Solutions: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Solutions  
 
(January 2026) 
 
Q1: Consider the following aqueous solutions: 
I. 2.2 g Glucose in 125 mL of solution 
II. 1.9 g Calcium chloride in 250 mL of solution 
III. 9.0 g Urea in 500 mL of solution 
IV. 20.5 g Aluminium sulphate in 750 mL of solution 
The correct increasing order of boiling point of these solutions will be: 
[Given: Molar mass in g mol ?¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and 
S = 32] 
A: III < I < II < IV 
B: II < III < IV < I 
C: II < III < I < IV 
D: I < II < III < IV 
Answer: D 
Explanation: 
?T
b
 = i · k
b
 · m 
For dilute solution (M ˜ m) 
We know that boiling point elevation depends on ?T
b
. 
Since k
b
 is the same for water in all cases, we compare only i × m (or i × M for dilute solutions). 
 
Now compare the values of i × m: 
(I) Glucose: 0.098 × 1 = 0.098 
(II) CaCl 2: 0.068 × 3 = 0.204 
(III) Urea: 0.3 × 1 = 0.3 
(IV) Aluminium sulphate: 0.08 × 5 = 0.4 
Order of ?T
b
 = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So order of BP = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So the answer will be I < II < III < IV. 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Solutions  
 
(January 2026) 
 
Q1: Consider the following aqueous solutions: 
I. 2.2 g Glucose in 125 mL of solution 
II. 1.9 g Calcium chloride in 250 mL of solution 
III. 9.0 g Urea in 500 mL of solution 
IV. 20.5 g Aluminium sulphate in 750 mL of solution 
The correct increasing order of boiling point of these solutions will be: 
[Given: Molar mass in g mol ?¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and 
S = 32] 
A: III < I < II < IV 
B: II < III < IV < I 
C: II < III < I < IV 
D: I < II < III < IV 
Answer: D 
Explanation: 
?T
b
 = i · k
b
 · m 
For dilute solution (M ˜ m) 
We know that boiling point elevation depends on ?T
b
. 
Since k
b
 is the same for water in all cases, we compare only i × m (or i × M for dilute solutions). 
 
Now compare the values of i × m: 
(I) Glucose: 0.098 × 1 = 0.098 
(II) CaCl 2: 0.068 × 3 = 0.204 
(III) Urea: 0.3 × 1 = 0.3 
(IV) Aluminium sulphate: 0.08 × 5 = 0.4 
Order of ?T
b
 = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So order of BP = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So the answer will be I < II < III < IV. 
 
Q2: At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of 
the ideal solution formed is 320 mm Hg. 
At this stage, one mole of A and one mole of B are added to the solution. The vapour 
pressure is now measured as 328.6 mm Hg. 
The vapour pressures (in mm Hg) of A and B are respectively: 
A: 400, 300 
B: 600, 400 
C: 500, 200 
D: 300, 200 
Answer: C 
Explanation: 
For an ideal solution, by Raoult’s law: 
 
Step 1: First mixture (2 mol A + 3 mol B) 
Total moles = 2 + 3 = 5 
 
Step 2: After adding 1 mol A and 1 mol B 
New moles: A = 3, B = 4, total = 7 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Solutions  
 
(January 2026) 
 
Q1: Consider the following aqueous solutions: 
I. 2.2 g Glucose in 125 mL of solution 
II. 1.9 g Calcium chloride in 250 mL of solution 
III. 9.0 g Urea in 500 mL of solution 
IV. 20.5 g Aluminium sulphate in 750 mL of solution 
The correct increasing order of boiling point of these solutions will be: 
[Given: Molar mass in g mol ?¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and 
S = 32] 
A: III < I < II < IV 
B: II < III < IV < I 
C: II < III < I < IV 
D: I < II < III < IV 
Answer: D 
Explanation: 
?T
b
 = i · k
b
 · m 
For dilute solution (M ˜ m) 
We know that boiling point elevation depends on ?T
b
. 
Since k
b
 is the same for water in all cases, we compare only i × m (or i × M for dilute solutions). 
 
Now compare the values of i × m: 
(I) Glucose: 0.098 × 1 = 0.098 
(II) CaCl 2: 0.068 × 3 = 0.204 
(III) Urea: 0.3 × 1 = 0.3 
(IV) Aluminium sulphate: 0.08 × 5 = 0.4 
Order of ?T
b
 = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So order of BP = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So the answer will be I < II < III < IV. 
 
Q2: At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of 
the ideal solution formed is 320 mm Hg. 
At this stage, one mole of A and one mole of B are added to the solution. The vapour 
pressure is now measured as 328.6 mm Hg. 
The vapour pressures (in mm Hg) of A and B are respectively: 
A: 400, 300 
B: 600, 400 
C: 500, 200 
D: 300, 200 
Answer: C 
Explanation: 
For an ideal solution, by Raoult’s law: 
 
Step 1: First mixture (2 mol A + 3 mol B) 
Total moles = 2 + 3 = 5 
 
Step 2: After adding 1 mol A and 1 mol B 
New moles: A = 3, B = 4, total = 7 
 
 
Q3: At 298 K, the mole percentage of N 2(g) in air is 80%. Water is in equilibrium with air at 
a pressure of 10 atm. 
What is the mole fraction of N 2(g) in water at 298 K? 
(K
H
 for N 2 is 6.5 × 10 7 mmHg) 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Solutions  
 
(January 2026) 
 
Q1: Consider the following aqueous solutions: 
I. 2.2 g Glucose in 125 mL of solution 
II. 1.9 g Calcium chloride in 250 mL of solution 
III. 9.0 g Urea in 500 mL of solution 
IV. 20.5 g Aluminium sulphate in 750 mL of solution 
The correct increasing order of boiling point of these solutions will be: 
[Given: Molar mass in g mol ?¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and 
S = 32] 
A: III < I < II < IV 
B: II < III < IV < I 
C: II < III < I < IV 
D: I < II < III < IV 
Answer: D 
Explanation: 
?T
b
 = i · k
b
 · m 
For dilute solution (M ˜ m) 
We know that boiling point elevation depends on ?T
b
. 
Since k
b
 is the same for water in all cases, we compare only i × m (or i × M for dilute solutions). 
 
Now compare the values of i × m: 
(I) Glucose: 0.098 × 1 = 0.098 
(II) CaCl 2: 0.068 × 3 = 0.204 
(III) Urea: 0.3 × 1 = 0.3 
(IV) Aluminium sulphate: 0.08 × 5 = 0.4 
Order of ?T
b
 = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So order of BP = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So the answer will be I < II < III < IV. 
 
Q2: At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of 
the ideal solution formed is 320 mm Hg. 
At this stage, one mole of A and one mole of B are added to the solution. The vapour 
pressure is now measured as 328.6 mm Hg. 
The vapour pressures (in mm Hg) of A and B are respectively: 
A: 400, 300 
B: 600, 400 
C: 500, 200 
D: 300, 200 
Answer: C 
Explanation: 
For an ideal solution, by Raoult’s law: 
 
Step 1: First mixture (2 mol A + 3 mol B) 
Total moles = 2 + 3 = 5 
 
Step 2: After adding 1 mol A and 1 mol B 
New moles: A = 3, B = 4, total = 7 
 
 
Q3: At 298 K, the mole percentage of N 2(g) in air is 80%. Water is in equilibrium with air at 
a pressure of 10 atm. 
What is the mole fraction of N 2(g) in water at 298 K? 
(K
H
 for N 2 is 6.5 × 10 7 mmHg) 
A: 9.35 × 10 ? 5 
B: 9.35 × 10 5 
C: 1.23 × 10 ? 7 
D: 1.17 × 10 ? 4 
Answer: A 
Explanation: 
P
n
2 = K_H · X
n
2 
P
n
2 = 0.8 × 10 = 8 atm 
8 × 760 = 6.5 × 10 7 × X
n
2 
 
X
n
2 = 9.35 × 10 ? 5 
 
Q4: Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour 
pressures of pure A and B are 55 and 15kNm
-2
 respectively. What is the mole fraction of A 
in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8? 
A: 0.5217 
B: 0.480 
C: 0.340 
D: 0.663  
Answer: A 
Explanation: 
For an ideal solution, Raoult’s law applies: 
 
Total pressure: 
 
 
Q5: A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute ‘A’ 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Solutions  
 
(January 2026) 
 
Q1: Consider the following aqueous solutions: 
I. 2.2 g Glucose in 125 mL of solution 
II. 1.9 g Calcium chloride in 250 mL of solution 
III. 9.0 g Urea in 500 mL of solution 
IV. 20.5 g Aluminium sulphate in 750 mL of solution 
The correct increasing order of boiling point of these solutions will be: 
[Given: Molar mass in g mol ?¹ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and 
S = 32] 
A: III < I < II < IV 
B: II < III < IV < I 
C: II < III < I < IV 
D: I < II < III < IV 
Answer: D 
Explanation: 
?T
b
 = i · k
b
 · m 
For dilute solution (M ˜ m) 
We know that boiling point elevation depends on ?T
b
. 
Since k
b
 is the same for water in all cases, we compare only i × m (or i × M for dilute solutions). 
 
Now compare the values of i × m: 
(I) Glucose: 0.098 × 1 = 0.098 
(II) CaCl 2: 0.068 × 3 = 0.204 
(III) Urea: 0.3 × 1 = 0.3 
(IV) Aluminium sulphate: 0.08 × 5 = 0.4 
Order of ?T
b
 = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So order of BP = Al 2(SO 4) 3 > Urea > CaCl 2 > Glucose 
So the answer will be I < II < III < IV. 
 
Q2: At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of 
the ideal solution formed is 320 mm Hg. 
At this stage, one mole of A and one mole of B are added to the solution. The vapour 
pressure is now measured as 328.6 mm Hg. 
The vapour pressures (in mm Hg) of A and B are respectively: 
A: 400, 300 
B: 600, 400 
C: 500, 200 
D: 300, 200 
Answer: C 
Explanation: 
For an ideal solution, by Raoult’s law: 
 
Step 1: First mixture (2 mol A + 3 mol B) 
Total moles = 2 + 3 = 5 
 
Step 2: After adding 1 mol A and 1 mol B 
New moles: A = 3, B = 4, total = 7 
 
 
Q3: At 298 K, the mole percentage of N 2(g) in air is 80%. Water is in equilibrium with air at 
a pressure of 10 atm. 
What is the mole fraction of N 2(g) in water at 298 K? 
(K
H
 for N 2 is 6.5 × 10 7 mmHg) 
A: 9.35 × 10 ? 5 
B: 9.35 × 10 5 
C: 1.23 × 10 ? 7 
D: 1.17 × 10 ? 4 
Answer: A 
Explanation: 
P
n
2 = K_H · X
n
2 
P
n
2 = 0.8 × 10 = 8 atm 
8 × 760 = 6.5 × 10 7 × X
n
2 
 
X
n
2 = 9.35 × 10 ? 5 
 
Q4: Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour 
pressures of pure A and B are 55 and 15kNm
-2
 respectively. What is the mole fraction of A 
in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8? 
A: 0.5217 
B: 0.480 
C: 0.340 
D: 0.663  
Answer: A 
Explanation: 
For an ideal solution, Raoult’s law applies: 
 
Total pressure: 
 
 
Q5: A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute ‘A’ 
of molar mass 60 g mol ?¹ and 0.9 g of a non-volatile non-electrolyte solute ‘B’ of molar 
mass 180 g mol ?¹ in 100 mL H 2O at 27°C. 
The osmotic pressure of the solution will be: 
[Given: R = 0.082 L atm K ?¹ mol ?¹] 
A: 2.46 atm 
B: 0.82 atm 
C: 1.47 atm 
D: 1.23 atm 
Answer: A 
Explanation:  
Moles of solute A: 
n
A
 = 0.3 / 60 = 0.005 mol 
Moles of solute B: 
n
B
 = 0.9 / 180 = 0.005 mol 
Total moles of solute particles (both are non-electrolytes, so no dissociation): 
n = n
A
 + n
B
 = 0.005 + 0.005 = 0.01 mol 
Volume of solution (dilute solution, take volume ˜ 100 mL = 0.1 L): 
V = 0.1 L 
Temperature: 
T = 27°C = 300 K 
Osmotic pressure: 
p = (n/V) RT = (0.01/0.1) × 0.082 × 300 
p = 0.1 × 24.6 = 2.46 atm 
Correct  A) 2.46 atm 
 
Q6: W g of a non-volatile electrolyte solid solute of molar mass M g mol ?¹, when 
dissolved in 100 mL water, decreases the vapour pressure of water from 640 mm Hg to 
600 mm Hg. 
If the aqueous solution of the electrolyte boils at 375 K and K
b
 for water = 0.52 K kg mol ?¹, 
then the mole fraction of the electrolyte solute (x 2) in the solution can be expressed as: 
(Given: density of water = 1 g/mL and boiling point of water = 373 K) 
A: (16 / 2.6) × (W / M) 
B: (1.3 / 8) × (M / W) 
C: (2.6 / 16) × (M / W) 
D: (1.3 / 8) × (W / M) 
Answer: D 
Explanation: 
Boiling point elevation: 
?T
b
 = 375 - 373 = 2 K 
Mass of water = 100 mL × 1 g mL ?¹ = 100 g = 0.1 kg 
Molality: