### Related Formula %S = frac32233 times fractextMass of BaSO_4textMass of organic compound times 100$%S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of organic compound}} \times 100$ ### Core Logic Let's substitute the given values into the formula: textMass of BaSO_4 = 0.40text g$\text{Mass of } BaSO_4 = 0.40\text{ g}$ textMass of organic compound = 0.20text g$\text{Mass of organic compound} = 0.20\text{ g}$ textMolar mass of BaSO_4 = 137 + 32 + (4 times 16) = 233text g/mol$\text{Molar mass of } BaSO_4 = 137 + 32 + (4 \times 16) = 233\text{ g/mol}$ %S = frac32233 times frac0.400.20 times 100 = frac32 times 2 times 100233 approx 27.468%$%S = \frac{32}{233} \times \frac{0.40}{0.20} \times 100 = \frac{32 \times 2 \times 100}{233} approx 27.468%$ ### Step 1: Match with the Question Layout Rounding to the standard value given in the official key: %S = 27.5% = 275 times 10^-1% implies x = 275$%S = 27.5% = 275 \times 10^{-1}% implies x = 275$ ### Pattern Recognition Carius method calculations depend heavily on standard conversion factors. The constant factor for sulphur gravimetry is frac32233$\frac{32}{233}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic To evaluate carbocation stability, apply the priority rules: Aromaticity > Resonance > Hyperconjugation. - **C:** Represents a cyclopropenyl cation derivative which achieves full aromatic stabilization due to its planar cyclic conjugated system satisfying Huckel's rule (2pi$2\pi$ electrons). This makes it the most stable. - **A:** Stabilized by extended resonance from multiple phenyl groups. - **B:** Contains fewer phenyl rings participating in active cross-conjugation relative to A. - **D:** Stabilized solely by simple aliphatic hyperconjugation, making it the least stable. Visual alignment chart:

Hence, the correct stability hierarchy is: mathrmC > mathrmA > mathrmB > mathrmD$\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$ ### Pattern Recognition Sees: Mixed aromatic, benzylic, and aliphatic carbocations. Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Organic compounds react with sodium bicarbonate (mathrmNaHCO_3$\mathrm{NaHCO}_3$) to liberate mathrmCO_2$\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid (mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$). Evaluating the structures: - **A:** Benzoic acid, which is significantly more acidic than carbonic acid. - **C:** Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$. - **D:** Benzenesulfonic acid, a highly strong mineral-like organic acid. - **B & E:** Standard phenols or weakly substituted phenols, which are less acidic than carbonic acid and do not liberate mathrmCO_2$\mathrm{CO}_2$. Therefore, structures A, C, and D give a positive test result. ### Pattern Recognition Sees: Sodium bicarbonate test for organic systems. Shortcut: Only carboxylic acids, sulfonic acids, and highly nitrated phenols like picric acid possess sufficient proton acidity to displace mathrmCO_2$\mathrm{CO}_2$ from bicarbonate ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us check the statements step-by-step: * Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached on either side of the divalent polyfunctional carbonyl group (-textCO-$-\text{CO}-$). Hence, they are metamers.

* **Statement B:** Cyanides (-textCN$-\text{CN}$) and Isocyanides (-textNC$-\text{NC}$) contain distinct functional groups, so they are functional isomers.

* **Statement C:** Phenol structures containing a methyl substituent at positions 2 and 3 are structural position isomers. * **Statement D:** The given structures represent members of a homologous series because they differ sequentially by a -textCH_2-$-\text{CH}_2-$ unit. ### Step 1: Verification Evaluating according to standard multi-choice options, statements A and B are perfectly validated. ### Pattern Recognition Shortcut: Metamers require variable alkyl distribution across a polyvalent heteroatom group. Functional isomers require changes like -textCN$-\text{CN}$ vs -textNC$-\text{NC}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us check the conjugate bases formed upon losing a proton: * Compounds (A), (B), and (C) generate conjugate bases stabilized by resonance through the aromatic ring or strong electron-withdrawing groups. * Compound (D) represents an ethynyl group in a terminal alkyne structure (EtO_2C-Cequiv CH$EtO_2C-Cequiv CH$). Its conjugate base features a localized negative charge on an sp$sp$-hybridized carbon. Because there is no resonance stabilization present for this anion, it is significantly less stable than the conjugate bases of the other functional groups. ### Step 1: Conclusion Since a less stable conjugate base implies a weaker parent acid, the terminal alkyne compound (D) is the least acidic. ### Pattern Recognition Shortcut: A resonance-stabilized anion is always more stable than a localized one. Look for the alkyne carbon versus oxygen/aromatic-centered acids. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques