In Dumas' method 292text mg$292\text{ mg}$ of an organic compound released 50text mL$50\text{ mL}$ of nitrogen gas (textN_2)$(\text{N}_2)$ at 300text K$300\text{ K}$ temperature and 715text mm Hg$715\text{ mm Hg}$ pressure. [cite: 435, 451] The percentage composition of 'N' in the organic compound is dots$\dots$ % (Nearest integer) [cite: 452, 455] (Aqueous tension at 300text K = 15text mm Hg$300\text{ K} = 15\text{ mm Hg}$)

### Core Logic * **Statement I is true:** In the conjugated system (CH_3-CH=CH-CH=O$CH_3-CH=CH-CH=O$), extended resonance delocalization occurs, shifting electron density toward the carbonyl oxygen: CH_3-oversetoplusCH-CH=CH-oversetominusO$$CH_3-\overset{\oplus}{C}H-CH=CH-\overset{\ominus}{O}$$ This configuration increases both the partial charge separation q$q$ and the dipole distance d, resulting in a significantly larger net dipole moment mu = q times d$\mu = q \times d$ compared to the non-conjugated saturated aldehyde. * **Statement II is false:** Due to resonance conjugation, the single bond between C_1$C_1$ and C_2$C_2$ acquires partial double-bond character. This double-bond character shortens the bond length, making it smaller than the standard single bond found in CH_3-CH_2-CH_2-CH=O$CH_3-CH_2-CH_2-CH=O$. ### Pattern Recognition Conjugation spreads partial charges across a longer carbon chain, expanding the charge separation distance to drive up the overall dipole moment mu$\mu$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula P_textdry N_2 = P_texttotal - P_textaqueous tension$$P_{\text{dry } N_2} = P_{\text{total}} - P_{\text{aqueous tension}}$$ PV = nRT implies n = fracPVRT$$PV = nRT \implies n = \frac{PV}{RT}$$ \% N = fractextMass of NitrogentextMass of Organic Compound times 100$$\% N = \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \times 100$$ ### Core Logic First, calculate the actual pressure exerted by the dry nitrogen gas: P_N_2 = 900 - 15 = 885mathrm~mmHg = frac885760mathrm~atm$$P_{N_2} = 900 - 15 = 885\mathrm{~mmHg} = \frac{885}{760}\mathrm{~atm}$$ Convert volume data to liters: V = 150mathrm~mL = 0.15mathrm~L$V = 150\mathrm{~mL} = 0.15\mathrm{~L}$. Using the ideal gas law to determine the moles of N_2$N_2$ collected: n = fracleft(frac885760right) times 0.150.0821 times 300 = frac1.1645 times 0.1524.63 approx 0.0071mathrm~moles$$n = \frac{\left(\frac{885}{760}\right) \times 0.15}{0.0821 \times 300} = \frac{1.1645 \times 0.15}{24.63} \approx 0.0071\mathrm{~moles}$$ Calculate the total mass of the liberated nitrogen gas: textMass = n times M_textmolar = 0.0071 times 28 = 0.1988mathrm~g$$\text{Mass} = n \times M_{\text{molar}} = 0.0071 \times 28 = 0.1988\mathrm{~g}$$ Determine the percentage composition relative to the initial 1mathrm~g$1\mathrm{~g}$ sample size: \% N = frac0.19881 times 100 = 19.88\% approx 20\%$$\% N = \frac{0.1988}{1} \times 100 = 19.88\% \approx 20\%$$ ### Pattern Recognition Always remember to subtract the aqueous tension value first. Failing to correct for water vapor pressure is the most common pitfall in Dumas' method calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula textPh-COOH + textNaOH ightarrow textPh-COO^-textNa^+ text (Water soluble salt) [cite: 884, 885]$$\text{Ph-COOH} + \text{NaOH} ightarrow \text{Ph-COO}^-\text{Na}^+ \text{ (Water soluble salt)} [cite: 884, 885]$$ ### Core Logic When aqueous textNaOH$\text{NaOH}$ is added to the mixture: 1. Benzoic acid (textPh-COOH$\text{Ph-COOH}$) reacts to form sodium benzoate (textPh-COONa$\text{Ph-COONa}$), which is highly water-soluble and moves completely into the aqueous layer. 2. Chlorobenzene (textPh-Cl$\text{Ph-Cl}$) and aniline (textPh-NH_2$\text{Ph-NH}_2$) are organic compounds that do not react with aqueous textNaOH$\text{NaOH}$ under normal conditions. Therefore, they remain unreacted in the organic layer (ethyl acetate). ### Step 1: Visual Reaction Progress The separation steps can be visualized tracking the structural components:

Thus, the ethyl acetate layer strictly contains chlorobenzene and aniline. ### Pattern Recognition Acid-base separation shortcut: Strong bases (textNaOH$\text{NaOH}$) pull organic acids into the aqueous phase as salts. Neutral compounds and basic amines stay behind in the organic layer unless a strong acid is added. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic To determine carbon atom hybridization within skeletal networks, evaluate the count of steric components (sigma$\sigma$-bonds attached to each carbon): * 4 text sigmatext-bonds ightarrow sp^3$4 \text{ } \sigma\text{-bonds} ightarrow sp^3$ * 3 text sigmatext-bonds ightarrow sp^2$3 \text{ } \sigma\text{-bonds} ightarrow sp^2$ (typically carbons forming one double bond like mathrmC=O$\mathrm{C=O}$ or mathrmC=C$\mathrm{C=C}$) * 2 text sigmatext-bonds ightarrow sp$2 \text{ } \sigma\text{-bonds} ightarrow sp$ (typically carbons forming a triple bond like -mathrmCequiv C-$-\mathrm{C\equiv C}-$ or -mathrmCequiv N$-\mathrm{C\equiv N}$) ### Step 1: Specific Atom Assignment Let's perform an audit across the skeletal sequence:

* sp^2$sp^2$ Carbons: 1. Carbonyl carbon (mathrmC=O$\mathrm{C=O}$) 2. Carbons sharing the first alkene motif (2 atoms) 3. Carbons sharing the second alkene motif (2 atoms) Total sp^2$sp^2$ carbons = 1 + 2 + 2 = 5$1 + 2 + 2 = 5$ * sp$sp$ Carbons: 1. Carbons bound in the central alkyne group (-mathrmCequiv C-$-\mathrm{C\equiv C}-$) (2 atoms) 2. Terminal nitrile carbon (-mathrmCequiv N$-\mathrm{C\equiv N}$) (1 atom) Total sp$sp$ carbons = 2 + 1 = 3$2 + 1 = 3$ ### Pattern Recognition Count all carbons involved in triple bonds (alkynes, nitriles) to find sp$sp$ items. Count all double-bonded carbons (alkenes, ketones) to quickly isolate the sp^2$sp^2$ population. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques