Solutions — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Solutions — colligative properties, relative lowering of vapour pressure, freezing point depression, osmotic pressure and mole fraction — with step-by-step solutions.
Solved JEE Main 2026 questions from the Solutions chapter — covering colligative properties, Raoult’s law, freezing-point depression, osmotic pressure and mole-fraction problems — each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Step 1 — Moles of solute from boiling point elevation.
$$\Delta T_b = T_b - T_b^\circ = 320 - 319.5 = 0.5\ \text{K}.$$$$\Delta T_b = K_b\, m \implies m = \frac{0.5}{0.3} = \frac{5}{3}\ \text{mol kg}^{-1}.$$With solvent mass $= 40\ \text{g} = 0.040\ \text{kg}$:
$$n_{\text{solute}} = m \times 0.040 = \frac{5}{3}\times 0.040 = 0.0667\ \text{mol}.$$Step 2 — Moles of solvent from relative lowering of vapour pressure.
$$\frac{\Delta p}{p^\circ} = x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}.$$$$\frac{760-750}{760} = \frac{10}{760} = \frac{1}{76} = \frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}.$$$$n_{\text{solute}}+n_{\text{solvent}} = 76\,n_{\text{solute}} \implies n_{\text{solvent}} = 75\,n_{\text{solute}}.$$$$n_{\text{solvent}} = 75 \times 0.0667 = 5.0\ \text{mol}.$$Answer: 5
Solution
By Raoult’s law for a non-volatile solute, the vapour pressure of the solution equals the mole fraction of the solvent times the pure-solvent vapour pressure:
$$p_{\text{soln}} = x_{\text{solvent}}\, p^\circ.$$$$x_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}}+n_{\text{solute}}} = \frac{1}{1+0.25} = \frac{1}{1.25} = 0.80.$$$$\frac{p_{\text{soln}}}{p^\circ} = 0.80 = 80\%.$$Answer: D (80%)
Solution
Step 1 — Molality of the acid.
$$n_{\text{acid}} = \frac{19.5}{78} = 0.25\ \text{mol},\qquad m = \frac{0.25}{0.500\ \text{kg}} = 0.5\ \text{mol kg}^{-1}.$$Step 2 — Van’t Hoff factor from freezing point depression.
$$\Delta T_f = i\,K_f\,m \implies i = \frac{\Delta T_f}{K_f\,m} = \frac{1}{1.86 \times 0.5} = \frac{1}{0.93} = 1.0753.$$Step 3 — Degree of dissociation. For a weak monobasic acid $\mathrm{HA \rightleftharpoons H^+ + A^-}$, $i = 1+\alpha$:
$$\alpha = i - 1 = 0.0753.$$Step 4 — Dissociation constant. With $C = m = 0.5\ \text{M}$:
$$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} = \frac{0.5 \times 0.00567}{0.9247} = \frac{0.002835}{0.9247} \approx 3.07\times10^{-3}.$$Answer: D ($3 \times 10^{-3}$)
Solution
Use $\pi = \dfrac{n}{V}RT$.
Solution A: $n_A = \dfrac{1}{50000} = 2\times10^{-5}\ \text{mol}$, $V_A = 0.5\ \text{L}$.
$$x = \frac{2\times10^{-5}}{0.5}(0.083)(300) = 4\times10^{-5}\times 24.9 = 9.96\times10^{-4}\ \text{bar}.$$Solution B: $n_B = \dfrac{2}{50000} = 4\times10^{-5}\ \text{mol}$, $V_B = 1\ \text{L}$.
$$y = \frac{4\times10^{-5}}{1}(0.083)(300) = 4\times10^{-5}\times 24.9 = 9.96\times10^{-4}\ \text{bar}.$$After mixing: total moles $= 6\times10^{-5}$, total volume $= 1.5\ \text{L}$.
$$z = \frac{6\times10^{-5}}{1.5}(0.083)(300) = 4\times10^{-5}\times 24.9 = 9.96\times10^{-4}\ \text{bar}.$$All three osmotic pressures equal $9.96\times10^{-4}$ bar.
Answer: A ($9.96 \times 10^{-4}$; $9.96 \times 10^{-4}$; $9.96 \times 10^{-4}$)
Solution
Take 100 g of solution: 10 g urea and 90 g water. Molar mass of urea $\mathrm{(NH_2)_2CO} = 2(14+2) + 12 + 16 = 60\ \text{g mol}^{-1}$.
$$n_{\text{urea}} = \frac{10}{60} = 0.1667\ \text{mol},\qquad n_{\text{water}} = \frac{90}{18} = 5\ \text{mol}.$$$$x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{urea}}} = \frac{5}{5 + 0.1667} = \frac{5}{5.1667} = 0.967.$$Answer: D ($0.967$)
Solution
Step 1 — Osmotic pressure from the height of the liquid column.
$$\pi = \rho g h = 1000 \times 10 \times (80.0\times10^{-3}) = 800\ \text{Pa} = 0.8\ \text{kPa}.$$Step 2 — Relate to molar mass via $\pi V = \dfrac{w}{M}RT$.
$$M = \frac{wRT}{\pi V} = \frac{20 \times 8.3 \times 300}{0.8 \times 1} = \frac{49800}{0.8} = 62250\ \text{g mol}^{-1}.$$$$M = 62.25\ \text{kg mol}^{-1} \approx 62\ \text{kg mol}^{-1}.$$Answer: 62
Solution
Statement I. In 100 g solution: 30 g methanol, 70 g CCl$_4$. Molar masses: $\mathrm{CH_3OH} = 32$, $\mathrm{CCl_4} = 12 + 4(35.5) = 154\ \text{g mol}^{-1}$.
$$n_{\text{methanol}} = \frac{30}{32} = 0.9375,\qquad n_{\mathrm{CCl_4}} = \frac{70}{154} = 0.4545.$$$$x_{\mathrm{CCl_4}} = \frac{0.4545}{0.9375 + 0.4545} = \frac{0.4545}{1.392} = 0.327 \approx 0.33.\ \checkmark$$Statement II. Pure methanol is strongly H-bonded. Adding non-polar CCl$_4$ breaks these H-bonds, weakening intermolecular attractions, so the escaping tendency (vapour pressure) exceeds the ideal value — a positive deviation from Raoult’s law. $\checkmark$
Both statements are true.
Answer: A (Both Statement I and Statement II are true)
Solution
For water, $K_b = 0.52\ \text{K kg mol}^{-1}$ and $K_f = 1.86\ \text{K kg mol}^{-1}$, so $K_f > K_b$.
- A. Claims $K_b > K_f$. Since $K_f > K_b$, statement A is not correct.
- B. For the same molality, $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$; because $K_f > K_b$, the freezing-point depression is larger, so B (boiling elevation larger) is not correct.
- C. Osmotic pressure is measurable at low concentrations and gives large, precise values, making it the preferred method for the high molar masses of proteins and polymers — correct.
- D. Benzoic acid dimerises in benzene through two O$\cdots$H–O hydrogen bonds between the carboxyl groups; the structure shown is correct.
Not correct: A and B.
Answer: A (A and B only)