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 4 Q46 jee_main_2025_24_jan_evening Abnormal Molar Masses and Van't Hoff Factor The observed and normal masses of compound mathrmMX_2$\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of mathrmMX_2$\mathrm{MX}_2$ is ____ %. (Nearest integer) Numerical Answer. Answer: 75 to 75 Solution ### Related Formula i = fractextNormal Molar MasstextObserved Molar Mass$$i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}}$$ i = 1 + (n - 1)alpha$$i = 1 + (n - 1)\alpha$$ ### Core Logic 1. Calculate the van 't Hoff factor (i$i$): i = frac16465.6 = 2.5$$i = \frac{164}{65.6} = 2.5$$ 2. Set up the dissociation equilibrium for the electrolyte mathrmMX_2$\mathrm{MX}_2$: mathrmMX_2 ightarrow mathrmM^2+ + 2mathrmX^-$$\mathrm{MX}_{2} ightarrow \mathrm{M}^{2+} + 2\mathrm{X}^{-}$$ Here, 1 molecule dissociates into n = 1 + 2 = 3$n = 1 + 2 = 3$ ions. 3. Relate i$i$ to the degree of ionization (alpha$\alpha$): i = 1 + (3 - 1)alpha = 1 + 2alpha$$i = 1 + (3 - 1)\alpha = 1 + 2\alpha$$ 2.5 = 1 + 2alpha implies 2alpha = 1.5 implies alpha = 0.75$$2.5 = 1 + 2\alpha \implies 2\alpha = 1.5 \implies \alpha = 0.75$$ 4. Convert to a percentage: textPercent dissociation = 0.75 cdot 100 = 75\%$$\text{Percent dissociation} = 0.75 \cdot 100 = 75\%$$ ### Pattern Recognition For a salt that dissociates into three ions (like mathrmMX_2$\mathrm{MX}_2$), the relationship simplifies to i = 1 + 2alpha$i = 1 + 2\alpha$. Calculating i$i$ from the ratio of the molar masses lets you find alpha$\alpha$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q42 jee_main_2025_24_jan_morning Depression in Freezing Point Consider the given plots of vapour pressure (VP) vs temperature (T/K) Which amongst the following options is correct graphical representation showing Delta T_mathrmf$\Delta T_{\mathrm{f}}$ , depression in the freezing point of solvent in a solution? A. Plot (1) B. Plot (2) C. Plot (3) D. Plot (4) Solution ### Related Formula Delta T_f = T_f^0 - T_f$$\Delta T_f = T_f^0 - T_f$$ ### Core Logic Dissolving a non-volatile solute lower the vapor pressure of the solution relative to the pure solvent across all temperature thresholds. The freezing point is defined as the temperature at which the vapor pressure of the liquid phase matches that of its solid phase. Because the solution's vapor pressure curve lies below that of the pure liquid solvent, its intersection with the frozen solvent curve occurs at a lower temperature (T_f < T_f^0$T_f < T_f^0$). This shift creates the characteristic freezing point depression step: Delta T_f = T_f^0 - T_f$\Delta T_f = T_f^0 - T_f$. Plot (3) correctly displays this thermodynamic behavior. ### Pattern Recognition The vapor pressure curve for the solution always runs lower than that of the pure solvent, shifting the freezing intersection to the left. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q42 jee_main_2025_28_jan_evening Osmosis and Osmotic Pressure Assume a living cell with 0.9\%$0.9\%$ (omega/omega)$(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will: A. Shrink since solution is 0.5\%\ (omega/omega)$0.5\%\ (\omega/\omega)$ B. Shrink since solution is 0.45\%\ (omega/omega)$0.45\%\ (\omega/\omega)$ as a result of association of glucose molecules (due to hydrogen bonding) C. Swell up since solution is 1\%$1\%$ D. Show no change in volume since solution is 0.9\%\ (omega/omega)$0.9\%\ (\omega/\omega)$ Solution ### Related Formula Mass percentage from mole fraction calculation: \%text w/w = fracx_1 cdot M_1x_1 cdot M_1 + x_2 cdot M_2 times 100$$\%\text{ w/w} = \frac{x_1 \cdot M_1}{x_1 \cdot M_1 + x_2 \cdot M_2} \times 100$$ ### Core Logic Inside the living cell, glucose concentration is 0.9\%text w/w$0.9\%\text{ w/w}$. The surrounding solution has equal mole fractions of glucose and water (x_textglucose = 0.5$x_{\text{glucose}} = 0.5$, x_textwater = 0.5$x_{\text{water}} = 0.5$). Let's calculate the mass percentage of the outer solution: - Mass of glucose component = 0.5 times 180 = 90mathrm\ g$0.5 \times 180 = 90\mathrm{\ g}$ - Mass of water component = 0.5 times 18 = 9mathrm\ g$0.5 \times 18 = 9\mathrm{\ g}$ - Total solution mass = 90 + 9 = 99mathrm\ g$90 + 9 = 99\mathrm{\ g}$ ### Step 1: Concentration Determination and Osmosis Profile Outer mass percentage: \%text w/w = frac9099 times 100 approx 90.9\%$$\%\text{ w/w} = \frac{90}{99} \times 100 \approx 90.9\%$$ Because the external environment is highly concentrated (hypertonic) compared to the inner cell (0.9\%$0.9\%$), water flows out of the cell via exosmosis, causing the **cell to shrink**. Note: Because the reasoning establishes shrinkage but the quantitative figures in the options are highly mismatched, this question is officially designated as a **Bonus** question. ### Pattern Recognition An equal mole fraction solution of a high-molar-mass solute (glucose, 180mathrm\ g/mol$180\mathrm{\ g/mol}$) and a low-molar-mass solvent (water, 18mathrm\ g/mol$18\mathrm{\ g/mol}$) is always extremely concentrated by mass. Placing a standard living cell into such a hypertonic solution inevitably causes fluid loss and cellular shrinkage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q jee_main_2025_29_jan_morning Abnormal Molar Mass and van't Hoff Factor 1.24 mathrm~g$1.24 \mathrm{~g}$ of mathrmAX_2$\mathrm{AX}_{2}$ (molar mass 124 mathrm~g mathrm~mol^-1$124 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in 1 mathrm~kg$1 \mathrm{~kg}$ of water to form a solution with boiling point of 100.0156^circ mathrmC$100.0156^{\circ} \mathrm{C}$ , while 25.4 mathrm~g$25.4 \mathrm{~g}$ of mathrmAY_2$\mathrm{AY}_{2}$ (molar mass 250 mathrm~g mathrm~mol^-1$250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in 2 mathrm~kg$2 \mathrm{~kg}$ of water constitutes a solution with a boiling point of 100.0260^circ mathrmC$100.0260^{\circ} \mathrm{C}$ . mathrmK_b(H_2O) = 0.52 \, K \, kg \, mol^-1$\mathrm{K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}}$ Which of the following is correct? A. mathrmAX2$\mathrm{AX}2$ and mathrmAY_2$\mathrm{AY}_2$ (both) are completely unionised. B. mathrmAX_2$\mathrm{AX}_2$ and mathrmAY_2$\mathrm{AY}_2$ (both) are fully ionised. C. mathrmAX_2$\mathrm{AX}_2$ is completely unionised while mathrmAY_2$\mathrm{AY}_2$ is fully ionised. D. mathrmAX_2$\mathrm{AX}_2$ is fully ionised while mathrmAY_2$\mathrm{AY}_2$ is completely unionised. Solution ### Formulas Used Elevation in boiling point formula involving the van't Hoff factor (i$i$): Delta T_b = i cdot K_b cdot m$$\Delta T_b = i \cdot K_b \cdot m$$ Where: * Delta T_b = T_b - T_b^circ$\Delta T_b = T_b - T_b^\circ$ (Boiling point elevation) * K_b = textEbullioscopic constant$K_b = \text{Ebullioscopic constant}$ * m = textMolality of the solution left(fractextmoles of solutetextmass of solvent in kgright)$m = \text{Molality of the solution } \left(\frac{\text{moles of solute}}{\text{mass of solvent in kg}}\right)$ ### Core Logic **Step 1: Evaluate solution system mathrmAX_2$\mathrm{AX}_2$** Delta T_b = 100.0156^circmathrmC - 100.0000^circmathrmC = 0.0156^circmathrmC$$\Delta T_b = 100.0156^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0156^\circ\mathrm{C}$$ textMolality m_1 = frac1.24text g / 124text g mol^-11text kg = 0.01text mol/kg$$\text{Molality } m_1 = \frac{1.24\text{ g} / 124\text{ g mol}^{-1}}{1\text{ kg}} = 0.01\text{ mol/kg}$$ Using the elevation formula: 0.0156 = i_mathrmAX_2 cdot 0.52 cdot 0.01$$0.0156 = i_{\mathrm{AX}_2} \cdot 0.52 \cdot 0.01$$ i_mathrmAX_2 = frac0.01560.0052 = 3$$i_{\mathrm{AX}_2} = \frac{0.0156}{0.0052} = 3$$ Since theoretical dissociation of mathrmAX_2 rightarrow mathrmA^2+ + 2mathrmX^-$\mathrm{AX}_2 \rightarrow \mathrm{A}^{2+} + 2\mathrm{X}^-$ produces 3$3$ particles, i = 3$i = 3$ implies that **mathrmAX_2$\mathrm{AX}_2$ is fully ionised**. --- **Step 2: Evaluate solution system mathrmAY_2$\mathrm{AY}_2$** Delta T_b = 100.0260^circmathrmC - 100.0000^circmathrmC = 0.0260^circmathrmC$$\Delta T_b = 100.0260^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0260^\circ\mathrm{C}$$ textMolality m_2 = frac25.4text g / 250text g mol^-12text kg = 0.0508text mol/kg$$\text{Molality } m_2 = \frac{25.4\text{ g} / 250\text{ g mol}^{-1}}{2\text{ kg}} = 0.0508\text{ mol/kg}$$ Using the elevation formula: 0.0260 = i_mathrmAY_2 cdot 0.52 cdot 0.0508$$0.0260 = i_{\mathrm{AY}_2} \cdot 0.52 \cdot 0.0508$$ i_mathrmAY_2 = frac0.02600.0264 approx 1$$i_{\mathrm{AY}_2} = \frac{0.0260}{0.0264} \approx 1$$ Since i approx 1$i \approx 1$, it behaves as a non-electrolyte, meaning **mathrmAY_2$\mathrm{AY}_2$ is completely unionised**. Thus, **mathrmAX_2$\mathrm{AX}_2$ is fully ionised while mathrmAY_2$\mathrm{AY}_2$ is completely unionised**. ### Pattern Recognition A van't Hoff factor matching the complete stoichiometric ion count (i = 3$i = 3$ for mathrmAX_2$\mathrm{AX}_2$) confirms complete ionisation, whereas a factor near unity (i = 1$i = 1$) indicates no dissociation into separate ions. **Correct Option:** **(D)** 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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