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 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 PQ_2$PQ_{2}$. When 1 g of PQ is dissolved in 50 g of solvent ‘A’. Delta T_b$\Delta T_{b}$ was 1.176 K while when 1 g of PQ_2$PQ_{2}$ is dissolved in 50 g of solvent ‘A’, Delta T_b$\Delta T_{b}$ was 0.689 K. (K_b$K_{b}$ of ‘A’ = 5text K kg mol^-1$5\text{ K kg mol}^{-1}$). The molar masses of elements P and Q (in textg 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 Delta T_b = K_b times m$$\Delta T_b = K_b \times m$$ m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight 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: (Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m$$(\Delta \mathrm{T_b})_{\mathrm{PQ}} = \mathrm{K_b} \cdot m$$ 1.176 = 5 times frac1mathrmM_1 times frac100050$$1.176 = 5 \times \frac{1}{\mathrm{M}_1} \times \frac{1000}{50}$$ mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol$$\mathrm{M}_1 = \frac{5 \times 20}{1.176} = 85.03 \text{ g/mol}$$ For compound PQ_2$PQ_2$: (Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689$$(\Delta \mathrm{T_b})_{\mathrm{PQ}_2} = 5 \times \frac{1}{\mathrm{M}_2} \times \frac{1000}{50} = 0.689$$ mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol$$\mathrm{M}_2 = \frac{5 \times 20}{0.689} = 145.13 \text{ g/mol}$$ Let molar mass of P & Q be mathrmM_P$\mathrm{M}_P$ and mathrmM_Q$\mathrm{M}_Q$ respectively: mathrmM_P + mathrmM_Q = 85.03 quad text--- (1)$$\mathrm{M}_P + \mathrm{M}_Q = 85.03 \quad \text{--- (1)}$$ mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2)$$\mathrm{M}_P + 2\mathrm{M}_Q = 145.13 \quad \text{--- (2)}$$ Subtracting (1) from (2): mathrmM_Q = 145.13 - 85.03 = 60.1 approx 60 text g/mol$$\mathrm{M}_Q = 145.13 - 85.03 = 60.1 \approx 60 \text{ g/mol}$$ Substituting back into (1): mathrmM_P + 60.1 = 85.03 implies mathrmM_P = 24.93 approx 25 text g/mol$$\mathrm{M}_P + 60.1 = 85.03 \implies \mathrm{M}_P = 24.93 \approx 25 \text{ g/mol}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q37 jee_main_2025_02_april_evening Molarity and Temperature Dependency 'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1^circmathrmC$1^{\circ}\mathrm{C}$ to 25^circmathrmC$25^{\circ}\mathrm{C}$. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range] A. textPlot (1)$\text{Plot (1)}$ B. textPlot (2)$\text{Plot (2)}$ C. textPlot (3)$\text{Plot (3)}$ D. textPlot (4)$\text{Plot (4)}$ Solution ### Related Formula textMolarity (M) = fracn_textsoluteV_textsolution (mathrmL)$$\text{Molarity } (M) = \frac{n_{\text{solute}}}{V_{\text{solution}} (\mathrm{L})}$$ ### Core Logic Since solubility of mathrmNaCl$\mathrm{NaCl}$ remains unchanged, the number of dissolved moles of mathrmNaCl$\mathrm{NaCl}$ solute (n_textsolute$n_{\text{solute}}$) remains strictly constant. Thus, molarity M$M$ is strictly dependent on the volume of water (solvent) as temperature changes: M propto frac1V_textsolution$$M \propto \frac{1}{V_{\text{solution}}}$$ ### Step 1: Understand Water's Anomalous Expansion Water exhibits unique anomalous density behavior near freezing: - From 1^circmathrmC$1^{\circ}\mathrm{C}$ to 4^circmathrmC$4^{\circ}\mathrm{C}$, the density of water **increases** to a maximum. This contraction means the volume (V$V$) of water **decreases**. - From 4^circmathrmC$4^{\circ}\mathrm{C}$ to 25^circmathrmC$25^{\circ}\mathrm{C}$, the density of water **decreases** due to standard thermal expansion. Consequently, the volume (V$V$) **increases**. ### Step 2: Relate Volume to Molarity Because volume is in the denominator of the molarity equation: - From 1^circmathrmC$1^{\circ}\mathrm{C}$ to 4^circmathrmC$4^{\circ}\mathrm{C}$: Volume decreases implies$\implies$ Molarity **increases**. - At 4^circmathrmC$4^{\circ}\mathrm{C}$: Volume is minimum implies$\implies$ Molarity reaches a **maximum**. - From 4^circmathrmC$4^{\circ}\mathrm{C}$ to 25^circmathrmC$25^{\circ}\mathrm{C}$: Volume increases implies$\implies$ Molarity **decreases**. This behavior is perfectly represented by **Plot (2)**, which features a distinct peak around 4^circmathrmC$4^{\circ}\mathrm{C}$. ### Pattern Recognition Water is at its densest (and occupies minimum volume) at exactly 3.98^circmathrmC$3.98^\circ\mathrm{C}$ (4^circmathrmC$4^\circ\mathrm{C}$). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q49 jee_main_2025_02_april_evening Elevation of Boiling Point and Molar Mass Determination When 1~mathrmg$1~\mathrm{g}$ each of compounds AB and mathrmAB_2$\mathrm{AB}_2$ are dissolved in 15~mathrmg$15~\mathrm{g}$ of water separately, they increased the boiling point of water by 2.7~mathrmK$2.7~\mathrm{K}$ and 1.5~mathrmK$1.5~\mathrm{K}$ respectively. The atomic mass of A (in amu) is times 10^-1$\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is 0.5~mathrmK~kg~mol^-1$0.5~\mathrm{K~kg~mol^{-1}}$) Numerical Answer. Answer: 25 to 25 Solution ### Related Formula Delta T_mathrmb = K_mathrmb cdot m = K_mathrmb cdot left( fracw_textsoluteM_textsolute cdot frac1000w_textsolvent right)$$\Delta T_{\mathrm{b}} = K_{\mathrm{b}} \cdot m = K_{\mathrm{b}} \cdot \left( \frac{w_{\text{solute}}}{M_{\text{solute}}} \cdot \frac{1000}{w_{\text{solvent}}} \right)$$ ### Core Logic Both dissolved compounds are non-electrolytes (van 't Hoff factor i = 1$i = 1$). We calculate the molar masses of mathrmAB$\mathrm{AB}$ and mathrmAB_2$\mathrm{AB_2}$ individually, then solve for the individual atomic masses of elements A and B. ### Step 1: Determine Molar Mass of AB Given Delta T_mathrmb = 2.7~mathrmK$\Delta T_{\mathrm{b}} = 2.7~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5~mathrmK~kg~mol^-1$K_{\mathrm{b}} = 0.5~\mathrm{K~kg~mol^{-1}}$: 2.7 = 0.5 times frac1M_mathrmAB times frac100015$$2.7 = 0.5 \times \frac{1}{M_{\mathrm{AB}}} \times \frac{1000}{15}$$ M_mathrmAB = frac0.5 times 100015 times 2.7 = frac50040.5 approx 12.3457~mathrmg~mol^-1$$M_{\mathrm{AB}} = \frac{0.5 \times 1000}{15 \times 2.7} = \frac{500}{40.5} \approx 12.3457~\mathrm{g~mol^{-1}}$$ ### Step 2: Determine Molar Mass of AB_2 Given Delta T_mathrmb = 1.5~mathrmK$\Delta T_{\mathrm{b}} = 1.5~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5$K_{\mathrm{b}} = 0.5$: 1.5 = 0.5 times frac1M_mathrmAB_2 times frac100015$$1.5 = 0.5 \times \frac{1}{M_{\mathrm{AB_2}}} \times \frac{1000}{15}$$ M_mathrmAB_2 = frac0.5 times 100015 times 1.5 = frac50022.5 approx 22.2222~mathrmg~mol^-1$$M_{\mathrm{AB_2}} = \frac{0.5 \times 1000}{15 \times 1.5} = \frac{500}{22.5} \approx 22.2222~\mathrm{g~mol^{-1}}$$ ### Step 3: Solve for Atomic Mass of A Let the atomic masses of elements A and B be a$a$ and b$b$ respectively: a + b = 12.3457 quad text--- (1)$$a + b = 12.3457 \quad \text{--- (1)}$$ a + 2b = 22.2222 quad text--- (2)$$a + 2b = 22.2222 \quad \text{--- (2)}$$ Subtracting equation (1) from (2): b = 22.2222 - 12.3457 = 9.8765~mathrmamu$$b = 22.2222 - 12.3457 = 9.8765~\mathrm{amu}$$ Substituting b$b$ back into equation (1): a = 12.3457 - 9.8765 = 2.4692~mathrmamu$$a = 12.3457 - 9.8765 = 2.4692~\mathrm{amu}$$ Expressing a$a$ in the requested format (times 10^-1$\times 10^{-1}$): a = 24.692 times 10^-1 approx 25 times 10^-1$$a = 24.692 \times 10^{-1} \approx 25 \times 10^{-1}$$ ### Pattern Recognition Mathematical consistency checks: Since mathrmAB_2$\mathrm{AB_2}$ has more atoms than mathrmAB$\mathrm{AB}$ of identical constituent mass, its molar mass should be higher, leading to a smaller elevation of boiling point for the same mass dissolved, which perfectly matches our 2.7~mathrmK rightarrow 1.5~mathrmK$2.7~\mathrm{K} \rightarrow 1.5~\mathrm{K}$ change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q jee_main_2025_02_april_morning Henry's Law Constant and Temperature Dependance Which of the following graph correctly represents the plots of mathrmK_H$\mathrm{K_H}$ at 1 bar gases in water versus temperature? A. B. C. D. Solution ### Related Formula Henry's Law formula connects partial pressure to solubility component: p = K_mathrmH cdot x$$p = K_{\mathrm{H}} \cdot x$$ ### Core Logic As temperature increases, gas dissolution is typically exothermic, meaning solubility initially drops, causing the Henry's constant K_mathrmH$K_{\mathrm{H}}$ to curve upward dynamically before varying at extreme points. For standard non-reactive noble/molecular gases at regular ranges, the magnitude order follows: K_mathrmH(mathrmHe) > K_mathrmH(mathrmN_2) > K_mathrmH(mathrmCH_4)$$K_{\mathrm{H}}(\mathrm{He}) > K_{\mathrm{H}}(\mathrm{N_2}) > K_{\mathrm{H}}(\mathrm{CH_4})$$ ### Step 1: Selection Graph (4) illustrates the correct relative order and curved profile properly across the given temperature frame. ### Pattern Recognition Higher K_mathrmH$K_{\mathrm{H}}$ value implies lower solubility of that gas at a given pressure. Helium is notoriously insoluble in water compared to organic or polarizable molecules like methane, hence its plot line must live at the top. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q32 jee_main_2025_02_april_morning Raoult's Law and Vapour Pressure A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mathrm~mmHg$200 \mathrm{~mmHg}$ and that of the solution is 500 mathrm~mmHg$500 \mathrm{~mmHg}$. The vapour pressure of pure B and the least volatile component of the solution, respectively, are : A. (1)\ 1400\ mathrmmmHg,\ textA$(1)\ 1400\ \mathrm{mmHg},\ \text{A}$ B. (2)\ 1400\ mathrmmmHg,\ textB$(2)\ 1400\ \mathrm{mmHg},\ \text{B}$ C. (3)\ 600\ mathrmmmHg,\ textB$(3)\ 600\ \mathrm{mmHg},\ \text{B}$ D. (4)\ 600\ mathrmmmHg,\ textA$(4)\ 600\ \mathrm{mmHg},\ \text{A}$ Solution ### Related Formula Raoult's Law for a multi-component solution mixtures: P_mathrmS = P_mathrmA^0 cdot X_mathrmA + P_mathrmB^0 cdot X_mathrmB$$P_{\mathrm{S}} = P_{\mathrm{A}}^0 \cdot X_{\mathrm{A}} + P_{\mathrm{B}}^0 \cdot X_{\mathrm{B}}$$ ### Core Logic Let's determine the mole fractions first based on the input molar amounts: X_mathrmA = frac11+3 = frac14, quad X_mathrmB = frac31+3 = frac34$$X_{\mathrm{A}} = \frac{1}{1+3} = \frac{1}{4}, \quad X_{\mathrm{B}} = \frac{3}{1+3} = \frac{3}{4}$$ Substitute the known properties into Raoult's equation block: 500 = 200 times frac14 + P_mathrmB^0 times frac34$$500 = 200 \times \frac{1}{4} + P_{\mathrm{B}}^0 \times \frac{3}{4}$$ 500 = 50 + P_mathrmB^0 times frac34$$500 = 50 + P_{\mathrm{B}}^0 \times \frac{3}{4}$$ 450 = P_mathrmB^0 times frac34 implies P_mathrmB^0 = 600 mathrm~mmHg$$450 = P_{\mathrm{B}}^0 \times \frac{3}{4} \implies P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$$ Comparing pure state components pressures: P_mathrmA^0 = 200 mathrm~mmHg$P_{\mathrm{A}}^0 = 200 \mathrm{~mmHg}$ and P_mathrmB^0 = 600 mathrm~mmHg$P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$. Lower vapor pressure indicates stronger intermolecular cohesion, making A the least volatile component. ### Step 1: Finalization Thus, the vapour pressure of pure B is 600mathrm~mmHg$600\mathrm{~mmHg}$ and the least volatile chemical is A. ### Pattern Recognition Volatility is directly proportional to pure vapor pressure (P^0$P^0$). Don't mix up 'least volatile' with 'lowest mole fraction'—always evaluate based solely on the isolated values of P^0$P^0$. ### 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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