Q.74 Chemistry ⭐ 360° Option Elimination Active

NEET 2023 Chemistry Question 74 Solution

Topic: Hydrocarbons · Official NTA Booklet Code G5

Question

Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is:

(1)18
(2)16
(3)32
(4)30

✓

Verified Against NTA Final Answer Key

Option (3) : 32

📚 Hydrocarbons Easy

📋 Given Parameters & Problem Setup

Reactant Sodium ethanoate (CH₃COONa)

Reagent Soda-lime (NaOH + CaO in 3:1 ratio, Δ)

Target quantity Weight (in grams) of TWO MOLES of the organic product

⚡

The 30-Second Exam Hall Shortcut: ⚡ Decarboxylation Product & 2-Mole Mass Hack

⏱️ Save ~30 seconds

CH₃COONa + NaOH ⟶ CH₄ + Na₂CO₃. The organic compound is methane (CH₄, molar mass = 16 g/mol). Two moles weigh 2 × 16 = 32 g in 10 seconds! (Option 3)

1 Identify the reaction: Heating sodium ethanoate with soda-lime (NaOH + CaO) is the classic decarboxylation reaction.

2 The carboxyl group is cleaved as Na₂CO₃: CH₃COONa + NaOH ⟶ CH₄ + Na₂CO₃.

3 The organic compound formed is methane (CH₄).

4 Calculate the molar mass of methane: M(CH₄) = 12 + 4(1) = 16 g mol⁻¹.

5 The question specifically asks for the weight of TWO MOLES of the compound:

6 Mass = 2 mol × 16 g mol⁻¹ = 32 g.

7 Lock Option (3) 32 immediately!

🛡️

Why You Can 100% Trust This Shortcut: Organic Syntheses · Decarboxylation of Carboxylic Acids (Sodium salts of carboxylic acids heated with soda-lime lose carbon dioxide to form alkanes containing one carbon less than the carboxylic acid).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Equation for Soda-Lime Decarboxylation

When sodium ethanoate is heated with soda-lime (a dry mixture of sodium hydroxide and calcium oxide in an approximate 3:1 mass ratio), it undergoes decarboxylation to produce an alkane with one less carbon atom than the parent carboxylate salt:

\text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO, } \Delta} \text{CH}_4(g) + \text{Na}_2\text{CO}_3(s)

Step 2

Identification of the Organic Product

Of the two products formed, Na₂CO₃ is an inorganic salt, so the organic compound obtained is methane (CH₄).

\text{Organic Product} = \text{Methane } (\text{CH}_4)

Step 3

Molar Mass Calculation of Methane

Calculate the molecular weight of methane from standard atomic masses (C = 12 u, H = 1 u):

M(\text{CH}_4) = 12 + 4(1) = 16\text{ g mol}^{-1}

Step 4

Evaluation of the Mass of Two Moles

The problem explicitly asks for the weight in grams of TWO MOLES of this organic compound:

\text{Mass} = n \times M = 2\text{ mol} \times 16\text{ g mol}^{-1} = 32\text{ g}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Molar mass of water (H₂O), completely irrelevant to methane.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate confused the product with water.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Mass of ONE mole of methane (CH₄), forgetting that the question asks for TWO moles.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate solved for 1 mole of CH₄ (16 g) and forgot to multiply by 2.

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate weight of 2 moles of methane: 2 × 16 g = 32 g.

💡

Why this is airtight: None. Candidate accounted for the "two moles" multiplier.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Molar mass of ethane (C₂H₆), which would form in Kolbe’s electrolysis, not soda-lime decarboxylation.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate confused soda-lime decarboxylation with Kolbe’s electrolytic coupling.

⚠️

NTA Trap Alert: Reporting the Molar Mass (16 g) Instead of Two Moles (32 g)

The most frequent mistake on this question is correctly identifying methane (CH₄, 16 g/mol) and immediately circling 16 (Option 2). The question prompt explicitly starts with "Weight (g) of TWO MOLES..."!

🛡️ How to avoid the -1 mark: Always reread the question stem: "two moles" means 2 × 16 = 32 g.

📖

Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Hydrocarbons Standard Benchmark

Core Scientific Principle Tested:

"Decarboxylation of sodium ethanoate yields methane: CH₃COONa + NaOH ⟶ CH₄ + Na₂CO₃. 2 moles of CH₄ weigh 32 g."

Syllabus Focus: Alkanes – Preparation from Carboxylic Acids (Decarboxylation)

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2023 CBT Test →

← Previous (Q.73) All Chemistry Questions Grid Next (Q.75) →

Sequential Reaction: Anti-Markovnikov HBr Addition, Cyanide Substitution & Mendius Reduction

View Solution →

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2023 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2023 Chemistry Question 74 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Decarboxylation Product & 2-Mole Mass Hack

Step-by-Step Rigorous Workflow & Mathematical Proof

Equation for Soda-Lime Decarboxylation

Identification of the Organic Product

Molar Mass Calculation of Methane

Evaluation of the Mass of Two Moles

360° Option Elimination Engine Key Feature

NTA Trap Alert: Reporting the Molar Mass (16 g) Instead of Two Moles (32 g)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Hydrocarbons

Grignard Reagent Acid-Base Reaction with Primary Amines

Stereospecific Electrophilic & Syn Additions to Alkenes (TAM & CSM Rules)

Stereoselective Hydrogenation of Alkynes (Lindlar vs Birch Reduction)

Sequential Reaction: Anti-Markovnikov HBr Addition, Cyanide Substitution & Mendius Reduction

Solve 4 NEET Questions a Day — One per Subject