### Related Formula \% text Bromine = fractextMolar Mass of BrtextMolar Mass of AgBr cdot fractextMass of AgBr formedtextMass of Organic Sample cdot 100$\% \text{ Bromine} = \frac{\text{Molar Mass of Br}}{\text{Molar Mass of AgBr}} \cdot \frac{\text{Mass of AgBr formed}}{\text{Mass of Organic Sample}} \cdot 100$ ### Core Logic 1. Calculate the molar mass of silver bromide (mathrmAgBr$\mathrm{AgBr}$): textMolar Mass of AgBr = 108 + 80 = 188text g/mol$\text{Molar Mass of AgBr} = 108 + 80 = 188\text{ g/mol}$ 2. Substitute the given values into the Carius quantitative formula: * textMass of AgBr = 0.15text g$\text{Mass of AgBr} = 0.15\text{ g}$ * textMass of sample = 0.25text g$\text{Mass of sample} = 0.25\text{ g}$ \% mathrmBr = frac80188 cdot frac0.150.25 cdot 100$\% \mathrm{Br} = \frac{80}{188} \cdot \frac{0.15}{0.25} \cdot 100$ \% mathrmBr = frac80188 cdot 0.6 cdot 100 = frac48188 cdot 100 = 25.5319\%$\% \mathrm{Br} = \frac{80}{188} \cdot 0.6 \cdot 100 = \frac{48}{188} \cdot 100 = 25.5319\%$ 3. Convert the percentage to match the requested output units (times 10^-1\%$\times 10^{-1}\%$): 25.5319\% = 255.319 cdot 10^-1\% approx 255 cdot 10^-1\%$25.5319\% = 255.319 \cdot 10^{-1}\% \approx 255 \cdot 10^{-1}\%$ Rounding to the nearest integer gives 255. ### Pattern Recognition Pay close attention to the final multiplier units requested in the blank (times 10^-1\%$\times 10^{-1}\%$). Always calculate the raw percentage first, then adjust the decimal place to match the required format. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Statement I is fully accurate: Dumas method is standardly applied for estimating elemental nitrogen across structural compounds. Statement II is incorrect: The reaction leading to ammonium sulphate generation by intense thermal heating alongside concentrated mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ describes the **Kjeldahl method**, not the Dumas strategy. The Dumas process instead relies on burning carbonaceous compounds explicitly with copper oxide to convert nitrogen cleanly into free gas (N_2$N_2$). ### Pattern Recognition Dumas method collects element gas N_2$N_2$ via volumetric analysis; Kjeldahl maps digestions via standard (mathrmNH_4)_2mathrmSO_4$(\mathrm{NH}_4)_2\mathrm{SO}_4$ pathways. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Direct phase transition without passing through an intermediate liquid phase defines sublimation: textSolid rightleftharpoons textVapour$\text{Solid} \rightleftharpoons \text{Vapour}$ ### Core Logic The schematic diagram represents a solid turning directly into vapor on heating, which then reverts back to a solid phase upon cooling. This distinct behavior isolates sublimable solids from non-sublimable impurities. ### Step 1: Identification This transformation perfectly defines the laboratory purification process known as **Sublimation**. ### Pattern Recognition Look for the skipping of the liquid state entirely: Solid rightarrow$\rightarrow$ Vapour rightarrow$\rightarrow$ Solid. Common examples include camphor, naphthalene, benzoic acid, and iodine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Isomers share an identical molecular formula but differ in structural arrangement or functional groups: textSame M_f neq textSame structural layout$\text{Same } M_f \neq \text{Same structural layout}$ ### Core Logic Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C_3H_6O$C_3H_6O$. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^circ$1^\circ$ amine) and N-methylethanamine (2^circ$2^\circ$ amine) both share the molecular formula C_3H_9N$C_3H_9N$. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. ### Step 1: Conclusion Match Since both structural statements are valid, both Statement I and Statement II are true.

### Pattern Recognition Always remember that 1^circ$1^\circ$, 2^circ$2^\circ$, and 3^circ$3^\circ$ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques