In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32text g mol^-1$32\text{ g mol}^{-1}$). Molar mass of barium sulphate is 233text g mol^-1$233\text{ g mol}^{-1}$.
A.4.55\%$4.55\%$
B.10.30\%$10.30\%$
C.21.97\%$21.97\%$
D.16.48\%$16.48\%$
Solution & Explanation
### 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
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#Carius method#percentage of sulphur#JEE Main 2026 Morning Q58#Organic Chemistry - Some Basic Principles and Techniques JEE Main 2026
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 6
Q43jee_main_2025_28_jan_morningAcidity of Organic Compounds
The compounds that produce mathrmCO_2$\mathrm{CO}_{2}$ with aqueous mathrmNaHCO_3$\mathrm{NaHCO}_{3}$ 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.textA and C only$\text{A and C only}$
B.textA, B and E only$\text{A, B and E only}$
C.textA, C and D only$\text{A, C and D only}$
D.textA and B only$\text{A and B only}$
Solution
### Core Logic
Organic compounds react with sodium bicarbonate (mathrmNaHCO_3$\mathrm{NaHCO}_3$) to liberate mathrmCO_2$\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid (mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$).
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 mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$.
- **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 mathrmCO_2$\mathrm{CO}_2$.
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 mathrmCO_2$\mathrm{CO}_2$ from bicarbonate ions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Identify the correct statements from the following:
A. textCH_3textCH_2textCOCH_2textCH_3$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ and textCH_3textCOCH_2textCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$ are metamers
B. textCH_3textCH_2textCH_2textCN$\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$ and textCH_3textCH_2textCH_2textNC$\text{CH}_3\text{CH}_2\text{CH}_2\text{NC}$ are functional isomers
C. 2-methylphenol and 3-methylphenol are position isomers
D. textCH_3textCH_2textNH_2$\text{CH}_3\text{CH}_2\text{NH}_2$ and textCH_3textCH_2textCH_2textNH_2$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ are homologous
Choose the correct answer from the options given below.
A. C & D only
B. B & C only
C. A & B only
D. A, B & C only
Solution
### Core Logic
Let us check the statements step-by-step:
* Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached on either side of the divalent polyfunctional carbonyl group (-textCO-$-\text{CO}-$). Hence, they are metamers. Metamerism illustration for Q32 - JEE Main 2025 Morning
* **Statement B:** Cyanides (-textCN$-\text{CN}$) and Isocyanides (-textNC$-\text{NC}$) contain distinct functional groups, so they are functional isomers. Metamerism illustration for Q32 - JEE Main 2025 Morning
* **Statement C:** Phenol structures containing a methyl substituent at positions 2 and 3 are structural position isomers.
* **Statement D:** The given structures represent members of a homologous series because they differ sequentially by a -textCH_2-$-\text{CH}_2-$ unit.
### Step 1: Verification
Evaluating according to standard multi-choice options, statements A and B are perfectly validated.
### Pattern Recognition
Shortcut: Metamers require variable alkyl distribution across a polyvalent heteroatom group. Functional isomers require changes like -textCN$-\text{CN}$ vs -textNC$-\text{NC}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q44jee_main_2025_03_april_morningAcidic Strength of Organic Compounds
The least acidic compound, among the following is:
A. Compound (D)
B. Compound (A)
C. Compound (B)
D. Compound (C)
Solution
### Core Logic
Let us check the conjugate bases formed upon losing a proton:
* Compounds (A), (B), and (C) generate conjugate bases stabilized by resonance through the aromatic ring or strong electron-withdrawing groups.
* Compound (D) represents an ethynyl group in a terminal alkyne structure (EtO_2C-Cequiv CH$EtO_2C-Cequiv CH$). Its conjugate base features a localized negative charge on an sp$sp$-hybridized carbon. Because there is no resonance stabilization present for this anion, it is significantly less stable than the conjugate bases of the other functional groups.
### Step 1: Conclusion
Since a less stable conjugate base implies a weaker parent acid, the terminal alkyne compound (D) is the least acidic.
### Pattern Recognition
Shortcut: A resonance-stabilized anion is always more stable than a localized one. Look for the alkyne carbon versus oxygen/aromatic-centered acids.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
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)
Numerical Answer.Answer: 111 to 111
Solution
### 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$86\text{ g/mol}$. 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 = 0.42text 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
Q47jee_main_2025_03_april_morningQuantitative Analysis - Estimation of Carbon
0.5 g of an organic compound on combustion gave 1.46 g of CO_2$CO_{2}$ and 0.9 g of H_2O$H_{2}O$. The percentage of carbon in the compound is _____. (Nearest integer)
[Given: Molar mass (in textg mol^-1$\text{g mol}^{-1}$) C: 12, H: 1, O: 16]
Numerical Answer.Answer: 80 to 80
Solution
### 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
More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2026_21_jan_morning
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 "Dumas method quantitative, Molar weight tracking, Nitrogen gas evolution")
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