (chlorine oxidises I? to I2, which is violet in CHCl3). 
So, the silver halide precipitate in Carius method is AgI. 
Calculation of % iodine 
Molar mass of AgI = 108 + 127 = 235 g mol?¹ 
Mass of AgI obtained = 0.12 g 
Mass of iodine in 0.12 g of AgI: 
m(I) = 0.12 × 127/235 = 0.06485 g 
Mass of compound (Y) = 0.15 g