### Related Formula Using the Principle of Atom Conservation (POAC) for Nitrogen: n_textcompound times textatoms of N per molecule = 2 times n_N_2$n_{\text{compound}} \times \text{atoms of N per molecule} = 2 \times n_{N_2}$ ### Core Logic The molecular weight of the given heterocyclic amine organic structure X is calculated as 86text g/mol$86\text{ g/mol}$.

Given mass of compound = 0.42text g$= 0.42\text{ g}$: textMoles of compound X = frac0.4286 $\text{Moles of compound } X = \frac{0.42}{86} $ ### Step 1: Calculating STP Volume Using POAC on Nitrogen atoms: n_N_2 = frac0.4286 $ n_{N_2} = \frac{0.42}{86} $ textVolume of N_2text at STP = n_N_2 times 22400text mL = frac0.4286 times 22400 approx 110.88text mL $ \text{Volume of } N_2\text{ at STP} = n_{N_2} \times 22400\text{ mL} = \frac{0.42}{86} \times 22400 \approx 110.88\text{ mL} $ Rounding to the nearest integer gives 111text mL$111\text{ mL}$. ### Pattern Recognition Shortcut: Always identify the molecular formula from the skeletal grid first. Once M = 86$M = 86$ and total textN = 2$\text{N} = 2$ atoms are established, use the stoichiometric ratio directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula The percentage of carbon via combustion details is found using: % text C = frac1244 times fractextMass of CO_2textMass of organic compound times 100$ % \text{ C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of organic compound}} \times 100$ ### Core Logic Let us substitute the parameters: * Mass of organic compound = 0.5text g$= 0.5\text{ g}$ * Mass of CO_2$CO_2$ collected = 1.46text g$= 1.46\text{ g}$ ### Step 1: Numerical Calculation \% text C = frac1244 times frac1.460.5 times 100$\% \text{ C} = \frac{12}{44} \times \frac{1.46}{0.5} \times 100$ % text C = frac12 times 1.4622 times 100 approx 79.63%$% \text{ C} = \frac{12 \times 1.46}{22} \times 100 \approx 79.63%$ Rounding to the nearest integer gives 80$80$. ### Pattern Recognition Shortcut: frac1244 approx 0.2727$\frac{12}{44} \approx 0.2727$. Multiply 0.2727 times 1.46$0.2727 \times 1.46$ to find the carbon mass (0.398text g$0.398\text{ g}$). Since 0.398text g$0.398\text{ g}$ out of 0.5text g$0.5\text{ g}$ is practically frac45$\frac{4}{5}$, the value is \right around 80\%$80\%$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### 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