Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.8% of Chemistry.
This question is from Quantitative Analysis - Dumas Method.
During estimation of nitrogen by Dumas' method of compound X (0.42 g):
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
mL of N2$N{2}$ gas will be liberated at STP. (nearest integer)
(Given molar mass in g mol: C: 12, H: 1, N: 14)
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Numerical Answer Type:
Enter a numerical valueAnswer: 111 to 111+4 marks
Solution & Explanation
Related Formula
Using the Principle of Atom Conservation (POAC) for Nitrogen:
ncompound × (atoms of N per molecule) = 2 × nN₂$$n_{\text{compound}} \times (\text{atoms of N per molecule}) = 2 \times n_{\text{N}_2}$$
Core Logic
The molecular weight of the given heterocyclic amine organic structure X$X$ (piperazine, C₄H₁₀N₂$\text{C}_4\text{H}_{10}\text{N}_2$) is calculated as:
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Given mass of compound = 0.42 g$= 0.42\text{ g}$:
Moles of compound X = (0.42)/(86) mol$$\text{Moles of compound } X = \frac{0.42}{86}\text{ mol}$$
Step 1: Calculating STP Volume
Since each molecule contains 2$2$ nitrogen atoms, 1 mol$1\text{ mol}$ of compound produces 1 mol$1\text{ mol}$ of N₂$\text{N}_2$ gas:
Using standard molar volume at STP (22700 mL/mol$22700\text{ mL/mol}$ per IUPAC convention, or 22400 mL/mol$22400\text{ mL/mol}$ in traditional calculations):
Volume of N₂ at STP = (0.42)/(86) × 22700 mL ≈ 110.86 mL ≈ 111 mL$$\text{Volume of } \text{N}_2\text{ at STP} = \frac{0.42}{86} \times 22700\text{ mL} \approx 110.86\text{ mL} \approx 111\text{ mL}$$
(Note: If calculated using 22400 mL/mol$22400\text{ mL/mol}$, Volume = (0.42)/(86) × 22400 ≈ 109.4 mL ≈ 109 mL$\text{Volume} = \frac{0.42}{86} \times 22400 \approx 109.4\text{ mL} \approx 109\text{ mL}$.)
Rounding to the nearest integer gives 111 (official accepted range: 109 to 111).
Pattern Recognition
Shortcut: Determine the molar mass (M = 86 g/mol$M = 86\text{ g/mol}$) and nitrogen atom count (2 N atoms 1 mol N₂ per mol of compound$2\text{ N atoms} \implies 1\text{ mol } \text{N}_2\text{ per mol of compound}$). Multiply moles directly by molar volume at STP to find the liberated gas volume.
Evaluation Rubric / Model Answer
111
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Keywords:#Dumas method quantitative#Molar weight tracking#Nitrogen gas evolution
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 9
Given below are two statements:
Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support.
Statement (II): In paper chromatography, the material of paper acts as a stationary phase.
In the light of the above statements, choose the correct answer from the options given below:
A. Both Statement I and Statement II are false
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are true
D. Statement I is false but Statement II is true
Solution
Core Logic
Statement I is true: In partition chromatography, the stationary phase is indeed a thin film of liquid held on the surface of an inert solid support.
Statement II is false: In paper chromatography, the water molecules trapped inside the cellulose network of the paper act as the stationary phase, not the paper material itself.
Pattern Recognition
Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q38jee_main_2025_29_jan_eveningSigma and Pi Bond Counting
Total number of sigma (sigma)$(sigma)$ and pi(pi)$pi(pi)$ bonds respectively present in hex-1-en-4-yne are:
Sigma and Pi Bond Counting diagram for Q38 - JEE Main 2025 Evening
Number of pi$pi$ bonds: 1$1$ from double bond + 2$2$ from triple bond = 3$3$pi$pi$ bonds.
Pattern Recognition
Every single bond is 1sigma$1sigma$, every double bond contains 1sigma + 1pi$1sigma + 1pi$, and every triple bond contains 1sigma + 2pi$1sigma + 2pi$.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q49jee_main_2025_29_jan_eveningQuantitative Estimation of Sulphur
In the sulphur estimation, 0.20 g$0.20\text{ g}$ of a pure organic compound gave 0.40 g$0.40\text{ g}$ of barium sulphate.
The percentage of sulphur in the compound is x × 10⁻¹%$x \times 10^{-1}\%$, where x$x$ = ________.
(Molar mass: O=16$O=16$, S=32$S=32$, Ba=137 in g mol⁻¹$Ba=137\text{ in g mol}^{-1}$)
Numerical Answer.Answer: 275 to 275
Solution
Related Formula
%S = (32)/(233) × Mass of BaSO₄Mass of organic compound × 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:
Mass of BaSO₄ = 0.40 g$$\text{Mass of } BaSO_4 = 0.40\text{ g}$$Mass of organic compound = 0.20 g$$\text{Mass of organic compound} = 0.20\text{ g}$$Molar mass of BaSO₄ = 137 + 32 + (4 × 16) = 233 g/mol$$\text{Molar mass of } BaSO_4 = 137 + 32 + (4 \times 16) = 233\text{ g/mol}$$%S = (32)/(233) × (0.40)/(0.20) × 100 = (32 × 2 × 100)/(233) 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:
The correct order of stability of following carbocations is :
The images show different structural models labeled A, B, C, and D for evaluating stability variations.
A
The images show different structural models labeled A, B, C, and D for evaluating stability variations.
B
The images show different structural models labeled A, B, C, and D for evaluating stability variations.
C
The images show different structural models labeled A, B, C, and D for evaluating stability variations.
D
A.A > B > C > D$\mathrm{A} > \mathrm{B} > \mathrm{C} > \mathrm{D}$
B.B > C > A > D$\mathrm{B} > \mathrm{C} > \mathrm{A} > \mathrm{D}$
C.C > B > A > D$\mathrm{C} > \mathrm{B} > \mathrm{A} > \mathrm{D}$
D.C > A > B > D$\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$
Solution
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 (2π$2\pi$ 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:
The images show different structural models labeled A, B, C, and D for evaluating stability variations.
Hence, the correct stability hierarchy is:
C > A > B > D$$\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$$
Pattern Recognition
Sees: Mixed aromatic, benzylic, and aliphatic carbocations.
Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q43jee_main_2025_28_jan_morningAcidity of Organic Compounds
The compounds that produce CO₂$\mathrm{CO}_{2}$ with aqueous NaHCO₃$\mathrm{NaHCO}_{3}$ solution are:
A. The prompt lists five structures labeled A through E evaluating structural acidities.
B. The prompt lists five structures labeled A through E evaluating structural acidities.
C. The prompt lists five structures labeled A through E evaluating structural acidities.
D. The prompt lists five structures labeled A through E evaluating structural acidities.
E. The prompt lists five structures labeled A through E evaluating structural acidities.
Choose the correct answer from the options given below:
A.A and C only$\text{A and C only}$
B.A, B and E only$\text{A, B and E only}$
C.A, C and D only$\text{A, C and D only}$
D.A and B only$\text{A and B only}$
Solution
Core Logic
Organic compounds react with sodium bicarbonate (NaHCO₃$\mathrm{NaHCO}_3$) to liberate CO₂$\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid (H₂CO₃$\mathrm{H}_2\mathrm{CO}_3$).
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 H₂CO₃$\mathrm{H}_2\mathrm{CO}_3$.
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 CO₂$\mathrm{CO}_2$.
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 CO₂$\mathrm{CO}_2$ from bicarbonate ions.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
 "Dumas method quantitative, Molar weight tracking, Nitrogen gas evolution")
 "Dumas method quantitative, Molar weight tracking, Nitrogen gas evolution")
