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 Q59 jee_main_2026_21_jan_morning Elevation of Boiling Point Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ₂$PQ_{2}$. When 1 g of PQ is dissolved in 50 g of solvent ‘A’. Δ Tb$\Delta T_{b}$ was 1.176 K while when 1 g of PQ₂$PQ_{2}$ is dissolved in 50 g of solvent ‘A’, Δ Tb$\Delta T_{b}$ was 0.689 K. (Kb$K_{b}$ of ‘A’ = 5 K kg mol⁻¹$5\text{ K kg mol}^{-1}$). The molar masses of elements P and Q (in g mol⁻¹$\text{g mol}^{-1}$) respectively, are : A. 70, 110$70, 110$ B. 65, 145$65, 145$ C. 60, 25$60, 25$ D. 25, 60$25, 60$ Solution Related Formula Δ Tb = Kb × m$$\Delta T_b = K_b \times m$$ m = Weight of solute (g)Molar mass of solute × 1000Weight of solvent (g)$$m = \frac{\text{Weight of solute (g)}}{\text{Molar mass of solute}} \times \frac{1000}{\text{Weight of solvent (g)}}$$ Core Logic For compound PQ: (Δ Tb)PQ = Kb · m$$(\Delta \mathrm{T_b})_{\mathrm{PQ}} = \mathrm{K_b} \cdot m$$ 1.176 = 5 × 1M₁ × (1000)/(50)$$1.176 = 5 \times \frac{1}{\mathrm{M}_1} \times \frac{1000}{50}$$ M₁ = (5 × 20)/(1.176) = 85.03 g/mol$$\mathrm{M}_1 = \frac{5 \times 20}{1.176} = 85.03 \text{ g/mol}$$ For compound PQ₂$PQ_2$: (Δ Tb)PQ₂ = 5 × 1M₂ × (1000)/(50) = 0.689$$(\Delta \mathrm{T_b})_{\mathrm{PQ}_2} = 5 \times \frac{1}{\mathrm{M}_2} \times \frac{1000}{50} = 0.689$$ M₂ = (5 × 20)/(0.689) = 145.13 g/mol$$\mathrm{M}_2 = \frac{5 \times 20}{0.689} = 145.13 \text{ g/mol}$$ Let molar mass of P & Q be MP$\mathrm{M}_P$ and MQ$\mathrm{M}_Q$ respectively: MP + MQ = 85.03 --- (1)$$\mathrm{M}_P + \mathrm{M}_Q = 85.03 \quad \text{--- (1)}$$ MP + 2MQ = 145.13 --- (2)$$\mathrm{M}_P + 2\mathrm{M}_Q = 145.13 \quad \text{--- (2)}$$ Subtracting (1) from (2): MQ = 145.13 - 85.03 = 60.1 ≈ 60 g/mol$$\mathrm{M}_Q = 145.13 - 85.03 = 60.1 \approx 60 \text{ g/mol}$$ Substituting back into (1): MP + 60.1 = 85.03 MP = 24.93 ≈ 25 g/mol$$\mathrm{M}_P + 60.1 = 85.03 \implies \mathrm{M}_P = 24.93 \approx 25 \text{ g/mol}$$ Chapter Mix Class 12 Chemistry: Solutions Q72 jee_main_2026_21_jan_evening Osmotic Pressure and Isotonic Solutions The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ____ g L⁻¹$\_\_\_\_ \text{ g L}^{-1}$. (Nearest integer) Given: R = 0.08 L atm K⁻¹ mol⁻¹$R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$ Assume complete dissociation of NaCl$\text{NaCl}$ (Given: Molar mass of Na and Cl are 23 and 35.5 g mol⁻¹$35.5 \text{ g mol}^{-1}$ respectively.) Numerical Answer. Answer: 15 to 15 Solution Related Formula π = i C R T$\pi = i C R T$ Core Logic Given: π = 12 atm$\pi = 12 \text{ atm}$ T = 300 K$T = 300 \text{ K}$ i = 2$i = 2$ (for complete dissociation of NaCl$\text{NaCl}$) R = 0.08 L atm K⁻¹ mol⁻¹$R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$ 12 = 2 × C × 0.08 × 300 12 = 48C C = 0.25 mol/L$$12 = 2 \times C \times 0.08 \times 300 \implies 12 = 48C \implies C = 0.25 \text{ mol/L}$$ Step 1: Calculating Strength Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol$\text{NaCl} = 23 + 35.5 = 58.5 \text{ g/mol}$. Strength = 0.25 × 58.5 = 14.625 g/L ≈ 15 g/L$= 0.25 \times 58.5 = 14.625 \text{ g/L} \approx 15 \text{ g/L}$. Pattern Recognition Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor i = 2$i = 2$ for NaCl$\text{NaCl}$. Chapter Mix Class 12 Chemistry: Solutions Q73 jee_main_2026_21_jan_evening Elevation in Boiling Point and Vapour Pressure Lowering A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol⁻¹$= 300 \text{ g mol}^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is ____ × 10⁻²$\_\_\_\_ \times 10^{-2}$. (Nearest integer) [Given: Kb$K_b$ of the solvent = 5.0 K kg mol⁻¹$= 5.0 \text{ K kg mol}^{-1}$] Assume the solution to be dilute and no association or dissociation of X takes place in solution. Numerical Answer. Answer: 3 to 3 Solution Related Formula Δ Tb = Kb · m, P° - PₛP° = χsolute$$\Delta T_b = K_b \cdot m, \quad \frac{P^{\circ} - P_s}{P^{\circ}} = \chi_{\text{solute}}$$ Core Logic Using boiling point elevation: Δ Tb = Kb × m 0.5 = 5.0 × m m = 0.1 mol/kg$$\Delta T_b = K_b \times m \implies 0.5 = 5.0 \times m \implies m = 0.1 \text{ mol/kg}$$ Moles of solvent Y = (150)/(300) = 0.5 mol$= \frac{150}{300} = 0.5 \text{ mol}$. Moles of solute X = 0.1 × 0.150 = 0.015 mol$= 0.1 \times 0.150 = 0.015 \text{ mol}$. Relative lowering of vapour pressure = χsolute = (0.015)/(0.015 + 0.5) ≈ (0.015)/(0.5) = 0.03 = 3 × 10⁻²$= \chi_{\text{solute}} = \frac{0.015}{0.015 + 0.5} \approx \frac{0.015}{0.5} = 0.03 = 3 \times 10^{-2}$. Step 1: Final Calculation Thus, the nearest integer value is 3. Pattern Recognition Sees: boiling point elevation combined with relative lowering of vapour pressure calculation. Trap: Confusing solvent mass with solution mass when calculating mole fraction. Chapter Mix Class 12 Chemistry: Solutions Q57 jee_main_2026_22_january_morning Henrys Law Consider a solution of CO₂(g)$CO_{2}(g)$ dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO₂$CO_{2}$ in vapour phase above water) [y-axis] with log (mole fraction of CO₂$CO_{2}$ in water) [x-axis] at 25°C$25^{\circ}C$? A. Graph 1$\text{Graph 1}$ B. Graph 2$\text{Graph 2}$ C. Graph 3$\text{Graph 3}$ D. Graph 4$\text{Graph 4}$ Solution Related Formula P = KH · X$$P = K_{H} \cdot X$$ Taking logarithm on both sides: P = KH + X$$\log P = \log K_{H} + \log X$$ Core Logic From Henry's Law: P(g) = X(g) + KH$$\log P(g) = \log X(g) + \log K_{H}$$ This equation represents a straight line of the form y = mx + c$y = mx + c$, where: y = P(g)$y = \log P(g)$ x = X(g)$x = \log X(g)$ Slope (m$m$) = +1 y-intercept (c$c$) = KH$\log K_{H}$ (which is positive). The plot of P$\log P$ versus X$\log X$ is a straight line with a positive slope and a positive intercept on the y-axis. Step 1: Conclusion Graph 3 corresponds to a straight line with a positive slope and a positive y-intercept. Pattern Recognition When variables are multiplied (P = KH X$P = K_{H} X$), their log plot is always a straight line with a slope of +1 and an intercept equal to the log of the constant. Chapter Mix Class 12 Chemistry: Solutions Q70 jee_main_2026_22_january_morning Henrys Law Given below are two statements: Statement I: The Henry's law constant KH$K_{H}$ is constant with respect to variations in solution's concentration over the range for which the solutions is ideally dilute. Statement II: KH$K_{H}$ does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options. A. Statement I is false but Statement II is true.$\text{Statement I is false but Statement II is true.}$ B. Statement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$ C. Both Statement I and Statement II are true.$\text{Both Statement I and Statement II are true.}$ D. Both Statement I and Statement II are false.$\text{Both Statement I and Statement II are false.}$ Solution Core Logic Statement I: KH$K_H$ is independent of concentration as long as the solution behaves ideally (is very dilute). This matches the physical definition of Henry's Law constant. (True) Statement II: Henry's Law constant (KH$K_H$) depends on the nature of the gas, the nature of the solvent, and the temperature. Therefore, it absolutely differs for the same gas dissolved in different solvents. (False) Step 1: Final Conclusion Statement I is true but Statement II is false. Pattern Recognition The constants for solubility laws (like KH$K_H$) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent. 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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