Q.82 Chemistry · Section A ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 82 Solution

Topic: Some Basic Concepts of Chemistry · Official NTA Booklet Code T1

Question

1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

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Verified Against NTA Final Answer Key

Option (4) : 250 mg

📚 Some Basic Concepts of Chemistry Easy

📋 Given Parameters & Problem Setup

Mass of NaOH initially taken 1 gram = 1000 mg

Volume of HCl solution 25 mL = 0.025 L

Molarity of HCl solution 0.75 M

Molar mass of NaOH 23 + 16 + 1 = 40 g/mol

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The 30-Second Exam Hall Shortcut: ⚡ Millimole Subtraction Shortcut

⏱️ Save ~45 seconds

NaOH taken = 1000 mg / 40 = 25 mmol. HCl added = 25 mL × 0.75 M = 18.75 mmol. Unreacted NaOH = 25 - 18.75 = 6.25 mmol -> 6.25 × 40 = 250 mg!

1 Calculate millimoles of NaOH initially present: mmol = (1.0 g / 40 g/mol) × 1000 = 25.0 mmol.

2 Calculate millimoles of HCl added: mmol = V(mL) × M = 25 mL × 0.75 M = 18.75 mmol.

3 1:1 reaction stoichiometry: NaOH + HCl -> NaCl + H2O consumes 18.75 mmol of NaOH.

4 Millimoles of NaOH remaining unreacted = 25.0 - 18.75 = 6.25 mmol.

5 Mass of unreacted NaOH = 6.25 mmol × 40 mg/mmol = 250 mg.

6 Match Option (4) in 10 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Some Basic Concepts of Chemistry (Stoichiometric calculations in solution and limiting reagent calculations).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Balanced Chemical Equation of Neutralization

Sodium hydroxide and hydrochloric acid undergo an equimolar 1:1 acid-base neutralization reaction:

\text{NaOH} + \text{HCl} \longrightarrow \text{NaCl} + \text{H}_2\text{O}

Step 2

Calculation of Initial Moles of Reactants

Computing the moles of each reactant: - Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol - Initial moles of NaOH: n(NaOH) = 1.0 g / 40 g/mol = 0.025 mol = 25.0 mmol - Initial moles of HCl: n(HCl) = Molarity × Volume(L) = 0.75 mol/L × 0.025 L = 0.01875 mol = 18.75 mmol

n_{\text{NaOH, initial}} = 25.0\;\text{mmol}, \quad n_{\text{HCl, initial}} = 18.75\;\text{mmol}

Step 3

Determining Limiting Reagent and Unreacted Substance

Because the stoichiometric ratio is 1:1 and n(HCl) < n(NaOH), hydrochloric acid (HCl) is the limiting reagent and is completely consumed. Moles of NaOH reacting = 18.75 mmol. Moles of NaOH left unreacted:

n_{\text{NaOH, unreacted}} = 25.0\;\text{mmol} - 18.75\;\text{mmol} = 6.25\;\text{mmol} = 6.25 \times 10^{-3}\;\text{mol}

Step 4

Conversion to Milligrams (mg)

Converting the remaining moles of sodium hydroxide to mass in milligrams:

m = (6.25 \times 10^{-3}\;\text{mol}) \times (40\;\text{g/mol}) = 0.250\;\text{g} = 250\;\text{mg}

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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:

Assumes complete stoichiometric neutralization, which is false since NaOH is in excess.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming all titrations are equimolar.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Arithmetic error in rounding or molar mass.

⚠️

Student Fallacy (Why candidates pick this trap): Using incorrect molar mass for NaOH.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

This represents the mass of NaOH that reacted (18.75 mmol × 40 = 750 mg), not the mass unreacted!

⚠️

Student Fallacy (Why candidates pick this trap): Calculating the mass reacted instead of the mass remaining.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Exact unreacted mass: 1000 mg initial - 750 mg reacted = 250 mg remaining.

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Why this is airtight: None.

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NTA Trap Alert: Option (3) Reacted vs Unreacted Trap

750 mg of NaOH reacts with the HCl. Students often compute 750 mg and immediately select Option (3), forgetting that the question specifically asks for the mass LEFT UNREACTED (1000 - 750 = 250 mg)!

🛡️ How to avoid the -1 mark: Underline the word "UNREACTED": Unreacted = Total Initial - Reacted.

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Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Some Basic Concepts of Chemistry Standard Benchmark

Core Scientific Principle Tested:

"In acid-base reactions, the excess reagent remaining is calculated by subtracting stoichiometric consumption by the limiting reagent from the initial quantity."

Syllabus Focus: Stoichiometry in Solutions: Limiting Reagents

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

A thousand milligrams entered, seven-fifty fought: two-fifty left behind unbought!

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NEET 2024 Chemistry Question 82 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Millimole Subtraction Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Balanced Chemical Equation of Neutralization

Calculation of Initial Moles of Reactants

Determining Limiting Reagent and Unreacted Substance

Conversion to Milligrams (mg)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Option (3) Reacted vs Unreacted Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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