JEE Main Previous Year Questions (2025): Solutions

 Page 1


JEE Main Previous Year Questions 
(2025): Solutions 
Q1: If ?? ?? ?? is ???? % ionised in an aqueous solution, then the value of van't Hoff factor 
(i) is ____ × ????
-?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 16 
Solution: 
Percent ionisation of ?? 2
?? = 30% 
The dissociation of ?? 2
?? in aqueous solution can be represented as 
?? 2
?? ? 2?? +
+ ?? 2-
 
1 mole of ?? 2
?? produces 2 moles of ?? +
and 1 mole of ?? 2-
 ions. 
The total number of moles of ions produced ( ?? ) from the dissociation is 
?? = 2 + 1 = 3 
The degree of dissociation given that ?? 2
?? is 30% ionised, the degree of dissociation ( ?? ) or 
degree of ionisation 
?? =
30
100
= 0.3 
Van't Hoff factor (i) can be calculated using the formula, 
?? = 1 + ( ?? - 1) ?? 
Substitute the values of ?? and ?? as, 
?? = 1 + ( 3 - 1)× 0.3 
= 1 + 2 × 0.3 
= 1 + 0.6 
= 1.6 
= 16 × 10
-1
 
Q2: When 1 g each of compounds AB and ????
?? are dissolved in 15 g of water 
separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. 
The atomic mass of ?? (in ?????? ) is ____ × ????
-?? (Nearest integer) 
(Given : Molal boiling point elevation constant is ?? . ?? ?? ???? ?????? -?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 25 
Solution: 
For AB 
?T
b
= 2.7 K
2.7 = 1 × 0.5 × m
 m =
27
5
 
Page 2


JEE Main Previous Year Questions 
(2025): Solutions 
Q1: If ?? ?? ?? is ???? % ionised in an aqueous solution, then the value of van't Hoff factor 
(i) is ____ × ????
-?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 16 
Solution: 
Percent ionisation of ?? 2
?? = 30% 
The dissociation of ?? 2
?? in aqueous solution can be represented as 
?? 2
?? ? 2?? +
+ ?? 2-
 
1 mole of ?? 2
?? produces 2 moles of ?? +
and 1 mole of ?? 2-
 ions. 
The total number of moles of ions produced ( ?? ) from the dissociation is 
?? = 2 + 1 = 3 
The degree of dissociation given that ?? 2
?? is 30% ionised, the degree of dissociation ( ?? ) or 
degree of ionisation 
?? =
30
100
= 0.3 
Van't Hoff factor (i) can be calculated using the formula, 
?? = 1 + ( ?? - 1) ?? 
Substitute the values of ?? and ?? as, 
?? = 1 + ( 3 - 1)× 0.3 
= 1 + 2 × 0.3 
= 1 + 0.6 
= 1.6 
= 16 × 10
-1
 
Q2: When 1 g each of compounds AB and ????
?? are dissolved in 15 g of water 
separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. 
The atomic mass of ?? (in ?????? ) is ____ × ????
-?? (Nearest integer) 
(Given : Molal boiling point elevation constant is ?? . ?? ?? ???? ?????? -?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 25 
Solution: 
For AB 
?T
b
= 2.7 K
2.7 = 1 × 0.5 × m
 m =
27
5
 
Let molar mass of ???? = ?? . 
So 
1/?? 15
× 1000=
27
5
 
?? = 12.34 
For AB
2
 
?T
b
= 1.5 K
1.5 = 1 × 0.5 × m
 m = 3
 
Let molar mass of AB
2
= y 
So 
1/?? 15
× 1000= 3 
?? =
1000
45
?? = 22.22
 
Now let a and b be atomic masses of A and B respectively, then 
?? + ?? = 12.34
?? + 2?? = 22.22
?? = 22.22 - 12.34 = 9. (ii) 
 
Now a = 12.34 - 9.88 = 2.46 
= 24.6 × 10
-1
= 25 × 10
-1
 
Q3: Sea water, which can be considered as a 6 molar ( ???? ) solution of NaCl , has a 
density of ?? ?? ????
-?? . The concentration of dissolved oxygen ( ?? ?? ) in sea water is 5.8 
ppm . Then the concentration of dissolved oxygen ( ?? ?? ) in sea water, is ?? × ????
-?? ?? . 
?? = ____ . (Nearest integer) 
Given: Molar mass of ???????? is ???? . ?? ?? ?????? -?? 
Molar mass of ?? ?? is ???? ?? ?????? -?? 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 2 
Solution: 
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl . 
Page 3


JEE Main Previous Year Questions 
(2025): Solutions 
Q1: If ?? ?? ?? is ???? % ionised in an aqueous solution, then the value of van't Hoff factor 
(i) is ____ × ????
-?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 16 
Solution: 
Percent ionisation of ?? 2
?? = 30% 
The dissociation of ?? 2
?? in aqueous solution can be represented as 
?? 2
?? ? 2?? +
+ ?? 2-
 
1 mole of ?? 2
?? produces 2 moles of ?? +
and 1 mole of ?? 2-
 ions. 
The total number of moles of ions produced ( ?? ) from the dissociation is 
?? = 2 + 1 = 3 
The degree of dissociation given that ?? 2
?? is 30% ionised, the degree of dissociation ( ?? ) or 
degree of ionisation 
?? =
30
100
= 0.3 
Van't Hoff factor (i) can be calculated using the formula, 
?? = 1 + ( ?? - 1) ?? 
Substitute the values of ?? and ?? as, 
?? = 1 + ( 3 - 1)× 0.3 
= 1 + 2 × 0.3 
= 1 + 0.6 
= 1.6 
= 16 × 10
-1
 
Q2: When 1 g each of compounds AB and ????
?? are dissolved in 15 g of water 
separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. 
The atomic mass of ?? (in ?????? ) is ____ × ????
-?? (Nearest integer) 
(Given : Molal boiling point elevation constant is ?? . ?? ?? ???? ?????? -?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 25 
Solution: 
For AB 
?T
b
= 2.7 K
2.7 = 1 × 0.5 × m
 m =
27
5
 
Let molar mass of ???? = ?? . 
So 
1/?? 15
× 1000=
27
5
 
?? = 12.34 
For AB
2
 
?T
b
= 1.5 K
1.5 = 1 × 0.5 × m
 m = 3
 
Let molar mass of AB
2
= y 
So 
1/?? 15
× 1000= 3 
?? =
1000
45
?? = 22.22
 
Now let a and b be atomic masses of A and B respectively, then 
?? + ?? = 12.34
?? + 2?? = 22.22
?? = 22.22 - 12.34 = 9. (ii) 
 
Now a = 12.34 - 9.88 = 2.46 
= 24.6 × 10
-1
= 25 × 10
-1
 
Q3: Sea water, which can be considered as a 6 molar ( ???? ) solution of NaCl , has a 
density of ?? ?? ????
-?? . The concentration of dissolved oxygen ( ?? ?? ) in sea water is 5.8 
ppm . Then the concentration of dissolved oxygen ( ?? ?? ) in sea water, is ?? × ????
-?? ?? . 
?? = ____ . (Nearest integer) 
Given: Molar mass of ???????? is ???? . ?? ?? ?????? -?? 
Molar mass of ?? ?? is ???? ?? ?????? -?? 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 2 
Solution: 
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl . 
 mass of solution = Volume × density 
 = 1000× 2
 mass of solution = 2000 g
 ppm =
 mass of O
2
2000
× 10
6
 mass of O
2
= 5.8 × 2 × 10
-3
 = 1.16 × 10
-2
 g
 molality for O
2
=
1.16 × 10
-2
/32
( 2000- 6 × 58.5)
× 1000
 
=
1.16 × 10
32 × 1649
 
= 0.000219 
= 2.19 × 10
-4
 
Correct answer ? 2 
Q4: Arrange the following solutions in order of their increasing boiling points. 
(i) ????
-?? ?????????? 
(ii) ????
-?? ?? Urea 
(iii) ????
-?? ?????????? 
(iv) ????
-?? ?????????? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ( i ) < ( ii) < ( iii) < ( iv ) 
B. (ii) < (i) < (iii) < (iv) 
C. (iv) < (iii) < (i) < (ii) 
D. (ii) < (i) = (iii) < (iv) 
Ans: B 
Solution: 
Step 1: Identify the van't Hoff factor (i) for each solute 
???????? dissociates (ideally) into two ions: 
NaCl? Na
+
+ Cl
-
, 
so ?? ˜ 2. 
Urea ( CH
4
 N
2
O ) is a non-electrolyte (does not dissociate), so ?? = 1. 
Step 2: Effective molar concentration of particles 
The total particle concentration for each solution is approximately ( ?? × molarity). 
(i) 10
-4
?? NaCl 
Effective concentration = 2 × 10
-4
= 2 × 10
-4
. 
(ii) 10
-4
?? Urea 
Effective concentration = 1 × 10
-4
= 1 × 10
-4
. 
(iii) 10
-3
?? NaCl 
Page 4


JEE Main Previous Year Questions 
(2025): Solutions 
Q1: If ?? ?? ?? is ???? % ionised in an aqueous solution, then the value of van't Hoff factor 
(i) is ____ × ????
-?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 16 
Solution: 
Percent ionisation of ?? 2
?? = 30% 
The dissociation of ?? 2
?? in aqueous solution can be represented as 
?? 2
?? ? 2?? +
+ ?? 2-
 
1 mole of ?? 2
?? produces 2 moles of ?? +
and 1 mole of ?? 2-
 ions. 
The total number of moles of ions produced ( ?? ) from the dissociation is 
?? = 2 + 1 = 3 
The degree of dissociation given that ?? 2
?? is 30% ionised, the degree of dissociation ( ?? ) or 
degree of ionisation 
?? =
30
100
= 0.3 
Van't Hoff factor (i) can be calculated using the formula, 
?? = 1 + ( ?? - 1) ?? 
Substitute the values of ?? and ?? as, 
?? = 1 + ( 3 - 1)× 0.3 
= 1 + 2 × 0.3 
= 1 + 0.6 
= 1.6 
= 16 × 10
-1
 
Q2: When 1 g each of compounds AB and ????
?? are dissolved in 15 g of water 
separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. 
The atomic mass of ?? (in ?????? ) is ____ × ????
-?? (Nearest integer) 
(Given : Molal boiling point elevation constant is ?? . ?? ?? ???? ?????? -?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 25 
Solution: 
For AB 
?T
b
= 2.7 K
2.7 = 1 × 0.5 × m
 m =
27
5
 
Let molar mass of ???? = ?? . 
So 
1/?? 15
× 1000=
27
5
 
?? = 12.34 
For AB
2
 
?T
b
= 1.5 K
1.5 = 1 × 0.5 × m
 m = 3
 
Let molar mass of AB
2
= y 
So 
1/?? 15
× 1000= 3 
?? =
1000
45
?? = 22.22
 
Now let a and b be atomic masses of A and B respectively, then 
?? + ?? = 12.34
?? + 2?? = 22.22
?? = 22.22 - 12.34 = 9. (ii) 
 
Now a = 12.34 - 9.88 = 2.46 
= 24.6 × 10
-1
= 25 × 10
-1
 
Q3: Sea water, which can be considered as a 6 molar ( ???? ) solution of NaCl , has a 
density of ?? ?? ????
-?? . The concentration of dissolved oxygen ( ?? ?? ) in sea water is 5.8 
ppm . Then the concentration of dissolved oxygen ( ?? ?? ) in sea water, is ?? × ????
-?? ?? . 
?? = ____ . (Nearest integer) 
Given: Molar mass of ???????? is ???? . ?? ?? ?????? -?? 
Molar mass of ?? ?? is ???? ?? ?????? -?? 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 2 
Solution: 
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl . 
 mass of solution = Volume × density 
 = 1000× 2
 mass of solution = 2000 g
 ppm =
 mass of O
2
2000
× 10
6
 mass of O
2
= 5.8 × 2 × 10
-3
 = 1.16 × 10
-2
 g
 molality for O
2
=
1.16 × 10
-2
/32
( 2000- 6 × 58.5)
× 1000
 
=
1.16 × 10
32 × 1649
 
= 0.000219 
= 2.19 × 10
-4
 
Correct answer ? 2 
Q4: Arrange the following solutions in order of their increasing boiling points. 
(i) ????
-?? ?????????? 
(ii) ????
-?? ?? Urea 
(iii) ????
-?? ?????????? 
(iv) ????
-?? ?????????? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ( i ) < ( ii) < ( iii) < ( iv ) 
B. (ii) < (i) < (iii) < (iv) 
C. (iv) < (iii) < (i) < (ii) 
D. (ii) < (i) = (iii) < (iv) 
Ans: B 
Solution: 
Step 1: Identify the van't Hoff factor (i) for each solute 
???????? dissociates (ideally) into two ions: 
NaCl? Na
+
+ Cl
-
, 
so ?? ˜ 2. 
Urea ( CH
4
 N
2
O ) is a non-electrolyte (does not dissociate), so ?? = 1. 
Step 2: Effective molar concentration of particles 
The total particle concentration for each solution is approximately ( ?? × molarity). 
(i) 10
-4
?? NaCl 
Effective concentration = 2 × 10
-4
= 2 × 10
-4
. 
(ii) 10
-4
?? Urea 
Effective concentration = 1 × 10
-4
= 1 × 10
-4
. 
(iii) 10
-3
?? NaCl 
Effective concentration = 2 × 10
-3
= 2 × 10
-3
. 
(iv) 10
-2
?? NaCl 
Effective concentration = 2 × 10
-2
= 2 × 10
-2
. 
Step 3: Compare to rank the boiling points 
A larger total particle concentration (and hence larger colligative effect) corresponds to a higher 
boiling point. Arrange from lowest to highest: 
Lowest: 10
-4
?? Urea [1 × 10
-4
] 
Next: 10
-4
?? NaCl [2 × 10
-4
] 
Next: 10
-3
?? NaCl [2 × 10
-3
] 
Highest: 10
-2
?? NaCl [2 × 10
-2
] 
Hence, in the format (ii) < (i) < (iii) < (iv). 
Final Answer 
(ii) < (i) < (iii) < (iv) (Option B) 
Q5: Consider a binary solution of two volatile liquid components 1 and ?? . ?? ?? and ?? ?? 
are the mole fractions of component 1 in liquid and vapour phase, respectively. The 
slope and intercept of the linear plot of 
?? ?? ?? vs 
?? ?? ?? are given respectively as : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 
P
1
0
P
2
0
,
P
1
0
-P
2
0
P
2
0
 
B. 
P
2
0
P
1
0
,
P
2
0
-P
1
0
P
2
0
 
C. 
P
2
0
P
1
0
,
P
1
0
-P
2
0
P
2
0
 
D. 
P
1
0
P
2
0
,
P
2
0
-P
1
0
P
2
0
 
Ans: D 
Solution: 
For a binary solution of two volatile liquid components labeled 1 and 2 , let ?? 1
 and ?? 1
 represent 
the mole fractions of component 1 in the liquid and vapor phases, respectively. The linear 
relationship between the inverse of these mole fractions is plotted as 
1
?? 1
 versus 
1
?? 1
. 
To derive the slope and intercept of this linear plot, consider the following calculations: 
Using Raoult's Law for a Liquid Solution: 
For a liquid solution with volatile components 1 and 2 : 
P
1
= P
T
· ?? 1
= P
1
?? · ?? 1
 
Therefore: 
Page 5


JEE Main Previous Year Questions 
(2025): Solutions 
Q1: If ?? ?? ?? is ???? % ionised in an aqueous solution, then the value of van't Hoff factor 
(i) is ____ × ????
-?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 16 
Solution: 
Percent ionisation of ?? 2
?? = 30% 
The dissociation of ?? 2
?? in aqueous solution can be represented as 
?? 2
?? ? 2?? +
+ ?? 2-
 
1 mole of ?? 2
?? produces 2 moles of ?? +
and 1 mole of ?? 2-
 ions. 
The total number of moles of ions produced ( ?? ) from the dissociation is 
?? = 2 + 1 = 3 
The degree of dissociation given that ?? 2
?? is 30% ionised, the degree of dissociation ( ?? ) or 
degree of ionisation 
?? =
30
100
= 0.3 
Van't Hoff factor (i) can be calculated using the formula, 
?? = 1 + ( ?? - 1) ?? 
Substitute the values of ?? and ?? as, 
?? = 1 + ( 3 - 1)× 0.3 
= 1 + 2 × 0.3 
= 1 + 0.6 
= 1.6 
= 16 × 10
-1
 
Q2: When 1 g each of compounds AB and ????
?? are dissolved in 15 g of water 
separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. 
The atomic mass of ?? (in ?????? ) is ____ × ????
-?? (Nearest integer) 
(Given : Molal boiling point elevation constant is ?? . ?? ?? ???? ?????? -?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 25 
Solution: 
For AB 
?T
b
= 2.7 K
2.7 = 1 × 0.5 × m
 m =
27
5
 
Let molar mass of ???? = ?? . 
So 
1/?? 15
× 1000=
27
5
 
?? = 12.34 
For AB
2
 
?T
b
= 1.5 K
1.5 = 1 × 0.5 × m
 m = 3
 
Let molar mass of AB
2
= y 
So 
1/?? 15
× 1000= 3 
?? =
1000
45
?? = 22.22
 
Now let a and b be atomic masses of A and B respectively, then 
?? + ?? = 12.34
?? + 2?? = 22.22
?? = 22.22 - 12.34 = 9. (ii) 
 
Now a = 12.34 - 9.88 = 2.46 
= 24.6 × 10
-1
= 25 × 10
-1
 
Q3: Sea water, which can be considered as a 6 molar ( ???? ) solution of NaCl , has a 
density of ?? ?? ????
-?? . The concentration of dissolved oxygen ( ?? ?? ) in sea water is 5.8 
ppm . Then the concentration of dissolved oxygen ( ?? ?? ) in sea water, is ?? × ????
-?? ?? . 
?? = ____ . (Nearest integer) 
Given: Molar mass of ???????? is ???? . ?? ?? ?????? -?? 
Molar mass of ?? ?? is ???? ?? ?????? -?? 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 2 
Solution: 
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl . 
 mass of solution = Volume × density 
 = 1000× 2
 mass of solution = 2000 g
 ppm =
 mass of O
2
2000
× 10
6
 mass of O
2
= 5.8 × 2 × 10
-3
 = 1.16 × 10
-2
 g
 molality for O
2
=
1.16 × 10
-2
/32
( 2000- 6 × 58.5)
× 1000
 
=
1.16 × 10
32 × 1649
 
= 0.000219 
= 2.19 × 10
-4
 
Correct answer ? 2 
Q4: Arrange the following solutions in order of their increasing boiling points. 
(i) ????
-?? ?????????? 
(ii) ????
-?? ?? Urea 
(iii) ????
-?? ?????????? 
(iv) ????
-?? ?????????? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ( i ) < ( ii) < ( iii) < ( iv ) 
B. (ii) < (i) < (iii) < (iv) 
C. (iv) < (iii) < (i) < (ii) 
D. (ii) < (i) = (iii) < (iv) 
Ans: B 
Solution: 
Step 1: Identify the van't Hoff factor (i) for each solute 
???????? dissociates (ideally) into two ions: 
NaCl? Na
+
+ Cl
-
, 
so ?? ˜ 2. 
Urea ( CH
4
 N
2
O ) is a non-electrolyte (does not dissociate), so ?? = 1. 
Step 2: Effective molar concentration of particles 
The total particle concentration for each solution is approximately ( ?? × molarity). 
(i) 10
-4
?? NaCl 
Effective concentration = 2 × 10
-4
= 2 × 10
-4
. 
(ii) 10
-4
?? Urea 
Effective concentration = 1 × 10
-4
= 1 × 10
-4
. 
(iii) 10
-3
?? NaCl 
Effective concentration = 2 × 10
-3
= 2 × 10
-3
. 
(iv) 10
-2
?? NaCl 
Effective concentration = 2 × 10
-2
= 2 × 10
-2
. 
Step 3: Compare to rank the boiling points 
A larger total particle concentration (and hence larger colligative effect) corresponds to a higher 
boiling point. Arrange from lowest to highest: 
Lowest: 10
-4
?? Urea [1 × 10
-4
] 
Next: 10
-4
?? NaCl [2 × 10
-4
] 
Next: 10
-3
?? NaCl [2 × 10
-3
] 
Highest: 10
-2
?? NaCl [2 × 10
-2
] 
Hence, in the format (ii) < (i) < (iii) < (iv). 
Final Answer 
(ii) < (i) < (iii) < (iv) (Option B) 
Q5: Consider a binary solution of two volatile liquid components 1 and ?? . ?? ?? and ?? ?? 
are the mole fractions of component 1 in liquid and vapour phase, respectively. The 
slope and intercept of the linear plot of 
?? ?? ?? vs 
?? ?? ?? are given respectively as : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 
P
1
0
P
2
0
,
P
1
0
-P
2
0
P
2
0
 
B. 
P
2
0
P
1
0
,
P
2
0
-P
1
0
P
2
0
 
C. 
P
2
0
P
1
0
,
P
1
0
-P
2
0
P
2
0
 
D. 
P
1
0
P
2
0
,
P
2
0
-P
1
0
P
2
0
 
Ans: D 
Solution: 
For a binary solution of two volatile liquid components labeled 1 and 2 , let ?? 1
 and ?? 1
 represent 
the mole fractions of component 1 in the liquid and vapor phases, respectively. The linear 
relationship between the inverse of these mole fractions is plotted as 
1
?? 1
 versus 
1
?? 1
. 
To derive the slope and intercept of this linear plot, consider the following calculations: 
Using Raoult's Law for a Liquid Solution: 
For a liquid solution with volatile components 1 and 2 : 
P
1
= P
T
· ?? 1
= P
1
?? · ?? 1
 
Therefore: 
P
T
?? 1
=
P
1
o
?? 1
 
Rearranging the Equation: 
By substituting and rearranging, we have: 
P
2
o
+ ?? 1
( P
1
o
- P
2
a
)
?? 1
=
P
1
o
?? 1
 
Simplifying further: 
P
2
o
?? 1
+ ( P
1
o
- P
2
o
) =
P
1
o
?? 1
 
Expressing 
1
?? 1
 : 
Solving for 
1
?? 1
, we obtain: 
1
?? 1
= (
P
1
o
P
2
o
) (
1
?? 1
) + (
P
2
o
- P
1
o
P
2
o
) 
Determining the Slope and Intercept: 
The slope of the line is: 
Slope =
P
1
°
P
2
°
 
The intercept of the line is: 
Intercept =
P
2
o
-P
1
o
P
2
o
 
In summary, for the plot of 
1
?? 1
 against 
1
?? 1
, the slope is 
P
1
o
P
2
o
 and the intercept is 
P
2
o
-P
1
o
P
2
o
. 
Q6: When a non-volatile solute is added to the solvent, the vapour pressure of the 
solvent decreases by ???? mm of Hg . The mole fraction of the solute in the solution is 
0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 
20 mm of Hg ? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 0.2 
B. 0.4 
C. 0.8 
D. 0.6 
Ans: D 
Solution: 
When a non-volatile solute is added to a solvent, it causes the vapour pressure of the solvent to 
decrease. In this scenario, when the vapour pressure decreases by 10 mm of Hg , the mole 
fraction of the solute in the solution is 0.2 .