Solutions JEE Main previous year questions with solutions

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Colligative Properties and Abnormal Molecular Masses
   Which of the following statements are not correct? A. For water, magnitude of $K_b$ is more than the magnitude of $K_f$. B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point. C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers. D. The dimerised form of benzoic acid in benzene is $C_6H_5-\underset{O}{\overset{O}{\|}}C-OH \cdots\cdots O=\underset{OH}{\overset{|}{C}}-C_6H_5$ Choose the correct answer from the options given below:
   1. A.A and B only
   2. B.A and D only
   3. C.A, B and D only
   4. D.A, C and D only
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   Answer: (C)

   For water, the molal elevation constant $K_b = 0.52 \text{ K kg mol}^{-1}$ and the molal depression constant $K_f = 1.86 \text{ K kg mol}^{-1}$. Thus, $K_b \lt K_f$. Statement A is incorrect. For a solution with a given molality $m$ of a non-volatile solute, the elevation in boiling point is $\Delta T_b = K_b m$ and the depression in freezing point is $\Delta T_f = K_f m$. Since $K_b \lt K_f$, it follows that $\Delta T_b \lt \Delta T_f$. Statement B is incorrect. Osmotic pressure measurement is preferred for determining the molar mass of macromolecules such as proteins and polymers because the measurement is done at room temperature and the magnitude of osmotic pressure is significant even for very dilute solutions. Statement C is correct. In benzene, benzoic acid undergoes dimerization through intermolecular hydrogen bonding, forming a cyclic dimer with two hydrogen bonds. The structure given in statement D represents a linear association with only one hydrogen bond. Statement D is incorrect. Therefore, statements A, B, and D are incorrect. Answer: A, B and D only
2. Q2JEE Main 2025 (28 Jan, Shift 2)Solubility and Concentration of Solutions
   Assume a living cell with $0.9 \%(\omega / \omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :
   1. A.show no change in volume since solution is $0.9 \%(\omega / \omega)$
   2. B.shrink since solution is $0.5 \%(\omega / \omega)$
   3. C.swell up since solution is $1 \%(\omega / \omega)$
   4. D.None of these
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   Answer: (D)

   Living cell $=0.9 \mathrm{gm}$ in 100 gm of solution $\% \mathrm{w} / \mathrm{w}=0.9$ Solution is have equal moles of glucose and water $=0.5$ Weight of solution $=0.5 \times 180+0.5 \times 18=99 \mathrm{gm}$ $\% \mathrm{w} / \mathrm{w} \simeq 90 \%$ Concentrated solution = Cell will shrink.
3. Q3JEE Main 2024 (09 Apr, Shift 2)Vapour pressure and raoult's law
   The vapour pressure of pure benzene and methyl benzene at $27^{\circ} \mathrm{C}$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _______ $\times 10^{-2}$ (nearest integer)
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   Answer: 23

   $\begin{aligned} \mathrm{X}_{\text {methylbenzene }} & =0.5 \\ \mathrm{Y}_{\text {methylbenzene }} & =\frac{\mathrm{P}_{\text {methylbenzene }}}{\mathrm{P}_{\text {total }}} \\ \mathrm{Y}_{\text {methylbenzene }} & =\frac{0.5 \times 24}{0.5 \times 80+0.5 \times 24} \\ & =\frac{12}{40+12}=0.23=23 \times 10^{-2}\end{aligned}$
4. Q4JEE Main 2022 (29 Jul, Shift 1)Henry's law
   If ${O}_{2}$ gas is bubbled through water at $303K$, the number of millimoles of ${O}_{2}$ gas that dissolve in $1$ litre of water is____(Nearest integer) (Given : Henry's Law constant for ${O}_{2}$ at $303K$ is $46.82k$ bar and partial pressure of ${O}_{2}=0.920$ bar) (Assume solubility of ${O}_{2}$ in water is too small, nearly negligible)
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   Answer: 1

   $p={K}_{H}\times x$ $0.920=46.82\times {10}^{3}$ bar $\times \dfrac{molof{O}_{2}}{molof{H}_{2}O}$ $0.920=46.82\times {10}^{3}\times \dfrac{molof{O}_{2}}{\dfrac{1000}{18}}$ $0.920=46.82\times 18\times {n}_{{O}_{2}}$ $\dfrac{0.920}{46.82\times 18}={n}_{{O}_{2}}$ $\Rightarrow 1.09\times {10}^{-3}={n}_{{0}_{2}}$ $\Rightarrow mmolof{O}_{2}=1millimole$
5. Q5JEE Main 2021 (27 Aug, Shift 1)Dual Behaviour of Matter and Heisenberg Uncertainty Principle
   $1kg$ of $0.75$ molal aqueous solution of sucrose can be cooled up to $-4^{\circ}C$ before freezing. The amount of ice (in $g$) that will be separated out is ___________. (Nearest integer) [Given : ${K}_{f}\left({H}_{2}O\right)=1.86Kkg{mol}^{-1}$]
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   Answer: 518

   Let mass of water initially present $=xg$ Mass of sucrose $=(1000-x)g$ Number of moles sucrose $=\dfrac{1000-x}{342}$ Molality $\left(M\right)=\dfrac{\dfrac{1000-x}{342}}{x/1000}=0.75$ $2565.5x={10}^{6}-1000x$ $x=795.86g$ no. of moles of sucrose$=0.5969$ ${\Delta T}_{f}={ik}_{f}m$ $4=1\times 1.86\times \dfrac{0.5969}{(795.86-a)g}\times 1000$ Ice separated$=518.3g$