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 2 Q49 jee_main_2025_07_april_morning Van't Hoff Factor The percentage dissociation of a salt (mathrmMX_3)$(\mathrm{MX}_3)$ solution at given temperature (van't Hoff factor texti = 2$\text{i} = 2$) is ______ %. (Nearest integer) Numerical Answer. Answer: 33 to 33 Solution ### Related Formula i = 1 + (n - 1)alpha$$i = 1 + (n - 1)\alpha$$ ### Core Logic For the salt mathrmMX_3$\mathrm{MX}_3$: mathrmMX_3 rightarrow mathrmM^3+ + 3mathrmX^-$$\mathrm{MX}_3 \rightarrow \mathrm{M}^{3+} + 3\mathrm{X}^-$$ - 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 (alpha$\alpha$): i = 1 + (4 - 1)alpha$$i = 1 + (4 - 1)\alpha$$ 2 = 1 + 3alpha implies 3alpha = 1 implies alpha = frac13 approx 0.3333$$2 = 1 + 3\alpha \implies 3\alpha = 1 \implies \alpha = \frac{1}{3} \approx 0.3333$$ Percentage dissociation: \% text dissociation = alpha times 100 = 33.33 \% approx 33 \%$$\% \text{ dissociation} = \alpha \times 100 = 33.33 \% \approx 33 \%$$ ### Pattern Recognition Shortcut: For MX_3$MX_3$ dissociating into 4$4$ ions, alpha = (i - 1) / 3$\alpha = (i - 1) / 3$. Since i = 2$i = 2$, alpha = 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. textH_2textO + textCH_3textCOC_2textH_5$\text{H}_2\text{O} + \text{CH}_3\text{COC}_2\text{H}_5$ B. textC_6textH_5textOH + textC_6textH_5textNH_2$\text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{NH}_2$ C. textCS_2 + textCH_3textCOCH_3$\text{CS}_2 + \text{CH}_3\text{COCH}_3$ D. textCH_3textOH + textCHCl_3$\text{CH}_3\text{OH} + \text{CHCl}_3$ 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 (C_6H_5OH$C_6H_5OH$) + Aniline (C_6H_5NH_2$C_6H_5NH_2$)**: The phenolic -textOH$-\text{OH}$ proton forms strong intermolecular hydrogen bonds with the lone pair of the -textNH_2$-\text{NH}_2$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q34 jee_main_2025_08_april_evening Colligative Properties For the weak acid dissociation equation: textHA(aq) rightleftharpoons textH^+text(aq) + textA^-text(aq)$$\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}$$ The freezing point depression of a 0.1 text m$0.1 \text{ m}$ aqueous solution of this monobasic weak acid HA is found to be 0.20^circ textC$0.20^\circ \text{C}$. The dissociation constant (K_a$K_a$) for the acid is: Given: K_f(textH_2textO) = 1.8 text K kg mol^-1$K_f(\text{H}_2\text{O}) = 1.8 \text{ K kg mol}^{-1}$, and assume molality approx$\approx$ molarity. A. 1.38 times 10^-3$1.38 \times 10^{-3}$ B. 1.1 times 10^-2$1.1 \times 10^{-2}$ C. 1.90 times 10^-3$1.90 \times 10^{-3}$ D. 1.89 times 10^-1$1.89 \times 10^{-1}$ Solution ### Related Formula Depression in freezing point colligative relation: Delta T_f = i cdot K_f cdot m$$\Delta T_f = i \cdot K_f \cdot m$$ Van 't Hoff factor for a weak acid dissociation: i = 1 + alpha$i = 1 + \alpha$ Dissociation constant formula: K_a = fracCalpha^21-alpha$$K_a = \frac{C\alpha^2}{1-\alpha}$$ ### Execution Step 1: Calculate the Van 't Hoff factor i$i$ using experimental freezing data: 0.20 = i times 1.8 times 0.1$$0.20 = i \times 1.8 \times 0.1$$ i = frac0.200.18 = frac2018 = frac109$$i = \frac{0.20}{0.18} = \frac{20}{18} = \frac{10}{9}$$ Step 2: Solve for degree of dissociation alpha$\alpha$: i = 1 + alpha implies frac109 = 1 + alpha$$i = 1 + \alpha \implies \frac{10}{9} = 1 + \alpha$$ alpha = frac109 - 1 = frac19$$\alpha = \frac{10}{9} - 1 = \frac{1}{9}$$ Step 3: Compute K_a$K_a$ substituting concentration C = 0.1 text M$C = 0.1 \text{ M}$ and alpha = frac19$\alpha = \frac{1}{9}$: K_a = frac0.1 times left(frac19right)^21 - frac19 = frac0.1 times frac181frac89 = frac0.181 times frac98 = frac0.172 = frac1720$$K_a = \frac{0.1 \times \left(\frac{1}{9}\right)^2}{1 - \frac{1}{9}} = \frac{0.1 \times \frac{1}{81}}{\frac{8}{9}} = \frac{0.1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720}$$ K_a approx 1.388 times 10^-3$$K_a \approx 1.388 \times 10^{-3}$$ ### Pattern Recognition When dealing with weak acids, always determine i$i$ first via colligative data, isolate alpha$\alpha$, and map directly to K_a = fracCalpha^21-alpha$K_a = \frac{C\alpha^2}{1-\alpha}$. Speed up computation by converting decimals into fractional fractions (0.2/0.18 = 10/9$0.2/0.18 = 10/9$) to maintain clean, mistake-free algebra. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Ionic Equilibrium Q32 jee_main_2025_29_jan_evening Depression of Freezing Point Given below are two statements: Statement (I): NaCl is added to the ice at 0^circC$0^{circ}C$, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0^circC$0^{circ}C$, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below: A. Statement I is false but Statement II is true B. Both Statement I and Statement II are true C. Both Statement I and Statement II are false D. Statement I is true but Statement II is false Solution ### Related Formula Delta T_f = i cdot K_f cdot m$$Delta T_f = i cdot K_f cdot m$$ ### Core Logic Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0^circC$0^{circ}C$, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. ### Pattern Recognition This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q40 jee_main_2025_28_jan_morning Colligative Properties - Freezing Point Depression What is the freezing point depression constant of a solvent, 50mathrmg$50\mathrm{g}$ of which contain 1mathrmg$1\mathrm{g}$ non-volatile solute (molar mass 256mathrmg\,mol^-1$256\mathrm{g\,mol}^{-1}$ ) and the decrease in freezing point is 0.40mathrmK$0.40\mathrm{K}$ ? A. 5.12mathrm\,K\,kg\,mol^-1$5.12\mathrm{\,K\,kg\,mol}^{-1}$ B. 4.43mathrm\,K\,kg\,mol^-1$4.43\mathrm{\,K\,kg\,mol}^{-1}$ C. 1.86mathrm~K~kg~mol^-1$1.86\mathrm{~K~kg~mol}^{-1}$ D. 3.72mathrm\,K\,kg\,mol^-1$3.72\mathrm{\,K\,kg\,mol}^{-1}$ Solution ### Related Formula Freezing point depression relationship: Delta T_f = K_f cdot m$$\Delta T_f = K_f \cdot m$$ where m$m$ is the molality defined as: m = fractextmoles of solutetextmass of solvent in kg$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$ ### Step 1: Compute Molality Moles of non-volatile solute: textmoles = frac1\,mathrmg256\,mathrmg\,mol^-1$$\text{moles} = \frac{1\,\mathrm{g}}{256\,\mathrm{g\,mol}^{-1}}$$ Mass of solvent in kg: textmass = 50\,mathrmg = 50 times 10^-3\,mathrmkg$$\text{mass} = 50\,\mathrm{g} = 50 \times 10^{-3}\,\mathrm{kg}$$ Therefore, molality values map to: m = frac1256 times 50 times 10^-3 = frac100012800 = frac564\,mathrmmol\,kg^-1$$m = \frac{1}{256 \times 50 \times 10^{-3}} = \frac{1000}{12800} = \frac{5}{64}\,\mathrm{mol\,kg}^{-1}$$ ### Step 2: Calculate K_f$K_f$ Substituting values into the core formula: 0.40 = K_f cdot left(frac564right)$$0.40 = K_f \cdot \left(\frac{5}{64}\right)$$ K_f = frac0.40 times 645 = 0.08 times 64 = 5.12\,mathrmK\,kg\,mol^-1$$K_f = \frac{0.40 \times 64}{5} = 0.08 \times 64 = 5.12\,\mathrm{K\,kg\,mol}^{-1}$$ ### Pattern Recognition Sees: Direct calculation of cryogenic context constant (K_f$K_f$). Shortcut: Isolate K_f = fracDelta T_f cdot M cdot W_textsolvent1000 cdot w_textsolute$K_f = \frac{\Delta T_f \cdot M \cdot W_{\text{solvent}}}{1000 \cdot w_{\text{solute}}}$. Substituting instantly returns 5.12. ### 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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