In Dumas method for estimation of nitrogen, 0.50 text g$0.50 \text{ g}$ of an organic compound gave 70 text mL$70 \text{ mL}$ of nitrogen collected at 300 text K$300 \text{ K}$ and 715 text mm$715 \text{ mm}$ pressure. The percentage of nitrogen in the organic compound is ____% (Aqueous tension at 300 text K$300 \text{ K}$ is 15 text mm$15 \text{ mm}$).

### Related Formula textPercentage of Sulphur = fractextMass of S text in BaSO_4textMolar mass of BaSO_4 times fractextMass of BaSO_4 text formedtextMass of organic compound times 100$$\text{Percentage of Sulphur} = \frac{\text{Mass of } S \text{ in } BaSO_4}{\text{Molar mass of } BaSO_4} \times \frac{\text{Mass of } BaSO_4 \text{ formed}}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Molar mass of BaSO_4$BaSO_4$ = 233 g/mol. Mass of Sulfur (S) in 1 mole of BaSO_4$BaSO_4$ = 32 g. Mass of BaSO_4$BaSO_4$ formed = 1.2 g. Mass of organic compound = 0.75 g. \% mathrmS = frac32233 times frac1.20.75 times 100$$\% \mathrm{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100$$ \% mathrmS = frac32 times 1.2 times 100233 times 0.75 = frac3840174.75 approx 21.97\%$$\% \mathrm{S} = \frac{32 \times 1.2 \times 100}{233 \times 0.75} = \frac{3840}{174.75} \approx 21.97\%$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic In structure 3, there are positive formal charges on two adjacent atoms (Oxygen and the Carbon adjacent to it). Like charges on adjacent atoms cause extreme electrostatic repulsion, making the structure highly unstable.

### Pattern Recognition Rules of resonance stability: 1. Complete octets are more stable. 2. More covalent bonds = more stable. 3. Least charge separation is more stable. 4. Like charges on adjacent atoms create massive destabilization (Least stable scenario). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic A. Propanal (CH_3CH_2CHO$CH_3CH_2CHO$) and propanone (CH_3COCH_3$CH_3COCH_3$) have different functional groups (aldehyde vs ketone) but the same molecular formula. They are functional isomers. (Correct) B. Ethoxyethane (C_2H_5-O-C_2H_5$C_2H_5-O-C_2H_5$) and methoxypropane (CH_3-O-C_3H_7$CH_3-O-C_3H_7$) differ in the alkyl chains attached to the polyvalent oxygen atom. They are metamers. (Correct) C. But-2-ene shows geometrical isomerism (cis-trans), but no optical isomerism as it lacks a chiral center. (Incorrect) D. But-1-ene and but-2-ene differ in the position of the double bond. They are position isomers, not functional isomers. (Incorrect) E. Pentane (straight chain) and 2,2-dimethyl propane (branched chain) have the same formula C_5H_12$C_5H_{12}$ but different carbon skeletons. They are chain isomers. (Correct) Correct statements: A, B, and E. ### Pattern Recognition Metamerism arises when there is a difference in the alkyl groups attached to a polyvalent functional group (e.g., -O-, -S-, -NH-, -CO-). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic - A. 2-Methylpropene and but-1-ene differ in carbon chain structure rightarrow$\rightarrow$ III. Chain isomers - B. Cis-but-2-ene and trans-but-2-ene differ in spatial arrangement rightarrow$\rightarrow$ I. Stereoisomers - C. 2-Butanol (alcohol) and diethyl ether (ether) have different functional groups rightarrow$\rightarrow$ II. Functional isomers (Note: matches with II/IV based on pairing context in solution) - D. But-1-ene and but-2-ene differ in position of double bond rightarrow$\rightarrow$ IV. Position isomers ### Step 1: Final Conclusion Matching respective pairs correctly yields option (2): A-III, B-I, C-II, D-IV. ### Pattern Recognition Sees: Match list of isomerism pairs. Trap: Mixing up position and chain isomers for alkenes. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Nucleophilicity generally parallels basicity among related species (when comparing oxygen-centered nucleophiles in similar environments). Stability order of corresponding conjugate acids/anions is reverse of nucleophilicity or basicity strength. Basicity/Nucleophilicity order: ^-textOH > textPhO^- > textCH_3textCOO^- > textClO_4^-$^-\text{OH} > \text{PhO}^- > \text{CH}_3\text{COO}^- > \text{ClO}_4^-$. ### Step 1: Final Conclusion Thus, option (4) represents the correct nucleophilicity order. ### Pattern Recognition Sees: Nucleophilic substitution rate and nucleophilicity order. Trap: Confusing leaving group ability with nucleophile strength. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques