JEE Main · Chemistry · Chapter 13

Solutions JEE Main PYQs (2020–2026)

141 questions from Solutions have appeared in JEE Main between 2020 and 2026, including 106 in the last five papers (2022–2026). The most-asked topic is Colligative Properties: Depression of Freezing Point & Osmotic Pressure (48 questions). Practise them all with instant answers and step-by-step solutions.

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141

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Numerical

Solved Solutions PYQs

3 recent questions with answers. The full set of 141 is free to practise.

JEE Main 2026EasyVapour Pressure of Liquid Solutions & Raoult's Law
Q1
When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x %$ of vapor pressure of pure solvent. What is $x %$ ?
- A50%
- B60%
- C70%
- D80%
Show answer & solution
Correct answer: (D)
Let the mole fraction of the solvent be $x_{1}$ and that of the solute be $x_{2}$ .

Given:

Moles of solute $n_{2} = 0.25$

Moles of solvent $n_{1} = 1$

$x_{1} = \frac{n _{1}}{n _{1} + n _{2}} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.8$

$x_{2} = 1 - x_{1} = 1 - 0.8 = 0.2$

For a non-volatile solute, the vapor pressure of the solution is given by:

$P_{solution} = x_{1} P^{0}$

where $P^{0}$ is the vapor pressure of the pure solvent.

$\frac{P _{solution}}{P ^{0}} \times 100 = x_{1} \times 100 = 0.8 \times 100 = 80%$

Final Answer:

$x = 80%$

Option D: 80%
JEE Main 2026MediumColligative Properties: RLVP & Elevation of Boiling Point
Q2
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 : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]
- AIII < I < II < IV
- BII < III < IV < I
- CII < III < I < IV
- DI < II < III < IV
Show answer & solution
Correct answer: (D)
$Δ T_{b} = i \cdot k_{b} \cdot m For dilute solution (M = m)$

We know that boiling point elevation depends on $Δ T_{b}$ .

Since $k_{b}$ is same for water in all cases, we compare only $i \times m$ (or $i \times M$ for dilute solutions).

$Molarity (I) M_{glucose} = \frac{2.2}{180} \times \frac{1000}{125} = 0.098 (II) M_{C a Cl_{2}} = \frac{1.9}{111} \times \frac{1000}{250} = 0.068 (III) M_{urea} = \frac{9}{60} \times \frac{1000}{500} = 0.3 (IV) M_{A l_{2} (SO_{4})_{2}} = \frac{20.5}{342} \times \frac{1000}{750} ≃ 0.08 i \times m 0.098 \times 1 0.068 \times 3 0.3 \times 1 0.08 \times 5$

Now compare the values of $i \times m$ :

Glucose: $0.098 \times 1 = 0.098$

$C a Cl_{2}$ : $0.068 \times 3 = 0.204$

Urea: $0.3 \times 1 = 0.3$

Aluminium sulphate: $0.08 \times 5 = 0.4$

Order of $Δ T_{b} = A l_{2} (SO_{4})_{3} >$ Urea $> C a Cl_{2} >$ Glucose

So order of BP $= A l_{2} (SO_{4})_{3} >$ Urea $> C a Cl_{2} >$ Glucose

So Answer will be I $<$ II $<$ III $<$ IV
JEE Main 2026MediumTypes of Solutions, Solubility & Henry's Law
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 $6.5 \times 1 0^{7} mm H g)$
- A $9.35 \times 1 0^{- 5}$
- B $9.35 \times 1 0^{5}$
- C $1.23 \times 1 0^{- 7}$
- D $1.17 \times 1 0^{- 4}$
Show answer & solution
Correct answer: (A)
$P_{N_{2}} = K_{H} \cdot X_{N_{2}} P_{N_{2}} = 0.8 \times 10 = 8 a t m 8 \times 760 = 6.5 \times 1 0^{7} \times X_{N_{2}} X_{N_{2}} = \frac{8 \times 760}{6.5 \times 1 0 ^{7}} X_{N_{2}} = 9.35 \times 1 0^{- 5}$