Solutions JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Solutions, free to read — no sign-in needed. The full chapter has 27 questions; sign in to attempt the remaining 22 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Passage: Two volatile liquids A and B form an ideal solution. Consider a $5$ molal solution of B in A inside a closed container having a total vapour pressure of $100$ mm Hg at $300$ K. The vapour pressure of pure A at $300$ K is $105$ mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Given: The gas constant $R = 0.08$ L atm K$^{-1}$ mol$^{-1}$ Molar mass of A is $50$ g mol$^{-1}$ Molar mass of B is $57$ g mol$^{-1}$ Density of liquid B at $300$ K is $0.5$ g/mL $1$ atm $= 760$ mm Hg Question: The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.
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   Answer: 0.16

   Molality of B in A is $5$ m, which means $5$ moles of B are dissolved in $1$ kg ($1000$ g) of solvent A. Number of moles of A, $n_A = \dfrac{1000}{50} = 20$ mol Mole fraction of A in the liquid phase, $x_A = \dfrac{n_A}{n_A + n_B} = \dfrac{20}{20 + 5} = 0.8$ Mole fraction of B in the liquid phase, $x_B = 1 - 0.8 = 0.2$ According to Raoult's law, the partial vapour pressure of A is: $P_A = P_A^0 x_A = 105 \times 0.8 = 84$ mm Hg Given total vapour pressure, $P_T = 100$ mm Hg Partial vapour pressure of B, $P_B = P_T - P_A = 100 - 84 = 16$ mm Hg Mole fraction of B in the vapour phase, $y_B = \dfrac{P_B}{P_T} = \dfrac{16}{100} = 0.16$ Answer: $0.16$
2. Q2JEE Advanced Adv 2026 (Paper 2)
   Passage: Two volatile liquids A and B form an ideal solution. Consider a $5$ molal solution of B in A inside a closed container having a total vapour pressure of $100$ mm Hg at $300$ K. The vapour pressure of pure A at $300$ K is $105$ mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Given: The gas constant $R = 0.08$ L atm K$^{-1}$ mol$^{-1}$ Molar mass of A is $50$ g mol$^{-1}$ Molar mass of B is $57$ g mol$^{-1}$ Density of liquid B at $300$ K is $0.5$ g/mL $1$ atm $= 760$ mm Hg Question: At $300$ K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____.
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   Answer: 2000

   For a $5$ molal solution of B in A, there are $5$ moles of B in $1000$ g of A. Moles of A $= \dfrac{1000}{50} = 20$ mol. Mole fraction of B, $x_B = \dfrac{5}{20 + 5} = 0.2$ Mole fraction of A, $x_A = 1 - 0.2 = 0.8$ Using Raoult's law for the total vapour pressure: $P_T = P_A^\circ x_A + P_B^\circ x_B$ $100 = 105 \times 0.8 + P_B^\circ \times 0.2$ $100 = 84 + 0.2 P_B^\circ$ $0.2 P_B^\circ = 16 \Rightarrow P_B^\circ = 80$ mm Hg The molar volume of pure B in the liquid phase ($V_{m,l}$) is: $V_{m,l} = \dfrac{\text{Molar mass of B}}{\text{Density of liquid B}} = \dfrac{57}{0.5} = 114$ mL/mol $= 0.114$ L/mol Assuming pure B behaves as an ideal gas in the vapour phase, its molar volume ($V_{m,v}$) at its vapour pressure $P_B^\circ$ is: $V_{m,v} = \dfrac{RT}{P_B^\circ} = \dfrac{0.08 \times 300}{\left(\dfrac{80}{760}\right)} = \dfrac{24 \times 760}{80} = 228$ L/mol The ratio of the molar volume of pure B in the vapour phase to its molar volume in the liquid phase is: $\text{Ratio} = \dfrac{V_{m,v}}{V_{m,l}} = \dfrac{228}{0.114} = 2000$ Answer: $2000$
3. Q3JEE Advanced Adv 2025 (Paper 2)
   At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ) where h is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is $2.00 \mathrm{~g} \mathrm{dm}^{-3}$, the molar mass of the macromolecule is calculated to be $\boldsymbol{X} \times 10^4 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $\boldsymbol{X}$ is $\qquad$ . Use : Universal gas constant $(\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} \mathrm{~s}^{-2}$
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   Answer: 2.49

   $\begin{aligned} & \pi=\rho g h=10^3 \times 10 \times 2 \times 10^{-2} \text { Pascal }=200 \text { Pascal } \\ & \pi=C R T \\ & 200=\frac{2}{M} \times 1000 \times 8.3 \times 300 \\ & M=24900=2.49 \times 10^4 \mathrm{~g} / \mathrm{mol} \\ & X=2.49\end{aligned}$
4. Q4JEE Advanced Adv 2024 (Paper 2)
   Vessel-1 contains $\mathbf{w}_2 \mathrm{~g}$ of a non-volatile solute $\mathbf{X}$ dissolved in $\mathbf{w}_1 \mathrm{~g}$ of water. Vessel- 2 contains $\mathbf{w}_2 \mathrm{~g}$ of another non-volatile solute $\mathbf{Y}$ dissolved in $\mathbf{w}_1 \mathrm{~g}$ of water. Both the vessels are at the same temperature and pressure. The molar mass of $\mathbf{X}$ is $80 \%$ of that of $\mathbf{Y}$. The van't Hoff factor for $\mathbf{X}$ is 1.2 times of that of $\mathbf{Y}$ for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is ________. $\%$ of the solution in Vessel-2.
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   Answer: 150

   Vessel-I $\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{I}}=\mathrm{i}_{\mathrm{x}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{x}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}$ $\mathrm{M}_{\mathrm{X}}=$ Molar mass of ' X ' Vessel-II $\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}=\mathrm{i}_{\mathrm{Y}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{Y}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}$ $M_Y=$ Molar mass of ' $Y$ ' $\begin{aligned} & \frac{\left(\Delta T_b\right)_{\mathrm{I}}}{\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}} \times 100=\frac{\mathrm{i}_{\mathrm{x}}}{\mathrm{i}_{\mathrm{Y}}} \cdot \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}} \times 100 \\ & =1.2 \times \frac{100}{80} \times 100 \\ & =150 \% \end{aligned}$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   $50mL$ of $0.2$ molal urea solution (density $=1.012g{mL}^{-1}$ at $300K$ ) is mixed with $250mL$ of a solution containing $0.06g$ of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in torr) of the resulting at $300K$ is [Use : Molar mass of urea $=60g{mol}^{-1}$; gas constant, $R=62L-torr{K}^{-1}{mol}^{-1}$; Assume, $\left.{\Delta }_{mix}H=0,{\Delta }_{\text{mix}}V=0\right]$
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   Answer: 682

   Mole of urea $=0.2$ Weight of urea = moles of urea $\times$ molar mass Weight of urea $=0.2\times 60=12g$ Weight of solvent $=1000g$ Weight of solution $=1012g$ Volume of solution $=\dfrac{\text{Weight of solution}}{\text{density of solution}}$ $∴$ Volume of solution $=\dfrac{1012}{1.012}=1000ml$ $∵1000ml$ solution contain $0.2$ mole $∴50ml$ solution contain $=\dfrac{0.2\times 50}{1000}=0.01$ Mole of urea in other solution $=\dfrac{0.06}{60}=0.001$ $∴$ Concentration of solution $=\dfrac{0.01+0.001}{\dfrac{300}{1000}}$ Now, osmotic pressure can be calculated as follows, $∴\pi =CRT$ $=0.0366\times 62\times 300$ $=682$