Grade 10 : Science - 4. Carbon Compounds (Set 1) | Concept of Maths

Question 1

1. Q1. Isomerism is a unique characteristic of carbon compounds. Calculate the total number of structural isomers possible for Hexane ($C_6H_{14}$) and Heptane ($C_7H_{16}$) respectively.

A) 5 and 9

B) 5 and 8

C) 6 and 9

D) 4 and 9

Answer

Correct Answer: A

Solution: - Hexane ($C_6H_{14}$) has 5 structural isomers: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. - Heptane ($C_7H_{16}$) has 9 structural isomers: n-heptane, 2-methylhexane, 3-methylhexane, 2,2-dimethylpentane, 2,3-dimethylpentane, 2,4-dimethylpentane, 3,3-dimethylpentane, 3-ethylpentane, and 2,2,3-trimethylbutane. Thus, the correct answer is 5 and 9 (Option A).

Question 2

2. Q2. An organic compound $A$ ($C_2H_4O_2$) reacts with $B$ ($C_2H_6O$) in the presence of an acid catalyst to produce a sweet-smelling ester $C$. When $C$ is treated with $NaOH$ solution (saponification), it yields ethanol and a sodium salt of compound $A$. If $0.1\text{ mol}$ of ester $C$ is fully saponified, what is the exact mass of the sodium salt formed? (Molar mass of $Na = 23\text{ g/mol}$, $C = 12\text{ g/mol}$, $O = 16\text{ g/mol}$, $H = 1\text{ g/mol}$)

A) $4.1\text{ g}$

B) $8.2\text{ g}$

C) $12.3\text{ g}$

D) $16.4\text{ g}$

Answer

Correct Answer: B

Solution: 1. Compound $A$ is ethanoic acid ($CH_3COOH$) and $B$ is ethanol ($C_2H_5OH$). 2. Reaction: $CH_3COOH + C_2H_5OH \xrightarrow{\text{Acid}} CH_3COOC_2H_5 \text{ (Ester } C) + H_2O$. 3. Saponification: $CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH$. 4. Molar mass of Sodium Acetate ($CH_3COONa$): $$M = (12 \times 2) + (1 \times 3) + (16 \times 2) + 23 = 24 + 3 + 32 + 23 = 82\text{ g/mol}$$ 5. Mass of $0.1\text{ mol}$ of $CH_3COONa$ = $0.1 \times 82 = 8.2\text{ g}$. Thus, the correct answer is $8.2\text{ g}$ (Option B).

Question 3

3. Q3. A gaseous hydrocarbon mixture containing $10\text{ mL}$ of a gaseous hydrocarbon $C_xH_y$ and $40\text{ mL}$ of oxygen gas is exploded. After cooling to room temperature, the residual gas volume is $30\text{ mL}$. When this remaining gas is treated with a potassium hydroxide ($KOH$) solution, the volume shrinks to $10\text{ mL}$ (representing unused oxygen). Identify the molecular formula of the hydrocarbon.

A) $C_2H_6$

B) $C_2H_4$

C) $C_3H_8$

D) $C_2H_2$

Answer

Correct Answer: B

Solution: 1. Reaction stoichiometry: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O(l)$. 2. Liquid water volume is negligible. Remaining gas ($30\text{ mL}$) contains $CO_2$ and unused $O_2$. 3. On adding $KOH$, $CO_2$ is absorbed. Decrease in volume = $30 - 10 = 20\text{ mL}$, which is the volume of $CO_2$ produced. 4. Since $10\text{ mL}$ of $C_xH_y$ gives $20\text{ mL}$ of $CO_2$, we have $10x = 20 \implies x = 2$. 5. Unused $O_2 = 10\text{ mL}$, so reacted $O_2 = 40 - 10 = 30\text{ mL}$. 6. Reacted $O_2$ stoichiometry: $10 \times (x + y/4) = 30 \implies 2 + y/4 = 3 \implies y = 4$. 7. Thus, the hydrocarbon is $C_2H_4$ (Ethene). Correct Option: $C_2H_4$ (Option B).

Question 4

4. Q4. In the photochemical chlorination of methane, which of the following reaction conditions should be preferred to maximize the selective yield of chloromethane ($CH_3Cl$) and minimize further substitution products like dichloromethane ($CH_2Cl_2$)?

A) Excess of chlorine gas ($Cl_2$)

B) Excess of methane gas ($CH_4$)

C) Equimolar ratio of chlorine and methane in dark

D) Addition of tetraethyl lead catalyst in cold conditions

Answer

Correct Answer: B

Solution: - If chlorine is in excess, free radicals continue to abstract hydrogen from $CH_3Cl$, yielding $CH_2Cl_2$, $CHCl_3$, and eventually $CCl_4$. - Using a large excess of methane ($CH_4$) ensures that chlorine radicals are far more likely to collide and react with unreacted $CH_4$ molecules rather than newly formed $CH_3Cl$, thereby maximizing the yield of monochloromethane ($CH_3Cl$). Thus, the correct answer is Excess of methane gas ($CH_4$) (Option B).

Question 5

5. Q5. An aliphatic acyclic organic compound with molecular formula $C_4H_6O$ has a certain degree of unsaturation. Which of the following functional groups / structural features must coexist in the compound to satisfy this formula?

A) One carbon-carbon double bond and one alcohol group

B) One carbon-carbon double bond and one aldehyde group (or ketone)

C) One triple bond and one ketone group

D) Two carbon-carbon double bonds and one carboxylic acid group

Answer

Correct Answer: B

Solution: 1. Let's find the Double Bond Equivalent (DBE) or Degree of Unsaturation for $C_4H_6O$: $$DBE = C + 1 - \frac{H}{2} = 4 + 1 - \frac{6}{2} = 5 - 3 = 2$$ 2. A DBE of 2 indicates either: 2 double bonds, 1 triple bond, 1 ring + 1 double bond, or 2 rings. 3. Let's analyze Option B: One $C=C$ double bond (DBE = 1) + one $C=O$ carbonyl group from the aldehyde/ketone (DBE = 1). Total DBE = 2. This perfectly satisfies the formula $C_4H_6O$ (e.g., but-2-enal: $CH_3-CH=CH-CHO$). Thus, the correct answer is Option B.

Question 6

6. Q6. An unknown hydrocarbon $X$ has the molecular formula $C_5H_8$. It rapidly decolourises bromine water and, upon treatment with ammoniacal silver nitrate solution, forms a characteristic white precipitate. What is the structural IUPAC configuration of $X$?

A) $CH_3-CH_2-CH_2-C\equiv CH$

B) $CH_3-CH_2-C\equiv C-CH_3$

C) $CH_2=CH-CH_2-CH=CH_2$

D) Cyclopentene

Answer

Correct Answer: A

Solution: - The molecular formula $C_5H_8$ has $DBE = 5 + 1 - 4 = 2$. It represents an alkyne, alkadiene, or cyclic alkene. - Decolourisation of bromine water indicates unsaturation ($C=C$ or $C\equiv C$). - Reaction with ammoniacal silver nitrate ($AgNO_3$) to form a white precipitate of silver acetylide is unique to terminal alkynes (alkynes with a hydrogen attached directly to the triple-bonded carbon, $-C\equiv C-H$). - $CH_3-CH_2-CH_2-C\equiv CH$ (Pent-1-yne) is a terminal alkyne, whereas Pent-2-yne is non-terminal and will not react. Thus, the correct answer is $CH_3-CH_2-CH_2-C\equiv CH$ (Option A).

Question 7

7. Q7. Which of the following statements is INCORRECT regarding micelle formation and the cleansing action of soaps?

A) Soap molecules consist of a hydrophilic ionic head and a hydrophobic long hydrocarbon tail.

B) In a water-based micelle, the hydrophobic tails align pointing towards the core, away from water.

C) In an organic solvent like ethanol, soap molecules form robust micelles with tails pointing outwards.

D) Soap micelles remain suspended as colloid because of mutual ion-ion repulsion of their charged heads.

Answer

Correct Answer: C

Solution: - Soap is highly soluble in ethanol because ethanol is an organic solvent that dissolves the long hydrocarbon tail perfectly. Hence, soap molecules do not aggregate to form micelles in ethanol. - Statement C is incorrect, which makes it the correct choice for this question. Thus, the correct option is C.

Question 8

8. Q8. Ethanol ($CH_3CH_2OH$) is treated with concentrated sulfuric acid ($H_2SO_4$) at different temperatures. What is the role of $H_2SO_4$ at $443\text{ K}$, and what is the major organic product if the reaction is executed at a lower temperature of $413\text{ K}$?

A) Oxidising agent; Ethene

B) Dehydrating agent; Diethyl ether ($C_2H_5-O-C_2H_5$)

C) Reducing agent; Ethane

D) Hydrating agent; Ethanoic acid

Answer

Correct Answer: B

Solution: 1. At $443\text{ K}$, concentrated sulfuric acid acts as a dehydrating agent, removing water from ethanol to yield ethene ($CH_2=CH_2$). 2. At a lower temperature ($413\text{ K}$), intermolecular dehydration occurs instead of intramolecular dehydration, yielding Diethyl ether ($C_2H_5-O-C_2H_5$) as the major product: $$2 C_2H_5OH \xrightarrow{\text{conc. } H_2SO_4, 413\text{ K}} C_2H_5-O-C_2H_5 + H_2O$$ Thus, the correct option is B.

Question 9

9. Q9. An organic compound $P$ is oxidized with alkaline $KMnO_4$ upon heating to form compound $Q$. Under acidic conditions, $P$ reacts with $Q$ to form a sweet-smelling compound $R$ ($C_4H_8O_2$). Identify the compounds $P$ and $Q$.

A) $P$ = Methanol, $Q$ = Methanoic acid

B) $P$ = Ethanol, $Q$ = Ethanoic acid

C) $P$ = Propanol, $Q$ = Propanoic acid

D) $P$ = Ethanol, $Q$ = Methanoic acid

Answer

Correct Answer: B

Solution: - Sweet-smelling compound $R$ with formula $C_4H_8O_2$ must be an ester. Since it is formed from an alcohol $P$ and a carboxylic acid $Q$, and $P$ oxidizes directly to $Q$, both $P$ and $Q$ must share the same number of carbons. - Splitting $4$ carbons equally gives $2$ carbons for the alcohol and $2$ carbons for the acid. - $P$ is Ethanol ($C_2H_5OH$) and $Q$ is Ethanoic acid ($CH_3COOH$). - Reaction: $CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5$ (Ethyl ethanoate, $C_4H_8O_2$) + $H_2O$. Thus, the correct answer is $P$ = Ethanol, $Q$ = Ethanoic acid (Option B).

Question 10

10. Q10. Consider the following assertions regarding a homologous series of carbon compounds: 1. The difference in molecular formula between any two successive members is $-CH_2-$, corresponding to a difference of $14\text{ u}$ in molecular mass. 2. Members of a series exhibit a gradual graduation in physical properties (e.g., boiling points rise with increase in molecular mass). 3. All members of a homologous series have identical melting and boiling points. 4. All members share similar chemical properties and can be represented by the same general formula. Select the combination containing only CORRECT statements:

A) 1, 2, and 4

B) 1, 3, and 4

C) 2 and 4 only

D) 1 and 2 only

Answer

Correct Answer: A

Solution: - Statement 1 is correct: successive members differ by $-CH_2-$ ($14\text{ u}$). - Statement 2 is correct: boiling and melting points rise regularly with increasing molecular mass due to enhanced van der Waals forces. - Statement 3 is incorrect: physical properties are NOT identical; they show a clear gradient. - Statement 4 is correct: because they share functional groups, their chemical properties are very similar. Thus, the correct combination is 1, 2, and 4 (Option A).

Grade 10 : Science - 4. Carbon Compounds (Set 1) | Concept of Maths