Match List-I with List-II

List-I	List-II	(A) Solution of chloroform and acetone	(I) Minimum boiling azeotrope	(B) Solution of ethanol and water	(II) Dimerizes	(C) Solution of benzene and toluene	(III) Maximum boiling azeotrope	(D) Solution of acetic acid in benzene	(IV) Δ Vmix=0$\Delta V{\text{mix}}=0$ Choose the correct answer from the options given below: A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)$\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}$ B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)$\text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}$ C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)$\text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}$ D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)$\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}$ Solution & Explanation Related Formula Negative Deviation from Raoult's Law Maximum Boiling Azeotrope$$\text{Negative Deviation from Raoult's Law} \implies \text{Maximum Boiling Azeotrope} $$ Positive Deviation from Raoult's Law Minimum Boiling Azeotrope$$\text{Positive Deviation from Raoult's Law} \implies \text{Minimum Boiling Azeotrope} $$ Ideal Solution Δ Vmix = 0$$\text{Ideal Solution} \implies \Delta V{\text{mix}} = 0 $$ Core Logic Evaluating molecular interaction behaviors: - (A) Chloroform and Acetone: Form intermolecular hydrogen bonds, showing negative deviation from Raoult's law, which forms a maximum boiling azeotrope arrow$\rightarrow$ (III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope arrow$\rightarrow$ (I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where Δ Vmix = 0$\Delta V{\text{mix}} = 0$ arrow$\rightarrow$ (IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize arrow$\rightarrow$ (II) Step 1: Alignment Selection Combining the pairs gives: (A)-(III), (B)-(I), (C)-(IV), (D)-(II). Pattern Recognition Liquid properties shortcut: Acetic acid in benzene is a classic textbook example of dimerization (i=0.5$i=0.5$). Benzene-toluene is the most famous ideal solution pair. Recognizing either instantly shortcuts the options list to the correct key. Chapter Mix Class 12 Chemistry: Solutions Reference Study Guides Solutions Keywords: #maximum boiling azeotrope example #JEE Main 2025 Evening Q38 #acetic acid dimerization benzene #ideal solutions delta V mix More Solutions Previous-Year Questions — Page 2 Q51 jee_main_2026_22_january_evening Mole Fraction and Concentration Terms At T(K)$T(\text{K})$, 100 g$100\text{ g}$ of 98% H₂SO₄ (w/w)$98\% \text{ H}_2\text{SO}_4\text{ (w/w)}$ aqueous solution is mixed with 100 g$100\text{ g}$ of 49% H₂SO₄ (w/w)$49\% \text{ H}_2\text{SO}_4\text{ (w/w)}$ aqueous solution. What is the mole fraction of H₂SO₄$\text{H}_2\text{SO}_4$ in the resultant solution? (Given: Atomic mass H = 1 u$\text{H} = 1\text{ u}$; S = 32 u$\text{S} = 32\text{ u}$; O = 16 u$\text{O} = 16\text{ u}$) (Assume that temperature after mixing remains constant) A. 0.9$0.9$ B. 0.1$0.1$ C. 0.337$0.337$ D. 0.663$0.663$ Solution Related Formula Mass of solute = Total mass of solution × Percentage (w/w)100$$\text{Mass of solute} = \text{Total mass of solution} \times \frac{\text{Percentage (w/w)}}{100}$$ Mole fraction (xA) = (nA)/(nA + nB)$$\text{Mole fraction } (x_A) = \frac{n_A}{n_A + n_B}$$ where, nA$n_A$ = moles of solute (H₂SO₄$\text{H}_2\text{SO}_4$) nB$n_B$ = moles of solvent (H₂O$\text{H}_2\text{O}$) Core Logic Step 1: Calculate total weight of H₂SO₄$\text{H}_2\text{SO}_4$: Weight = (100 × (98)/(100)) + (100 × (49)/(100)) = 98 + 49 = 147 g$$\text{Weight} = \left(100 \times \frac{98}{100}\right) + \left(100 \times \frac{49}{100}\right) = 98 + 49 = 147\text{ g}$$ Step 2: Calculate total weight of H₂O$\text{H}_2\text{O}$: Total weight of solution = 100 + 100 = 200 g$$\text{Total weight of solution} = 100 + 100 = 200\text{ g}$$ Weight of H₂O = 200 - 147 = 53 g$$\text{Weight of } \text{H}_2\text{O} = 200 - 147 = 53\text{ g}$$ Step 3: Calculate moles and mole fraction: nH₂SO₄ = (147)/(98) = 1.5$$n_{\text{H}_2\text{SO}_4} = \frac{147}{98} = 1.5$$ nH₂O = (53)/(18) ≈ 2.944$$n_{\text{H}_2\text{O}} = \frac{53}{18} \approx 2.944$$ xH₂SO₄ = (1.5)/(1.5 + 2.944) = 0.337$$x_{\text{H}_2\text{SO}_4} = \frac{1.5}{1.5 + 2.944} = 0.337$$ Pattern Recognition Sees: Mixing two solution concentrations (w/w). Shortcut: Sum the component masses directly to find total solute mass and total solvent mass, then apply standard mole fraction formula. Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry Q52 jee_main_2026_23_january_morning Non-Ideal Solutions and Raoult's Law Which one of the following graphs accurately represents the plot of partial pressure of CS₂$CS_{2}$ vs its mole fraction in a mixture of acetone and CS₂$CS_{2}$ at constant temperature? A. (1)$(1)$ B. (2)$(2)$ C. (3)$(3)$ D. (4)$(4)$ Solution Related Formula PCS₂ > PCS₂^° · XCS₂$$P_{CS_2} > P_{CS_2}^\circ \cdot X_{CS_2}$$ Core Logic A mixture of carbon disulfide (CS₂$CS_2$) and acetone (CH₃-CO-CH₃$CH_3-CO-CH_3$) shows a positive deviation from Raoult's Law. This happens because the dipole-dipole interactions between acetone and CS₂$CS_2$ are weaker than the attractive interactions between the pure molecules themselves. Step 1: Graph Interpretation Due to positive deviation, the actual partial pressure curve will bulge upwards (convex) relative to the ideal straight line (Raoult's Law). The first graph precisely depicts this upward curve starting from zero and reaching P^°CS₂$P^\circ_{CS_2}$. Pattern Recognition Acetone + Carbon disulfide = Positive Deviation. Positive deviation means vapor pressure is higher than expected, so the curve sags upwards above the straight-line prediction. Chapter Mix Class 12 Chemistry: Solutions Q72 jee_main_2026_23_january_evening Raoult's Law Two liquids A and B form an ideal solution. At 320 K$320 \text{ K}$, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg$500 \text{ mm Hg}$. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg$20 \text{ mm Hg}$. Vapour pressure (in mm Hg$\text{mm Hg}$) of B in the pure state is ____. (Nearest integer) Numerical Answer. Answer: 200 to 200 Solution Related Formula PTotal = PA° XA + PB° XB$$P_{\text{Total}} = P_A^{\circ} X_A + P_B^{\circ} X_B$$ Core Logic Case 1: Moles of A (nA$n_A$) = 3, Moles of B (nB$n_B$) = 1. Total moles = 3 + 1 = 4$3 + 1 = 4$. Mole fraction of A (XA$X_A$) = (3)/(4)$\frac{3}{4}$, Mole fraction of B (XB$X_B$) = (1)/(4)$\frac{1}{4}$. Vapour pressure (Pₛ$P_{\text{s}}$) = 500 mm Hg$500 \text{ mm Hg}$. Using Raoult's law: 500 = PA°((3)/(4)) + PB°((1)/(4))$$500 = P_A^{\circ}\left(\frac{3}{4}\right) + P_B^{\circ}\left(\frac{1}{4}\right)$$ 2000 = 3PA° + PB° --- (Equation 1)$$2000 = 3P_A^{\circ} + P_B^{\circ} \quad \text{--- (Equation 1)}$$ Step 1: Setting up Case 2 Case 2: 1 mol of A is further added. So, new moles of A (nA$n_A^{\prime}$) = 4. Moles of B remains 1. Total moles = 4 + 1 = 5$4 + 1 = 5$. Mole fraction of A (XA$X_A^{\prime}$) = (4)/(5)$\frac{4}{5}$, Mole fraction of B (XB$X_B^{\prime}$) = (1)/(5)$\frac{1}{5}$. Vapour pressure increases by 20 mm Hg$20 \text{ mm Hg}$, so new Pₛ = 500 + 20 = 520 mm Hg$P_{\text{s}} = 500 + 20 = 520 \text{ mm Hg}$. Using Raoult's law: 520 = PA°((4)/(5)) + PB°((1)/(5))$$520 = P_A^{\circ}\left(\frac{4}{5}\right) + P_B^{\circ}\left(\frac{1}{5}\right)$$ 2600 = 4PA° + PB° --- (Equation 2)$$2600 = 4P_A^{\circ} + P_B^{\circ} \quad \text{--- (Equation 2)}$$ Step 2: Solving for Pure Vapour Pressures Subtract Equation 1 from Equation 2: (4PA° + PB°) - (3PA° + PB°) = 2600 - 2000$$(4P_A^{\circ} + P_B^{\circ}) - (3P_A^{\circ} + P_B^{\circ}) = 2600 - 2000$$ PA° = 600 mm Hg$$P_A^{\circ} = 600 \text{ mm Hg}$$ Substitute PA°$P_A^{\circ}$ back into Equation 1: 3(600) + PB° = 2000$$3(600) + P_B^{\circ} = 2000$$ 1800 + PB° = 2000$$1800 + P_B^{\circ} = 2000$$ PB° = 200 mm Hg$$P_B^{\circ} = 200 \text{ mm Hg}$$ Pattern Recognition A standard two-equation Raoult's law system. Cross-multiply out the denominators immediately to get simple linear equations (Ax + By = C$Ax + By = C$) which allow straightforward elimination. Chapter Mix Class 12 Chemistry: Solutions Q55 jee_main_2026_24_january_morning Osmotic Pressure A solution is prepared by dissolving 0.3 g$0.3 \text{ g}$ of a non-volatile non-electrolyte solute 'A' of molar mass 60 g mol⁻¹$60 \text{ g mol}^{-1}$ and 0.9 g$0.9 \text{ g}$ of a non-volatile non-electrolyte solute 'B' of molar mass 180 g mol⁻¹$180 \text{ g mol}^{-1}$ in 100 mL$100 \text{ mL}$ H₂O$\mathrm{H}_2\mathrm{O}$ at 27°C$27^{\circ}\text{C}$. Osmotic pressure of the solution will be [Given: R = 0.082 L atm K⁻¹ mol⁻¹$R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$] A. 1.23 atm$1.23 \text{ atm}$ B. 2.46 atm$2.46 \text{ atm}$ C. 0.82 atm$0.82 \text{ atm}$ D. 1.47 atm$1.47 \text{ atm}$ Solution Related Formula π = Ctotal RT$$\pi = C_{\text{total}} RT$$ where, Ctotal = nA + nBVsolution (L)$C_{\text{total}} = \frac{n_A + n_B}{V_{\text{solution (L)}}}$ Core Logic Mass of solute 'A' = 0.3 g$0.3 \text{ g}$ Moles of solute 'A' = 0.3 g60 g/mol = (1)/(200) mol$\frac{0.3 \text{ g}}{60 \text{ g/mol}} = \frac{1}{200} \text{ mol}$ Mass of solute 'B' = 0.9 g$0.9 \text{ g}$ Moles of solute 'B' = 0.9 gm180 g/mol = (1)/(200) mol$\frac{0.9 \text{ gm}}{180 \text{ g/mol}} = \frac{1}{200} \text{ mol}$ Total moles = (1)/(200) + (1)/(200) = (2)/(200) mol$\frac{1}{200} + \frac{1}{200} = \frac{2}{200} \text{ mol}$ Volume of solution = 100 mL = 0.1 L$100 \text{ mL} = 0.1 \text{ L}$ Total molarity (C$C$) of all solutes: C = (2/200)/(100) × 1000 = (1)/(10) M$$C = \frac{2/200}{100} \times 1000 = \frac{1}{10} \text{ M}$$ Step 1: Calculate Osmotic Pressure π = C RT$\pi = C RT$ π = (1)/(10) × 0.082 × 300$$\pi = \frac{1}{10} \times 0.082 \times 300$$ π = 2.46 atm$$\pi = 2.46 \text{ atm}$$ Pattern Recognition For non-electrolytes, i=1$i=1$. Add the moles of all solutes present to find the total effective molarity. Use T = 300 K$T = 300 \text{ K}$ for 27°C$27^{\circ}\text{C}$. Chapter Mix Class 12 Chemistry: Solutions Q62 jee_main_2026_24_january_morning Colligative Properties 'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g mol⁻¹$g \text{ mol}^{-1}$ when dissolved in 100 mL$100 \text{ mL}$ water, decreases vapour pressure of water from 640 mm Hg$640 \text{ mm Hg}$ to 600 mm Hg$600 \text{ mm Hg}$. If aqueous solution of the electrolyte boils at 375 K$375 \text{ K}$ and Kb$K_b$ for water is 0.52 K kg mol⁻¹$0.52 \text{ K kg mol}^{-1}$, then the mole fraction of the electrolyte solute (X₂)$(X_2)$ in the solution can be expressed as (Given : density of water = 1 g/mL$1 \text{ g/mL}$ and boiling point of water = 373 K$373 \text{ K}$) A. (1.3)/(8) × (W)/(M)$\frac{1.3}{8} \times \frac{W}{M}$ B. (16)/(2.6) × (W)/(M)$\frac{16}{2.6} \times \frac{W}{M}$ C. (2.6)/(16) × (M)/(W)$\frac{2.6}{16} \times \frac{M}{W}$ D. (1.3)/(8) × (M)/(W)$\frac{1.3}{8} \times \frac{M}{W}$ Solution Related Formula Δ PP° = i · Xsolute$$\frac{\Delta P}{P^{\circ}} = i \cdot X_{\text{solute}}$$ Δ Tb = i · Kb · m$$\Delta T_b = i \cdot K_b \cdot m$$ Core Logic From Relative Lowering of Vapour Pressure: P° = 640 mm Hg$P^{\circ} = 640 \text{ mm Hg}$ Pₛ = 600 mm Hg$P_s = 600 \text{ mm Hg}$ Δ P = 40 mm Hg$\Delta P = 40 \text{ mm Hg}$ Moles of solute n = (W)/(M)$n = \frac{W}{M}$ Mole fraction Xsolute = Δ PP° · (1)/(i)$X_{\text{solute}} = \frac{\Delta P}{P^{\circ}} \cdot \frac{1}{i}$ (for dilute solutions, or properly i · Xsolute = Δ PP°$i \cdot X_{\text{solute}} = \frac{\Delta P}{P^{\circ}}$ as given in the pdf approach) Δ PP° = i · Xsolute Xsolute = (40)/(640) × (1)/(i)$\frac{\Delta P}{P^{\circ}} = i \cdot X_{\text{solute}} \implies X_{\text{solute}} = \frac{40}{640} \times \frac{1}{i}$ Now, from Boiling Point Elevation: Δ Tb = 375 - 373 = 2 K$\Delta T_b = 375 - 373 = 2 \text{ K}$ m = moles of solutemass of solvent in kg = (W/M)/(100/1000) = (W)/(M) × 10$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W/M}{100/1000} = \frac{W}{M} \times 10$ Δ Tb = i × Kb × m$\Delta T_b = i \times K_b \times m$ 2 = i × 0.52 × ( (W/M)/(100) × 1000 )$2 = i \times 0.52 \times \left( \frac{W/M}{100} \times 1000 \right)$ i = (2)/(5.2) × (M)/(W)$i = \frac{2}{5.2} \times \frac{M}{W}$ Step 1: Calculate Mole Fraction Substitute the value of i$i$ into the Xsolute$X_{\text{solute}}$ equation: Xsolute = (40)/(640) × (1)/(i)$$X_{\text{solute}} = \frac{40}{640} \times \frac{1}{i}$$ Xsolute = (1)/(16) × (5.2)/(2) × (W)/(M)$$X_{\text{solute}} = \frac{1}{16} \times \frac{5.2}{2} \times \frac{W}{M}$$ Xsolute = (2.6)/(16) × (W)/(M) = (1.3)/(8) × (W)/(M)$$X_{\text{solute}} = \frac{2.6}{16} \times \frac{W}{M} = \frac{1.3}{8} \times \frac{W}{M}$$ Pattern Recognition When an unknown electrolyte is involved, isolate the van't Hoff factor (i$i$) from one colligative property equation and substitute it into the other to eliminate i$i$ and solve for the target variable. Chapter Mix Class 12 Chemistry: Solutions More Solutions Questions — jee_main_2025_07_april_evening Practice all Solutions previous-year questions → XY is the membrane / partition between two chambers 1 A solution is made by mixing one mole of volatile Given below are two statements: Statement (I): NaCl is added Given below are two statements : Statement (I): Molal depression Which of the following binary mixtures does not show the For the weak acid dissociation equation: The freezing point depression Liquid and form an ideal solution. The vapour pressure of 'x' g of NaCl is added to water in a What is the freezing point depression constant of a solvent, Assume a living cell with of glucose solution (aqueous). This Sea water, which can be considered as a 6 molar The observed and normal masses of compound are 65.6 and When each of compounds AB and are dissolved in of The percentage dissociation of a salt solution at given temperature At , of aqueous solution is mixed with of aqueous Which one of the following graphs accurately represents the plot At T(K), moles of liquid A and moles of liquid Two liquids A and B form an ideal solution at A solution is prepared by dissolving of a non-volatile non-electrolyte Consider a solution of dissolved in water in a closed Elements P and Q form two types of non-volatile, non-ionizable 'W' g of a non-volatile electrolyte solid solute of molar Identify the mixture that shows positive deviations from Raoult's Law Consider the following aqueous solutions. I. Glucose in of solution. A solution of two miscible liquids showing negative deviation from At 298 K, the mole percentage of in air is Given below are two statements: Statement I: The Henry's law If a substance 'A' dissolves in solution of a mixture The osmotic pressure of a living cell is 12 atm Two liquids A and B form an ideal solution. At A substance 'X' (1.5 g) dissolved in 150 g of We have three aqueous solutions of labelled as 'A', 'B' What happens to freezing point of benzene when small quantity Identify the correct statements : (A) The molality of 2.5 The solution from the following with highest depression in freezing Mixture of chloroform and acetone forms a solution with negative A solution of is by mass and has a density The osmotic pressure of a dilute solution is at . The molarity of orthophosphoric acid ( ) having purity by The mass of sodium acetate required to prepare of aqueous SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Login Sign Up Dashboard Rankbit System Revise Mistakes Rankbit JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) Report Issue Question Issue Type Wrong Answer Wrong Solution Formula Error Typo / Language Issue Diagram Issue Option Mismatch Duplicate Question Out of Syllabus Other Optional Notes Cancel Submit Report Quick Review Question Accuracy Feedback Accurate Needs Fix Difficulty Feedback Easy Medium Hard Confusion Score Low High Optional Notes Cancel Submit Review Welcome back Continue with Google or Sign In Don't have an account? 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