Q.100 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 100 Solution

Topic: States of Matter · Official NTA Booklet Code M1

Question

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O₂ and 2 g H₂ confined in a total volume of one litre at 0°C is: [Given R = 0.082 L atm mol⁻¹ K⁻¹, T = 273 K]

✓

Verified Against NTA Final Answer Key

Option (3) : 25.18

📚 States of Matter Easy

📋 Given Parameters & Problem Setup

Mass of O2 4 g (M = 32 g/mol)

Mass of H2 2 g (M = 2 g/mol)

Volume (V) 1.0 L

Temperature (T) 0°C = 273 K

Gas Constant (R) 0.082 L atm mol^-1 K^-1

⚡

The 30-Second Exam Hall Shortcut: ⚡ Total Moles = 4/32 + 2/2 = 1.125 mol $\implies$ P = n·R·T / V = 25.18 atm

⏱️ Save ~20 seconds

$n(O_2) = \frac{4}{32} = 0.125\text{ mol}$ . $n(H_2) = \frac{2}{2} = 1.0\text{ mol}$ . Total moles $n_{\text{total}} = 1.125\text{ mol}$ . Total pressure $P = \frac{n_{\text{total}} R T}{V} = \frac{1.125 \times 0.082 \times 273}{1} = 1.125 \times 22.386 = 25.18\text{ atm}$ ! Direct 15-second calculation matching Option (3)!

1 Calculate moles of each gas component: -

n(O_2) = \frac{4\text{ g}}{32\text{ g/mol}} = 0.125\text{ mol}

n(H_2) = \frac{2\text{ g}}{2\text{ g/mol}} = 1.0\text{ mol}

2 Sum total moles:

n_{\text{total}} = 0.125 + 1.0 = 1.125\text{ mol}

3 Calculate molar gas volume at given conditions:

RT = 0.082 \times 273 = 22.386\text{ L atm/mol}

4 Calculate total pressure:

P_{\text{total}} = \frac{n_{\text{total}} RT}{V} = \frac{1.125 \times 22.386}{1.0} = 25.184\text{ atm} \approx 25.18\text{ atm}

5 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical ideal gas equation of state and Dalton's law of partial pressures.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating Moles of Oxygen and Hydrogen Gas

Using molar masses $M(O_2) = 32\text{ g mol}^{-1}$ and $M(H_2) = 2\text{ g mol}^{-1}$ : - Moles of oxygen: $n_{\text{O}_2} = \frac{m_{\text{O}_2}}{M_{\text{O}_2}} = \frac{4\text{ g}}{32\text{ g mol}^{-1}} = \frac{1}{8} = 0.125\text{ mol}$ - Moles of hydrogen: $n_{\text{H}_2} = \frac{m_{\text{H}_2}}{M_{\text{H}_2}} = \frac{2\text{ g}}{2\text{ g mol}^{-1}} = 1.0\text{ mol}$

Step 2

Determining Total Moles of Gas in the Mixture

Assuming no chemical reaction occurs at $0^\circ\text{C}$ in the absence of a spark or catalyst, Dalton's law states that each gas behaves independently: $n_{\text{total}} = n_{\text{O}_2} + n_{\text{H}_2} = 0.125 + 1.0 = 1.125\text{ mol}$

Step 3

Applying the Ideal Gas Equation

Given $V = 1.0\text{ L}$ , $T = 273\text{ K}$ , and $R = 0.082\text{ L atm mol}^{-1}\text{K}^{-1}$ : $P_{\text{total}} = \frac{n_{\text{total}} R T}{V} = \frac{1.125\text{ mol} \times 0.082\text{ L atm mol}^{-1}\text{K}^{-1} \times 273\text{ K}}{1.0\text{ L}}$

P_{\text{total}} = 1.125 \times 22.386 = 25.18425\text{ atm} \approx \mathbf{25.18\text{ atm}}

⭐

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:

Off by a factor of 10 ( $2.518$ instead of $25.18$ ), corresponding to volume $V = 10\text{ L}$ instead of $1\text{ L}$ .

⚠️

Student Fallacy (Why candidates pick this trap): Decimal position error.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of rounding and decimal shift error.

⚠️

Student Fallacy (Why candidates pick this trap): Arithmetic miscalculation.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact total pressure: $1.125 \times 0.082 \times 273 = 25.184\text{ atm} \approx 25.18\text{ atm}$ .

💡

Why this is airtight: None; textbook gaseous state calculation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Approximation using $22.4$ and rounding error.

⚠️

Student Fallacy (Why candidates pick this trap): Using imprecise molar volume.

⚠️

NTA Trap Alert: Molar Mass of Hydrogen (H vs H2) and Decimal Traps

Two traps exist here: 1. Using atomic weight of hydrogen ( $1\text{ g/mol}$ ) instead of molecular weight of hydrogen gas ( $H_2 = 2\text{ g/mol}$ ). 2. Decimal shift between Option (1) ( $2.518$ ) and Option (3) ( $25.18$ ). Since $1\text{ mol}$ of gas at STP occupies $22.4\text{ L}$ , cramming $1.125\text{ mol}$ into only $1\text{ L}$ MUST produce high pressure of over $20\text{ atm}$ ( $25.18\text{ atm}$ ).

🛡️ How to avoid the -1 mark: Sanity check:

1.125\text{ mol}

1\text{ L}

\implies P \approx 1.125 \times 22.4 \approx 25\text{ atm}

, ruling out

2.518\text{ atm}

📖

Core NEET Syllabus Standard

NCERT Chemistry · States of Matter Standard Benchmark

Core Scientific Principle Tested:

"The total pressure exerted by a mixture of non-reacting gases is equal to the sum of partial pressures: P_total = n_total R T / V."

Syllabus Focus: Ideal Gas Equation and Dalton's Law of Partial Pressures

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 100 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Total Moles = 4/32 + 2/2 = 1.125 mol ⟹ \implies⟹ P = n·R·T / V = 25.18 atm

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating Moles of Oxygen and Hydrogen Gas

Determining Total Moles of Gas in the Mixture

Applying the Ideal Gas Equation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Molar Mass of Hydrogen (H vs H2) and Decimal Traps

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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The 30-Second Exam Hall Shortcut: ⚡ Total Moles = 4/32 + 2/2 = 1.125 mol $\implies$ P = n·R·T / V = 25.18 atm