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 3 Q41 jee_main_2025_03_april_morning Elevation in Boiling Point 2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebullioscopic constant of water =0.52text K kg mol^-1$=0.52\text{ K kg mol}^{-1}$) A. 379.2 K B. 377.3 K C. 375.3 K D. 277.3 K Solution ### Related Formula The net boiling point elevation for multiple non-volatile solutes is given by: Delta T_b = (i_1 m_1 + i_2 m_2) K_b$$\Delta T_b = (i_1 m_1 + i_2 m_2) K_b$$ ### Core Logic Both ethylene glycol and glucose are non-electrolytes, so their van 't Hoff factors are equal to unity (i_1 = i_2 = 1$i_1 = i_2 = 1$). textTotal moles of solute = 2 + 2 = 4text moles $$\text{Total moles of solute} = 2 + 2 = 4\text{ moles} $$ textMass of solvent (water) = 500text g = 0.5text kg $$ \text{Mass of solvent (water)} = 500\text{ g} = 0.5\text{ kg} $$ textTotal molality (m) = frac4text mol0.5text kg = 8text mol/kg $$ \text{Total molality (m)} = \frac{4\text{ mol}}{0.5\text{ kg}} = 8\text{ mol/kg} $$ ### Step 1: Compute Elevation and Final Temperature Delta T_b = 8 times 0.52 = 4.16text K $$ Delta T_b = 8 \times 0.52 = 4.16\text{ K} $$ textBoiling point of solution = T_b^circ + Delta T_b = 373.15text K + 4.16text K = 377.31text K approx 377.3text K $$ \text{Boiling point of solution} = T_b^{\circ} + \Delta T_b = 373.15\text{ K} + 4.16\text{ K} = 377.31\text{ K} \approx 377.3\text{ K} $$ ### Pattern Recognition Shortcut: Since both are molecular non-dissociating solutes, simply \sum their moles (2 + 2 = 4$2 + 2 = 4$). Diluting 4 moles in 0.5text kg$0.5\text{ kg}$ gives an effective concentration of 8text m$8\text{ m}$. Multiplying 8 times 0.52$8 \times 0.52$ gives a shift value of 4.16text K$4.16\text{ K}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q32 jee_main_2025_04_april_evening Colligative Properties Given below are two statements : Statement (I): Molal depression constant K_f$K_f$ is given by fracM_lRT_fDelta S_fus$\frac{M_lRT_f}{\Delta S_{fus}}$, where symbols have their usual meaning. Statement (II): K_f$K_f$ for benzene is less than the K_f$K_f$ for water. In the light of the above statements, choose the most appropriate answer from the options given below: A. Statement I is incorrect but Statement II is correct B. Both Statement I and Statement II are incorrect. C. Both Statement I and Statement II are correct D. Statement I is correct but Statement II is incorrect Solution ### Related Formula K_f = fracM_1 R T_f^2Delta H_fus = fracM_1 R T_fleft(fracDelta H_fusT_fright) = fracM_1 R T_fDelta S_fus$$K_f = \frac{M_1 R T_f^2}{\Delta H_{fus}} = \frac{M_1 R T_f}{\left(\frac{\Delta H_{fus}}{T_f}\right)} = \frac{M_1 R T_f}{\Delta S_{fus}}$$ ### Core Logic - **Statement I is correct:** Substituting Delta S_fus = fracDelta H_fusT_f$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$ directly matches the given structural relationship formula. - **Statement II is incorrect:** Standard cryoscopic constants are: - For Benzene: K_f approx 5.12 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 5.12 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$ - For Water: K_f approx 1.86 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 1.86 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$ Therefore, K_f$K_f$ for benzene is greater than that of water, making Statement II false. ### Pattern Recognition Keep numerical benchmarks for common solvent colligative constants (K_b, K_f$K_b, K_f$ for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated K_f$K_f$ value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q46 jee_main_2025_04_april_evening Concentration Terms Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of 2mathrm~g~mL^-1$2\mathrm{~g~mL}^{-1}$ . The concentration of dissolved oxygen left(mathrmO_2right)$\left(\mathrm{O}_2\right)$ in sea water is 5.8mathrm~ppm$5.8\mathrm{~ppm}$ . Then the concentration of dissolved oxygen left(mathrmO_2right)$\left(\mathrm{O}_2\right)$ in sea water, is mathrmx times 10^-4mathrmm$\mathrm{x} \times 10^{-4}\mathrm{m}$ . mathrmx =$\mathrm{x} =$ _______. (Nearest integer) Given: Molar mass of NaCl is 58.5mathrm~g~mol^-1$58.5\mathrm{~g~mol}^{-1}$ Molar mass of mathrmO_2$\mathrm{O}_2$ is 32mathrm~g~mol^-1$32\mathrm{~g~mol}^{-1}$ Numerical Answer. Answer: 1.9 to 2.1 Solution ### Related Formula textppm = fractextmass of solutetextmass of solution times 10^6$$\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$$ textMolality (m) = fractextmoles of solutetextmass of solvent in kg$$\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$ ### Core Logic 1. Consider 1000 mathrm~mL$1000 \mathrm{~mL}$ of seawater solution: textMass of solution = textVolume times textdensity = 1000 times 2 = 2000 mathrm~g$$\text{Mass of solution} = \text{Volume} \times \text{density} = 1000 \times 2 = 2000 \mathrm{~g}$$ textMass of NaCl = 6 text moles times 58.5 = 351 mathrm~g$$\text{Mass of NaCl} = 6 \text{ moles} \times 58.5 = 351 \mathrm{~g}$$ textMass of solvent (water) = 2000 - 351 = 1649 mathrm~g = 1.649 mathrm~kg$$\text{Mass of solvent (water)} = 2000 - 351 = 1649 \mathrm{~g} = 1.649 \mathrm{~kg}$$ 2. Compute the mass and moles of dissolved O_2$O_2$ using the ppm value: textppm = 5.8 = fractextmass of O_22000 times 10^6 implies textmass of O_2 = 1.16 times 10^-2 mathrm~g$$\text{ppm} = 5.8 = \frac{\text{mass of } O_2}{2000} \times 10^6 \implies \text{mass of } O_2 = 1.16 \times 10^{-2} \mathrm{~g}$$ textmoles of O_2 = frac1.16 times 10^-232 = 3.625 times 10^-4 text moles$$\text{moles of } O_2 = \frac{1.16 \times 10^{-2}}{32} = 3.625 \times 10^{-4} \text{ moles}$$ 3. Determine the molality (m$m$) of oxygen: textmolality = frac3.625 times 10^-41.649 approx 2.19 times 10^-4 mathrm~m$$\text{molality} = \frac{3.625 \times 10^{-4}}{1.649} \approx 2.19 \times 10^{-4} \mathrm{~m}$$ Matching the pattern mathbfx times 10^-4mathrmm$\mathbf{x} \times 10^{-4}\mathrm{m}$, we get mathbfx approx 2.19$\mathbf{x} \approx 2.19$. The nearest integer is **2**. ### Pattern Recognition For high concentration saline solutions, the mass of the solvent drops significantly below the total mass of the solution. Be careful to subtract the solute weight (351 mathrm~g$351 \mathrm{~g}$) before computing molality. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q26 jee_main_2025_04_april_morning Reverse Osmosis XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration c_1$c_1$ and c_2$c_2$ (c_1 > c_2$c_1 > c_2$) mathrmmol~L^-1$\mathrm{mol~L^{-1}}$. For the reverse osmosis to take place identify the correct condition (Here p_1$p_1$ and p_2$p_2$ are pressures applied on chamber 1 and 2): A. text(B) and (D) only$\text{(B) and (D) only}$ B. text(A) and (D) only$\text{(A) and (D) only}$ C. text(A) and (C) only$\text{(A) and (C) only}$ D. text(C) only$\text{(C) only}$ Solution ### Related Formula pi = c R T$\pi = c R T$ where pi$\pi$ is the osmotic pressure of the solution. ### Core Logic Given that c_1 > c_2$c_1 > c_2$, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis. To achieve **reverse osmosis**, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that exceeds its osmotic pressure pi$\pi$. textCondition for Reverse Osmosis: p_1 > pi$$\text{Condition for Reverse Osmosis: } p_1 > \pi$$ Cellophane and parchment paper both act as suitable semi-permeable membranes for this setup. Thus, statements (A) and (C) are correct. ### Pattern Recognition Reverse osmosis always requires external pressure applied on the concentrated solution side (c_texthigh$c_{\text{high}}$) such that P_textapplied > pi$P_{\text{applied}} > \pi$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q34 jee_main_2025_07_april_evening Raoult's Law and Liquid-Vapour Composition Liquid textA$\text{A}$ and textB$\text{B}$ form an ideal solution. The vapour pressure of pure liquids textA$\text{A}$ and textB$\text{B}$ are 350 and 750text mm Hg$750\text{ mm Hg}$ respectively at the same temperature. If textx_textA$\text{x}_{\text{A}}$ and textx_textB$\text{x}_{\text{B}}$ are the mole fraction of textA$\text{A}$ and textB$\text{B}$ in solution while texty_textA$\text{y}_{\text{A}}$ and texty_textB$\text{y}_{\text{B}}$ are the mole fraction of textA$\text{A}$ and textB$\text{B}$ in vapour phase then: A. fractextx_textAtextx_textB < fractexty_textAtexty_textB$\frac{\text{x}_\text{A}}{\text{x}_\text{B}} < \frac{\text{y}_\text{A}}{\text{y}_\text{B}}$ B. fractextx_textAtextx_textB = fractexty_textAtexty_textB$\frac{\text{x}_\text{A}}{\text{x}_\text{B}} = \frac{\text{y}_\text{A}}{\text{y}_\text{B}}$ C. fractextx_textAtextx_textB > fractexty_textAtexty_textB$\frac{\text{x}_\text{A}}{\text{x}_\text{B}} > \frac{\text{y}_\text{A}}{\text{y}_\text{B}}$ D. (textx_textA - texty_textA) < (textx_textB - texty_textB)$(\text{x}_\text{A} - \text{y}_\text{A}) < (\text{x}_\text{B} - \text{y}_\text{B})$ Solution ### Related Formula P_textA = y_textA P_texttotal = x_textA P^0_textA$$P_\text{A} = y_\text{A} P_{\text{total}} = x_\text{A} P^0_\text{A}$$ P_textB = y_textB P_texttotal = x_textB P^0_textB$$P_\text{B} = y_\text{B} P_{\text{total}} = x_\text{B} P^0_\text{B}$$ Dividing both partial pressure formulations yields: fracy_textAy_textB = left(fracP^0_textAP^0_textB ight) cdot fracx_textAx_textB $$\frac{y_\text{A}}{y_\text{B}} = \left(\frac{P^0_\text{A}}{P^0_\text{B}} ight) \cdot \frac{x_\text{A}}{x_\text{B}} $$ ### Core Logic Given pure saturation thresholds: P^0_textA = 350text mm Hg, quad P^0_textB = 750text mm Hg $$P^0_\text{A} = 350\text{ mm Hg}, \quad P^0_\text{B} = 750\text{ mm Hg} $$ Comparing pure component volatility profiles: P^0_textA < P^0_textB implies fracP^0_textAP^0_textB < 1 $$P^0_\text{A} < P^0_\text{B} \implies \frac{P^0_\text{A}}{P^0_\text{B}} < 1 $$ Substituting this inequality into the ratio formula gives: fracy_textAy_textB < 1 cdot fracx_textAx_textB implies fracy_textAy_textB < fracx_textAx_textB $$\frac{y_\text{A}}{y_\text{B}} < 1 \cdot \frac{x_\text{A}}{x_\text{B}} \implies \frac{y_\text{A}}{y_\text{B}} < \frac{x_\text{A}}{x_\text{B}} $$ ### Step 1: Rearranging Ratio Forms Inverting the inequality expression fields safely yields: fracx_textAx_textB > fracy_textAy_textB $$\frac{x_\text{A}}{x_\text{B}} > \frac{y_\text{A}}{y_\text{B}} $$ ### Pattern Recognition Konovalov's Rule Shortcut: The vapour phase is always enriched with the more volatile component. Since component textB$\text{B}$ has a higher pure vapour pressure (750 > 350$750 > 350$), it will be preferentially enriched in the vapour phase, meaning y_textB/y_textA > x_textB/x_textA$y_\text{B}/y_\text{A} > x_\text{B}/x_\text{A}$. Reversing the fractions directly matches option (3). ### 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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