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 4 Q49 jee_main_2025_02_april_evening Elevation of Boiling Point and Molar Mass Determination When 1~g$1~\mathrm{g}$ each of compounds AB and AB₂$\mathrm{AB}_2$ are dissolved in 15~g$15~\mathrm{g}$ of water separately, they increased the boiling point of water by 2.7~K$2.7~\mathrm{K}$ and 1.5~K$1.5~\mathrm{K}$ respectively. The atomic mass of A (in amu) is × 10⁻¹$\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is 0.5~ K~kg~mol⁻¹$0.5~\mathrm{K~kg~mol^{-1}}$) Numerical Answer. Answer: 25 to 25 Solution Related Formula Δ Tb = Kb · m = Kb · ( wsoluteMsolute · 1000wsolvent )$$\Delta T_{\mathrm{b}} = K_{\mathrm{b}} \cdot m = K_{\mathrm{b}} \cdot \left( \frac{w_{\text{solute}}}{M_{\text{solute}}} \cdot \frac{1000}{w_{\text{solvent}}} \right)$$ Core Logic Both dissolved compounds are non-electrolytes (van 't Hoff factor i = 1$i = 1$). We calculate the molar masses of AB$\mathrm{AB}$ and AB₂$\mathrm{AB_2}$ individually, then solve for the individual atomic masses of elements A and B. Step 1: Determine Molar Mass of AB Given Δ Tb = 2.7~K$\Delta T_{\mathrm{b}} = 2.7~\mathrm{K}$, wsolute = 1~g$w_{\text{solute}} = 1~\mathrm{g}$, wsolvent = 15~g$w_{\text{solvent}} = 15~\mathrm{g}$, and Kb = 0.5~ K~kg~mol⁻¹$K_{\mathrm{b}} = 0.5~\mathrm{K~kg~mol^{-1}}$: 2.7 = 0.5 × 1MAB × (1000)/(15)$$2.7 = 0.5 \times \frac{1}{M_{\mathrm{AB}}} \times \frac{1000}{15}$$ MAB = (0.5 × 1000)/(15 × 2.7) = (500)/(40.5) ≈ 12.3457~ g~mol⁻¹$$M_{\mathrm{AB}} = \frac{0.5 \times 1000}{15 \times 2.7} = \frac{500}{40.5} \approx 12.3457~\mathrm{g~mol^{-1}}$$ Step 2: Determine Molar Mass of AB_2 Given Δ Tb = 1.5~K$\Delta T_{\mathrm{b}} = 1.5~\mathrm{K}$, wsolute = 1~g$w_{\text{solute}} = 1~\mathrm{g}$, wsolvent = 15~g$w_{\text{solvent}} = 15~\mathrm{g}$, and Kb = 0.5$K_{\mathrm{b}} = 0.5$: 1.5 = 0.5 × 1MAB₂ × (1000)/(15)$$1.5 = 0.5 \times \frac{1}{M_{\mathrm{AB_2}}} \times \frac{1000}{15}$$ MAB₂ = (0.5 × 1000)/(15 × 1.5) = (500)/(22.5) ≈ 22.2222~ g~mol⁻¹$$M_{\mathrm{AB_2}} = \frac{0.5 \times 1000}{15 \times 1.5} = \frac{500}{22.5} \approx 22.2222~\mathrm{g~mol^{-1}}$$ Step 3: Solve for Atomic Mass of A Let the atomic masses of elements A and B be a$a$ and b$b$ respectively: a + b = 12.3457 --- (1)$$a + b = 12.3457 \quad \text{--- (1)}$$ a + 2b = 22.2222 --- (2)$$a + 2b = 22.2222 \quad \text{--- (2)}$$ Subtracting equation (1) from (2): b = 22.2222 - 12.3457 = 9.8765~amu$$b = 22.2222 - 12.3457 = 9.8765~\mathrm{amu}$$ Substituting b$b$ back into equation (1): a = 12.3457 - 9.8765 = 2.4692~amu$$a = 12.3457 - 9.8765 = 2.4692~\mathrm{amu}$$ Expressing a$a$ in the requested format (× 10⁻¹$\times 10^{-1}$): a = 24.692 × 10⁻¹ ≈ 25 × 10⁻¹$$a = 24.692 \times 10^{-1} \approx 25 \times 10^{-1}$$ Pattern Recognition Mathematical consistency checks: Since AB₂$\mathrm{AB_2}$ has more atoms than AB$\mathrm{AB}$ of identical constituent mass, its molar mass should be higher, leading to a smaller elevation of boiling point for the same mass dissolved, which perfectly matches our 2.7~K arrow 1.5~K$2.7~\mathrm{K} \rightarrow 1.5~\mathrm{K}$ change. Chapter Mix Class 12 Chemistry: Solutions Q jee_main_2025_02_april_morning Henry's Law Constant and Temperature Dependance Which of the following graph correctly represents the plots of KH$\mathrm{K_H}$ at 1 bar gases in water versus temperature? A. B. C. D. Solution Related Formula Henry's Law formula connects partial pressure to solubility component: p = KH · x$$p = K_{\mathrm{H}} \cdot x$$ Core Logic As temperature increases, gas dissolution is typically exothermic, meaning solubility initially drops, causing the Henry's constant KH$K_{\mathrm{H}}$ to curve upward dynamically before varying at extreme points. For standard non-reactive noble/molecular gases at regular ranges, the magnitude order follows: KH(He) > KH(N₂) > KH(CH₄)$$K_{\mathrm{H}}(\mathrm{He}) > K_{\mathrm{H}}(\mathrm{N_2}) > K_{\mathrm{H}}(\mathrm{CH_4})$$ Step 1: Selection Graph (4) illustrates the correct relative order and curved profile properly across the given temperature frame. Pattern Recognition Higher KH$K_{\mathrm{H}}$ value implies lower solubility of that gas at a given pressure. Helium is notoriously insoluble in water compared to organic or polarizable molecules like methane, hence its plot line must live at the top. Chapter Mix Class 12 Chemistry: Solutions Q32 jee_main_2025_02_april_morning Raoult's Law and Vapour Pressure A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 ~mmHg$200 \mathrm{~mmHg}$ and that of the solution is 500 ~mmHg$500 \mathrm{~mmHg}$. The vapour pressure of pure B and the least volatile component of the solution, respectively, are : A. (1) 1400 mmHg, A$(1)\ 1400\ \mathrm{mmHg},\ \text{A}$ B. (2) 1400 mmHg, B$(2)\ 1400\ \mathrm{mmHg},\ \text{B}$ C. (3) 600 mmHg, B$(3)\ 600\ \mathrm{mmHg},\ \text{B}$ D. (4) 600 mmHg, A$(4)\ 600\ \mathrm{mmHg},\ \text{A}$ Solution Related Formula Raoult's Law for a multi-component solution mixtures: PS = PA⁰ · XA + PB⁰ · XB$$P_{\mathrm{S}} = P_{\mathrm{A}}^0 \cdot X_{\mathrm{A}} + P_{\mathrm{B}}^0 \cdot X_{\mathrm{B}}$$ Core Logic Let's determine the mole fractions first based on the input molar amounts: XA = (1)/(1+3) = (1)/(4), XB = (3)/(1+3) = (3)/(4)$$X_{\mathrm{A}} = \frac{1}{1+3} = \frac{1}{4}, \quad X_{\mathrm{B}} = \frac{3}{1+3} = \frac{3}{4}$$ Substitute the known properties into Raoult's equation block: 500 = 200 × (1)/(4) + PB⁰ × (3)/(4)$$500 = 200 \times \frac{1}{4} + P_{\mathrm{B}}^0 \times \frac{3}{4}$$ 500 = 50 + PB⁰ × (3)/(4)$$500 = 50 + P_{\mathrm{B}}^0 \times \frac{3}{4}$$ 450 = PB⁰ × (3)/(4) PB⁰ = 600 ~mmHg$$450 = P_{\mathrm{B}}^0 \times \frac{3}{4} \implies P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$$ Comparing pure state components pressures: PA⁰ = 200 ~mmHg$P_{\mathrm{A}}^0 = 200 \mathrm{~mmHg}$ and PB⁰ = 600 ~mmHg$P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$. Lower vapor pressure indicates stronger intermolecular cohesion, making A the least volatile component. Step 1: Finalization Thus, the vapour pressure of pure B is 600~mmHg$600\mathrm{~mmHg}$ and the least volatile chemical is A. Pattern Recognition Volatility is directly proportional to pure vapor pressure (P⁰$P^0$). Don't mix up 'least volatile' with 'lowest mole fraction'—always evaluate based solely on the isolated values of P⁰$P^0$. Chapter Mix Class 12 Chemistry: Solutions Q49 jee_main_2025_07_april_morning Van't Hoff Factor The percentage dissociation of a salt (MX₃)$(\mathrm{MX}_3)$ solution at given temperature (van't Hoff factor i = 2$\text{i} = 2$) is ______ %. (Nearest integer) Numerical Answer. Answer: 33 to 33 Solution Related Formula i = 1 + (n - 1)α$$i = 1 + (n - 1)\alpha$$ Core Logic For the salt MX₃$\mathrm{MX}_3$: MX₃ arrow M³⁺ + 3X^-$$\mathrm{MX}_3 \rightarrow \mathrm{M}^{3+} + 3\mathrm{X}^-$$ Number of ions formed per formula unit, n = 1 + 3 = 4$n = 1 + 3 = 4$. Given i = 2$i = 2$, let's find the degree of dissociation (α$\alpha$): i = 1 + (4 - 1)α$$i = 1 + (4 - 1)\alpha$$ 2 = 1 + 3α 3α = 1 α = (1)/(3) ≈ 0.3333$$2 = 1 + 3\alpha \implies 3\alpha = 1 \implies \alpha = \frac{1}{3} \approx 0.3333$$ Percentage dissociation: % dissociation = α × 100 = 33.33 % ≈ 33 %$$\% \text{ dissociation} = \alpha \times 100 = 33.33 \% \approx 33 \%$$ Pattern Recognition Shortcut: For MX₃$MX_3$ dissociating into 4$4$ ions, α = (i - 1) / 3$\alpha = (i - 1) / 3$. Since i = 2$i = 2$, α = 1/3 = 33%$\alpha = 1/3 = 33\%$ directly. Evaluation Rubric / Model Answer Simple formula application verifying the degree of dissociation to obtain a precise 33$33$ percent. Chapter Mix Class 12 Chemistry: Solutions Q33 jee_main_2025_08_april_evening Azeotropic Mixtures Which of the following binary mixtures does not show the behaviour of minimum boiling azeotropes? A. H₂O + CH₃COC₂H₅$\text{H}_2\text{O} + \text{CH}_3\text{COC}_2\text{H}_5$ B. C₆H₅OH + C₆H₅NH₂$\text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{NH}_2$ C. CS₂ + CH₃COCH₃$\text{CS}_2 + \text{CH}_3\text{COCH}_3$ D. CH₃OH + CHCl₃$\text{CH}_3\text{OH} + \text{CHCl}_3$ Solution Core Logic Let's connect deviation behaviors to azeotropic styles: Minimum Boiling Azeotropes: Formed by liquid binary solutions that display a strong positive deviation from Raoult's Law. In these mixtures, inter-molecular forces between components (A-B$A-B$) are weaker than pure self-interactions (A-A$A-A$ or B-B$B-B$). Maximum Boiling Azeotropes: Formed by liquid binary mixtures showing a notable negative deviation from Raoult's Law. Here, new inter-molecular interactions (A-B$A-B$) become significantly stronger. Analyzing Phenol (C₆H₅OH$C_6H_5OH$) + Aniline (C₆H₅NH₂$C_6H_5NH_2$): The phenolic -OH$-\text{OH}$ proton forms strong intermolecular hydrogen bonds with the lone pair of the -NH₂$-\text{NH}_2$ group of aniline. These new forces exceed the initial individual fluid bonds, lowering vapor pressure below ideal expectations (negative deviation) and creating a maximum boiling azeotrope. Pattern Recognition Phenol + Aniline, and Chloroform + Acetone are classic textbook models of strong negative deviation from Raoult's Law. Negative deviation explicitly pairs with maximum boiling azeotropes, standing out instantly against minimum boiling selections. 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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