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) Delta Vtextmix=0$\Delta V{\text{mix}}=0$ Choose the correct answer from the options given below: A. text(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}$ B. text(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$\text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}$ C. text(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$\text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}$ D. text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}$ Solution & Explanation ### Related Formula textNegative Deviation from Raoult's Law implies textMaximum Boiling Azeotrope $$\text{Negative Deviation from Raoult's Law} \implies \text{Maximum Boiling Azeotrope} $$ textPositive Deviation from Raoult's Law implies textMinimum Boiling Azeotrope $$\text{Positive Deviation from Raoult's Law} \implies \text{Minimum Boiling Azeotrope} $$ textIdeal Solution implies Delta Vtextmix = 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 ightarrow$ ightarrow$ (III) - (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope ightarrow$ ightarrow$ (I) - (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where Delta Vtextmix = 0$\Delta V{\text{mix}} = 0$ ightarrow$ ightarrow$ (IV) - (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize ightarrow$ ightarrow$ (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. ### Evaluation Rubric / Model Answer null ### 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 7 Q87 jee_main_2024_31_jan_evening Concentration Terms The molarity of 1text L$1\text{ L}$ orthophosphoric acid (H_3PO_4$H_3PO_4$) having 70\%$70\%$ purity by weight (specific gravity 1.54text g cm^-3$1.54\text{ g cm}^{-3}$) is ________ textM$\text{M}$. (Molar mass of H_3PO_4 = 98text g mol^-1$H_3PO_4 = 98\text{ g mol}^{-1}$) Numerical Answer. Answer: 11 to 11 Solution ### Related Formula M = frac\% text purity times textdensity times 10textMolar Mass$$M = \frac{\% \text{ purity} \times \text{density} \times 10}{\text{Molar Mass}}$$ ### Core Logic Specific gravity is numerically equivalent to density in textg/cm^3$\text{g/cm}^3$, so density = 1.54text g/mL$1.54\text{ g/mL}$. Volume of solution = 1text L = 1000text mL$1\text{ L} = 1000\text{ mL}$. Mass of solution = textVolume times textDensity = 1000 times 1.54 = 1540text g$\text{Volume} \times \text{Density} = 1000 \times 1.54 = 1540\text{ g}$. ### Step 1: Finding Solute Mass and Molarity Since the purity is 70\%$70\%$ by weight, the mass of H_3PO_4$H_3PO_4$ in the solution is: textMass of H_3PO_4 = 1540 times 0.70 = 1078text g$\text{Mass of } H_3PO_4 = 1540 \times 0.70 = 1078\text{ g}$. Moles of H_3PO_4 = frac107898 = 11text moles$H_3PO_4 = \frac{1078}{98} = 11\text{ moles}$. Since this is dissolved in 1text L$1\text{ L}$ of solution, the Molarity is: M = frac11text moles1text L = 11text M$$M = \frac{11\text{ moles}}{1\text{ L}} = 11\text{ M}$$ ### Pattern Recognition Shortcut formula directly substitutes the values: M = frac70 times 1.54 times 1098 = frac107898 = 11$M = \frac{70 \times 1.54 \times 10}{98} = \frac{1078}{98} = 11$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry Q63 jee_main_2024_31_jan_morning Non-Ideal Solutions Identify the mixture that shows positive deviations from Raoult's Law A. (CH_3)_2CO + C_6H_5NH_2$(CH_3)_2CO + C_6H_5NH_2$ B. CHCl_3 + C_6H_6$CHCl_3 + C_6H_6$ C. CHCl_3 + (CH_3)_2CO$CHCl_3 + (CH_3)_2CO$ D. (CH_3)_2CO + CS_2$(CH_3)_2CO + CS_2$ Solution ### Core Logic (CH_3)_2CO + CS_2$(CH_3)_2CO + CS_2$ exhibits positive deviations from Raoult's Law because the interactions between acetone and carbon disulphide molecules are weaker than the respective pure component interactions. ### Pattern Recognition Mixtures like Acetone + Aniline or Chloroform + Benzene/Acetone form stronger hydrogen bonds after mixing, showing negative deviation. Acetone + CS_2$CS_2$ or Ethanol + Acetone break existing strong interactions, leading to positive deviation. ### Evaluation Rubric / Model Answer null ### 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, 2 moles each of ethylene glycol and glucose are dissolved Consider the given plots of vapour pressure (VP) vs temperature 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 Elements P and Q form two types of non-volatile, non-ionizable Identify the mixture that shows positive deviations from Raoult's Law A solution of two miscible liquids showing negative deviation from If a substance 'A' dissolves in solution of a mixture We have three aqueous solutions of labelled as 'A', 'B' What happens to freezing point of benzene when small quantity The solution from the following with highest depression in freezing Molality of solution (density ) is ________ . A solution of is by mass and has a density The osmotic pressure of a dilute solution is at . 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