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 5 Q jee_main_2025_29_jan_morning van't Hoff Factor If mathrmA_2mathrmB$\mathrm{A}_2\mathrm{B}$ is 30\%$30\%$ ionised in an aqueous solution, then the value of van't Hoff factor (i) is ________ times 10^-1$\times 10^{-1}$. Numerical Answer. Answer: 16 to 16 Solution ### Related Formula i = 1 + (y - 1)alpha$$i = 1 + (y - 1)\alpha$$ ### Core Logic For electrolyte mathrmA_2mathrmB$\mathrm{A}_2\mathrm{B}$ undergoing dissociation: mathrmA_2B rightarrow 2mathrmA^+ + mathrmB^2-$$\mathrm{A_2B} \rightarrow 2\mathrm{A}^+ + \mathrm{B}^{2-}$$ Total count of ions produced per molecule y = 3$y = 3$. Given degree of dissociation alpha = 30\% = 0.3$\alpha = 30\% = 0.3$. Substituting into the formula: i = 1 + (3 - 1) cdot 0.3$$i = 1 + (3 - 1) \cdot 0.3$$ i = 1 + 2 cdot 0.3 = 1 + 0.6 = 1.6$$i = 1 + 2 \cdot 0.3 = 1 + 0.6 = 1.6$$ Expressing the value in the requested format: 1.6 = 16 times 10^-1$$1.6 = 16 \times 10^{-1}$$ Thus, the integer value to enter is 16. ### Pattern Recognition Always calculate total stoichiometric species y$y$ carefully before executing the linear factor combination to avoid basic arithmetic errors. ### Chapter Mix Class 12 Chemistry: Solutions Q75 jee_main_2024_01_february_morning Abnormal Molar Masses We have three aqueous solutions of NaCl$NaCl$ labelled as 'A', 'B' and 'C' with concentration 0.1 mathrm~M$0.1 \mathrm{~M}$, 0.01 mathrm~M$0.01 \mathrm{~M}$ & 0.001 mathrm~M$0.001 \mathrm{~M}$, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order. A. i_A < i_B < i_C$i_A < i_B < i_C$ B. i_A < i_C < i_B$i_A < i_C < i_B$ C. i_A = i_B = i_C$i_A = i_B = i_C$ D. i_A > i_B > i_C$i_A > i_B > i_C$ Solution ### Core Logic For a strong electrolyte like NaCl$NaCl$, the theoretical van 't Hoff factor i_theory$i_{theory}$ is 2 (Na^+$Na^+$ and Cl^-$Cl^-$). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed i$i$. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed i$i$ approaches the theoretical value. ### Step 1: Correlate Concentration with i$i$ Higher concentration implies$\implies$ more ion-pairing implies$\implies$ lower i$i$. Given concentrations: A = 0.1 mathrm~M$A = 0.1 \mathrm{~M}$ (highest concentration) B = 0.01 mathrm~M$B = 0.01 \mathrm{~M}$ C = 0.001 mathrm~M$C = 0.001 \mathrm{~M}$ (most dilute) Therefore, the actual i$i$ values follow the reverse order of concentration: i_A (0.1 mathrm~M) < i_B (0.01 mathrm~M) < i_C (0.001 mathrm~M)$i_A (0.1 \mathrm{~M}) < i_B (0.01 \mathrm{~M}) < i_C (0.001 \mathrm{~M})$. ### Execution Salt Values of i (for different conc. of a Salt) 0.1 M 0.01 M 0.001 M NaCl 1.87 1.94 1.97 i$i$ approaches 2 as the solution becomes very dilute. ### Pattern Recognition For strong electrolytes, effective dissociation (and thus i$i$) increases as dilution increases because ions interfere with each other less. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q84 jee_main_2024_29_january_evening Interconversion of Concentration Terms Molality of 0.8mathrmM$0.8mathrm{M}$ mathrmH_2mathrmSO_4$mathrm{H}_2mathrm{SO}_4$ solution (density 1.06mathrmgcm^-3$1.06mathrm{gcm}^{-3}$) is ________ times 10^-3mathrmm$\times 10^{-3}\mathrm{m}$. Numerical Answer. Answer: 815 to 815 Solution ### Related Formula m = fracM times 1000, (1000 times d) - (M times M_B)$$m = \frac{M \times 1000}, {(1000 \times d) - (M \times M_B)}$$ where, M$M$ = Molarity = 0.8text M$0.8\text{ M}$ d$d$ = Density = 1.06text g/cm^3$1.06\text{ g/cm}^3$ M_B$M_B$ = Molar mass of solute (H_2SO_4$H_2SO_4$) = 98text g/mol$98\text{ g/mol}$ ### Core Logic Substituting the given values into the equation: m = frac0.8 times 1000, (1000 times 1.06) - (0.8 times 98)$$m = \frac{0.8 \times 1000}, {(1000 \times 1.06) - (0.8 \times 98)}$$ Calculating the denominator parameters: textDenominator = 1060 - 78.4 = 981.6text g$$\text{Denominator} = 1060 - 78.4 = 981.6\text{ g}$$ ### Step 1: Final Resolution Solving for molality: m = frac800, 981.6 approx 0.815text m = 815 times 10^-3text m$$m = \frac{800}, {981.6} \approx 0.815\text{ m} = 815 \times 10^{-3}\text{ m}$$ Thus, the integer factor value is **815**. ### Pattern Recognition Ensure you explicitly subtract the mass of the solute from the total mass of the solution to correctly isolate the mass of the solvent needed for molality calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Q67 jee_main_2024_27_jan_morning Vapour Pressure and Deviations from Raoult's Law A solution of two miscible liquids showing negative deviation from Raoult's law will have: A. increased vapour pressure, increased boiling point B. increased vapour pressure, decreased boiling point C. decreased vapour pressure, decreased boiling point D. decreased vapour pressure, increased boiling point Solution ### Core Logic A system demonstrating a negative deviation from Raoult's law implies tighter molecular attractions between components (A-B$A-B$ interactions are stronger than A-A$A-A$ or B-B$B-B$). This decreases the aggregate escaping tendency, yielding a decreased total vapour pressure. Consequently, a higher thermal energy threshold is required to reach the boiling threshold, causing an increased boiling point. ### Pattern Recognition Negative deviation rightarrow$\rightarrow$ Vapour Pressure drops rightarrow$\rightarrow$ Boiling Point rises inversely. ### Chapter Mix Class 12 Chemistry: Solutions Q85 jee_main_2024_29_jan_morning Concentration Terms A solution of mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ is 31.4\%$31.4\%$ mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ by mass and has a density of 1.25mathrmg / mL$1.25\mathrm{g / mL}$ . The molarity of the mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ solution is \_\_\_\_\_\_ mathrmM$\_\_\_\_\_\_ \mathrm{M}$ (nearest integer) [Given molar mass of mathrmH_2mathrmSO_4 = 98mathrmg mol^-1$\mathrm{H}_2\mathrm{SO}_4 = 98\mathrm{g mol}^{-1}$ ] Numerical Answer. Answer: 4 to 4 Solution ### Related Formula textMolarity (M) = frac\% text by mass times 10 times dM_w$$\text{Molarity } (M) = \frac{\% \text{ by mass} \times 10 \times d}{M_w}$$ ### Core Logic Let's assume we have 100 g of the solution. Mass of H_2SO_4$H_2SO_4$ in 100 g solution = 31.4text g$31.4\text{ g}$. Moles of H_2SO_4$H_2SO_4$ (n_textsolute$n_{\text{solute}}$) = frac31.498text mol$\frac{31.4}{98}\text{ mol}$. Volume of the solution (V$V$) can be found using density: V = fractextMass of solutiontextDensity = frac1001.25text mL$$V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100}{1.25}\text{ mL}$$ ### Step 1: Calculating Molarity Molarity is defined as moles of solute per liter of solution: M = fracn_textsoluteV(textin mL) times 1000$$M = \frac{n_{\text{solute}}}{V(\text{in mL})} \times 1000$$ M = frac31.4 / 98100 / 1.25 times 1000$$M = \frac{31.4 / 98}{100 / 1.25} \times 1000$$ M = frac31.4 times 1.2598 times 100 times 1000$$M = \frac{31.4 \times 1.25}{98 \times 100} \times 1000$$ M = frac39.2598 times 10$$M = \frac{39.25}{98} \times 10$$ M = 0.4005 times 10$$M = 0.4005 \times 10$$ M = 4.005text M$$M = 4.005\text{ M}$$ Rounding off to the nearest integer gives 4. ### Pattern Recognition Whenever percentage by mass (w/w$w/w$) and density (d$d$ in g/mL) are given, use the direct formula: M = frac\%(w/w) times d times 10M_w$M = \frac{\%(w/w) \times d \times 10}{M_w}$. This saves enormous time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry 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 . The molarity of orthophosphoric acid ( ) having purity by The mass of sodium acetate required to prepare of aqueous SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Login Sign Up Dashboard Rankbit System Revise Mistakes Rankbit JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) Report Issue Question Issue Type Wrong Answer Wrong Solution Formula Error Typo / Language Issue Diagram Issue Option Mismatch Duplicate Question Out of Syllabus Other Optional Notes Cancel Submit Report Quick Review Question Accuracy Feedback Accurate Needs Fix Difficulty Feedback Easy Medium Hard Confusion Score Low High Optional Notes Cancel Submit Review Welcome back Continue with Google or Sign In Don't have an account? 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