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 3 Q54 jee_main_2026_24_january_evening Vapour Pressure of Liquid Solutions Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15kNm⁻²$15\mathrm{kNm}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8? A. 0.5217 B. 0.480 C. 0.663 D. 0.340 Solution Related Formula YAYB = PA⁰PB⁰ · XAXB$$\frac{\mathrm{Y}_A}{\mathrm{Y}_B} = \frac{\mathrm{P}_A^0}{\mathrm{P}_B^0} \cdot \frac{\mathrm{X}_A}{\mathrm{X}_B}$$ Where Y$Y$ indicates mole fraction in vapour phase and X$X$ indicates mole fraction in liquid phase. Core Logic Given data: PA⁰ = 55 ~kNm⁻²$P_A^0 = 55 \mathrm{~kNm}^{-2}$ PB⁰ = 15 ~kNm⁻²$P_B^0 = 15 \mathrm{~kNm}^{-2}$ YA = 0.8 YB = 1 - 0.8 = 0.2$Y_A = 0.8 \implies Y_B = 1 - 0.8 = 0.2$ Applying the ratio form of Raoult's and Dalton's laws: (0.8)/(0.2) = (55)/(15) × XAXB$$\frac{0.8}{0.2} = \frac{55}{15} \times \frac{\mathrm{X}_A}{\mathrm{X}_B}$$ 4 = (11)/(3) × XAXB$$4 = \frac{11}{3} \times \frac{\mathrm{X}_A}{\mathrm{X}_B}$$ XAXB = (12)/(11)$$\frac{\mathrm{X}_A}{\mathrm{X}_B} = \frac{12}{11}$$ Step 1: Calculate Mole Fraction in Liquid Since XA + XB = 1$X_A + X_B = 1$, we can express XA$X_A$ as: XA = (XA / XB)/(1 + XA / XB)$$X_A = \frac{X_A / X_B}{1 + X_A / X_B}$$ XA = (12/11)/(1 + 12/11) = (12)/(23)$$X_A = \frac{12/11}{1 + 12/11} = \frac{12}{23}$$ XA 0.5217$$X_A \simeq 0.5217$$ Pattern Recognition When dealing with equilibrium between liquid and vapour states, taking the ratio (YA)/(YB) = (PA)/(PB)$\frac{Y_A}{Y_B} = \frac{P_A}{P_B}$ bypasses finding the total pressure Ptotal$P_{\text{total}}$ directly and simplifies fraction algebra. Chapter Mix Class 12 Chemistry: Solutions Q68 jee_main_2026_24_january_evening Henry's Law At 298 K, the mole percentage of N₂(g)$N_{2}(g)$ in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N₂(g)$N_{2}(g)$ in water at 298 K? ( KH$K_{H}$ for N₂$N_{2}$ is 6.5 × 10⁷$6.5 \times 10^{7}$ mm Hg) A. 1.23 × 10⁻⁷$1.23 \times 10^{-7}$ B. 1.17 × 10⁻⁴$1.17 \times 10^{-4}$ C. 9.35 × 10⁵$9.35 \times 10^{5}$ D. 9.35 × 10⁻⁵$9.35 \times 10^{-5}$ Solution Related Formula Pgas = KH · Xgas$$P_{\mathrm{gas}} = K_H \cdot X_{\mathrm{gas}}$$ Core Logic Given data: Total pressure of air = 10 atm Mole percentage of N₂$N_2$ in air = 80% PN₂ = (mole fraction in air) × Ptotal = 0.8 × 10 = 8 atm$P_{\mathrm{N}_2} = (\text{mole fraction in air}) \times P_{\text{total}} = 0.8 \times 10 = 8 \text{ atm}$ Convert partial pressure to mm Hg because KH$K_H$ is given in mm Hg: PN₂ = 8 × 760 mm Hg$P_{\mathrm{N}_2} = 8 \times 760 \text{ mm Hg}$ Step 1: Apply Henry's Law PN₂ = KH · XN₂$$P_{\mathrm{N}_2} = K_H \cdot X_{\mathrm{N}_2}$$ 8 × 760 = 6.5 × 10⁷ × XN₂$$8 \times 760 = 6.5 \times 10^7 \times X_{\mathrm{N}_2}$$ XN₂ = (8 × 760)/(6.5 × 10⁷)$$X_{\mathrm{N}_2} = \frac{8 \times 760}{6.5 \times 10^7}$$ XN₂ = (6080)/(6.5 × 10⁷) 935.38 × 10⁻⁷$$X_{\mathrm{N}_2} = \frac{6080}{6.5 \times 10^7} \simeq 935.38 \times 10^{-7}$$ XN₂ = 9.35 × 10⁻⁵$$X_{\mathrm{N}_2} = 9.35 \times 10^{-5}$$ Pattern Recognition Always align the pressure units. Since KH$K_H$ dictates the unit ecosystem, convert Pgas$P_{\text{gas}}$ to match KH$K_H$ (e.g., atm to mm Hg via × 760$\times 760$). Chapter Mix Class 12 Chemistry: Solutions Q53 jee_main_2026_28_january_morning Raoults Law for Binary Mixtures At T(K), 2$2$ moles of liquid A and 3$3$ moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320~mm~Hg$320\mathrm{~mm~Hg}$. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6~mm~Hg$328.6\mathrm{~mm~Hg}$. The vapour pressure (in mm~Hg$\mathrm{mm~Hg}$) of A and B are respectively: A. 300, 200$300, 200$ B. 600, 400$600, 400$ C. 400, 300$400, 300$ D. 500, 200$500, 200$ Solution Related Formula PS = XA PA^° + XB PB^°$$P_S = X_A P_A^\circ + X_B P_B^\circ$$ Step 1: First Condition 2$2$ moles of A + 3$3$ moles of B (Total 5$5$ moles)\nXA = (2)/(5), XB = (3)/(5)$$X_A = \frac{2}{5}, \quad X_B = \frac{3}{5}$$\n320 = PA^° ((2)/(5)) + PB^° ((3)/(5))$$320 = P_A^\circ \left(\frac{2}{5}\right) + P_B^\circ \left(\frac{3}{5}\right)$$\n2 PA^° + 3 PB^° = 1600 (I)$$2 P_A^\circ + 3 P_B^\circ = 1600 \quad \dots (I)$$ Step 2: Second Condition Add 1$1$ mole of A & 1$1$ mole of B (Total 7$7$ moles)\nXA^ = (3)/(7), XB^ = (4)/(7)$$X_A^\prime = \frac{3}{7}, \quad X_B^\prime = \frac{4}{7}$$\n328.6 = PA^° ((3)/(7)) + PB^° ((4)/(7))$$328.6 = P_A^\circ \left(\frac{3}{7}\right) + P_B^\circ \left(\frac{4}{7}\right)$$\n3 PA^° + 4 PB^° = 2300.2 (II)$$3 P_A^\circ + 4 P_B^\circ = 2300.2 \quad \dots (II)$$ Step 3: Solve the Linear Equations Multiply eq (I) by 3 and eq (II) by 2:\n6 PA^° + 9 PB^° = 4800$$6 P_A^\circ + 9 P_B^\circ = 4800$$\n6 PA^° + 8 PB^° = 4600.4$$6 P_A^\circ + 8 P_B^\circ = 4600.4$$\nSubtracting the two yields:\nPB^° = 199.6 200~mm~Hg$$P_B^\circ = 199.6 \simeq 200\mathrm{~mm~Hg}$$\nSubstitute PB^°$P_B^\circ$ back into (I):\n2 PA^° + 3(200) = 1600 2 PA^° = 1000 PA^° 500~mm~Hg$$2 P_A^\circ + 3(200) = 1600 \implies 2 P_A^\circ = 1000 \implies P_A^\circ \simeq 500\mathrm{~mm~Hg}$$ Pattern Recognition Formulating two linear equations based on Raoult's Law mole fractions provides a fast algebraic elimination route to find pure vapour pressures. Chapter Mix Class 12 Chemistry: Solutions Q64 jee_main_2026_28_january_evening Colligative Properties Consider the following aqueous solutions. I. 2.2 g$2.2\text{ g}$ Glucose in 125 mL$125\text{ mL}$ of solution. II. 1.9 g$1.9\text{ g}$ Calcium chloride in 250 mL$250\text{ mL}$ of solution. III. 9.0 g$9.0\text{ g}$ Urea in 500 mL$500\text{ mL}$ of solution. IV. 20.5 g$20.5\text{ g}$ Aluminium sulphate in 750 mL$750\text{ mL}$ of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol⁻¹$\text{g mol}^{-1}$: H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32$H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27\text{ and }S=32$] A. (1) I < II < III < IV$(1)\text{ I < II < III < IV}$ B. (2) III < I < II < IV$(2)\text{ III < I < II < IV}$ C. (3) II < III < I < IV$(3)\text{ II < III < I < IV}$ D. (4) II < III < IV < I$(4)\text{ II < III < IV < I}$ Solution Related Formula Δ Tb = i · Kb · m$$\Delta T_b = i \cdot K_b \cdot m$$ For dilute solutions, Molarity (M) ≈$\approx$ Molality (m). Thus, Δ Tb ∝ i × M$\Delta T_b \propto i \times M$ Core Logic Solution Molarity (M = (W)/(Mw) × (1000)/(V)$M = \frac{W}{M_w} \times \frac{1000}{V}$) Effective Concentration (i × M$i \times M$) (I) Glucose (i=1$i=1$) (2.2)/(180) × (1000)/(125) = 0.098 M$\frac{2.2}{180} \times \frac{1000}{125} = 0.098\text{ M}$ 1 × 0.098 = 0.098$1 \times 0.098 = 0.098$ (II) CaCl₂$CaCl_2$ (i=3$i=3$) (1.9)/(111) × (1000)/(250) = 0.068 M$\frac{1.9}{111} \times \frac{1000}{250} = 0.068\text{ M}$ 3 × 0.068 = 0.204$3 \times 0.068 = 0.204$ (III) Urea (i=1$i=1$) (9)/(60) × (1000)/(500) = 0.3 M$\frac{9}{60} \times \frac{1000}{500} = 0.3\text{ M}$ 1 × 0.3 = 0.300$1 \times 0.3 = 0.300$ (IV) Al₂(SO₄)₃$Al_2(SO_4)_3$ (i=5$i=5$) (20.5)/(342) × (1000)/(750) = 0.080 M$\frac{20.5}{342} \times \frac{1000}{750} = 0.080\text{ M}$ 5 × 0.080 = 0.400$5 \times 0.080 = 0.400$ Step 1: Compare Effective Concentrations Comparing the values of i × M$i \times M$: 0.098 < 0.204 < 0.300 < 0.400$0.098 < 0.204 < 0.300 < 0.400$ ⇒ Glucose (I) < CaCl₂ (II) < Urea (III) < Al₂(SO₄)₃ (IV)$\Rightarrow \text{Glucose (I)} < CaCl_2 \text{ (II)} < \text{Urea (III)} < Al_2(SO_4)_3 \text{ (IV)}$ Step 2: Final Conclusion Order of boiling points corresponds directly to effective concentration. Thus: I < II < III < IV. Pattern Recognition For multiple salt mixtures, always multiply molarity by the Van't Hoff factor (i$i$). Do not compare just raw mass or raw molarity. Chapter Mix Class 12 Chemistry: Solutions Q37 jee_main_2025_02_april_evening Molarity and Temperature Dependency 'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1°C$1^{\circ}\mathrm{C}$ to 25°C$25^{\circ}\mathrm{C}$. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range] A. Plot (1)$\text{Plot (1)}$ B. Plot (2)$\text{Plot (2)}$ C. Plot (3)$\text{Plot (3)}$ D. Plot (4)$\text{Plot (4)}$ Solution Related Formula Molarity (M) = nsoluteVsolution (L)$$\text{Molarity } (M) = \frac{n_{\text{solute}}}{V_{\text{solution}} (\mathrm{L})}$$ Core Logic Since solubility of NaCl$\mathrm{NaCl}$ remains unchanged, the number of dissolved moles of NaCl$\mathrm{NaCl}$ solute (nsolute$n_{\text{solute}}$) remains strictly constant. Thus, molarity M$M$ is strictly dependent on the volume of water (solvent) as temperature changes: M ∝ 1Vsolution$$M \propto \frac{1}{V_{\text{solution}}}$$ Step 1: Understand Water's Anomalous Expansion Water exhibits unique anomalous density behavior near freezing: From 1°C$1^{\circ}\mathrm{C}$ to 4°C$4^{\circ}\mathrm{C}$, the density of water increases to a maximum. This contraction means the volume (V$V$) of water decreases. From 4°C$4^{\circ}\mathrm{C}$ to 25°C$25^{\circ}\mathrm{C}$, the density of water decreases due to standard thermal expansion. Consequently, the volume (V$V$) increases. Step 2: Relate Volume to Molarity Because volume is in the denominator of the molarity equation: From 1°C$1^{\circ}\mathrm{C}$ to 4°C$4^{\circ}\mathrm{C}$: Volume decreases $\implies$ Molarity increases. At 4°C$4^{\circ}\mathrm{C}$: Volume is minimum $\implies$ Molarity reaches a maximum. From 4°C$4^{\circ}\mathrm{C}$ to 25°C$25^{\circ}\mathrm{C}$: Volume increases $\implies$ Molarity decreases. This behavior is perfectly represented by Plot (2), which features a distinct peak around 4°C$4^{\circ}\mathrm{C}$. Pattern Recognition Water is at its densest (and occupies minimum volume) at exactly 3.98^$3.98^\circ\mathrm{C}$ (4^$4^\circ\mathrm{C}$). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature. 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? Sign up Enter the 6-digit code we emailed to your email Verify & Continue Didn't get it? Resend code ← Back Enter your account email and we'll send you a link to reset your password. Send reset link ← Back to sign in