Q.93 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 93 Solution

Topic: Solutions · Official NTA Booklet Code M1

Question

The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is: [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

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Verified Against NTA Final Answer Key

Option (3) : 336 mm of Hg

📚 Solutions Easy

📋 Given Parameters & Problem Setup

Solution Temperature 45°C

Molar Ratio (Benzene : Octane) 3 : 2

P°(Benzene) 280 mm Hg

P°(Octane) 420 mm Hg

⚡

The 30-Second Exam Hall Shortcut: ⚡ Raoult's Law: Ptotal = P°B · χB + P°O · χO

⏱️ Save ~20 seconds

Mole fractions: $\chi_B = \frac{3}{3+2} = 0.6$ and $\chi_O = \frac{2}{3+2} = 0.4$ . Total vapour pressure $P_{\text{total}} = 280(0.6) + 420(0.4) = 168 + 168 = 336\text{ mm Hg}$ ! Instant 15-second arithmetic matching Option (3)!

1 Calculate mole fractions from given molar ratio

3 : 2

: -

\chi_{\text{benzene}} = \frac{3}{5} = 0.6

\chi_{\text{octane}} = \frac{2}{5} = 0.4

2 Apply Raoult's Law for ideal solution:

P_{\text{total}} = P_B^\circ \chi_B + P_O^\circ \chi_O

3 Substitute:

P_{\text{total}} = (280 \times 0.6) + (420 \times 0.4)

4 Calculate partial pressures:

P_B = 168\text{ mm Hg}

P_O = 168\text{ mm Hg}

5 Sum:

P_{\text{total}} = 168 + 168 = 336\text{ mm Hg}

6 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Raoult's Law formulation for ideal volatile binary liquid mixtures.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining Component Mole Fractions

Let the number of moles of benzene ( $B$ ) be $n_B = 3x$ and octane ( $O$ ) be $n_O = 2x$ . The total moles in the solution are $n_{\text{total}} = 3x + 2x = 5x$ . - Mole fraction of benzene: $\chi_B = \frac{n_B}{n_{\text{total}}} = \frac{3x}{5x} = \frac{3}{5} = 0.6$ - Mole fraction of octane: $\chi_O = \frac{n_O}{n_{\text{total}}} = \frac{2x}{5x} = \frac{2}{5} = 0.4$

Step 2

Calculating Component Partial Pressures via Raoult's Law

For an ideal solution, the partial vapour pressure of each volatile component is equal to the vapour pressure of the pure component multiplied by its mole fraction in the liquid phase: - Partial pressure of benzene: $P_B = P_B^\circ \cdot \chi_B = 280\text{ mm Hg} \times 0.6 = 168\text{ mm Hg}$ - Partial pressure of octane: $P_O = P_O^\circ \cdot \chi_O = 420\text{ mm Hg} \times 0.4 = 168\text{ mm Hg}$

Step 3

Computing Total Vapour Pressure

Applying Dalton's Law of Partial Pressures:

P_{\text{total}} = P_B + P_O = 168\text{ mm Hg} + 168\text{ mm Hg} = 336\text{ mm Hg}

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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:

Arbitrary incorrect pressure.

⚠️

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

2 2

✗ Incorrect

Verdict & Scientific Reason:

This is the partial pressure of a single component ( $P_B$ or $P_O$ ), NOT the total vapor pressure of the mixture.

⚠️

Student Fallacy (Why candidates pick this trap): Reporting partial pressure instead of total pressure.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact total vapour pressure calculated by Raoult's law ( $168 + 168 = 336\text{ mm Hg}$ ).

💡

Why this is airtight: None; flawless solution physical chemistry execution.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Simple arithmetic average of pure vapour pressures: $\frac{280 + 420}{2} = 350\text{ mm Hg}$ (which assumes an equimolar $1:1$ ratio instead of $3:2$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming a 1:1 equimolar mixture.

⚠️

NTA Trap Alert: Partial Pressure vs. Total Pressure Trap (Option 2 vs. Option 3)

A student who computes $P_B = 280 \times 0.6 = 168\text{ mm Hg}$ might see $168\text{ mm Hg}$ listed as Option (2) and immediately mark it. Remember that Option (2) represents ONLY the partial vapor pressure of benzene. The question asks for the TOTAL vapor pressure: $P_{\text{total}} = P_B + P_O = 168 + 168 = 336\text{ mm Hg}$ .

🛡️ How to avoid the -1 mark: Always read the prompt carefully: "total pressure"

\implies P_{\text{total}} = P_B + P_O

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Core NEET Syllabus Standard

NCERT Chemistry · Solutions Standard Benchmark

Core Scientific Principle Tested:

"According to Raoult's law, for a solution of volatile liquids, the total vapour pressure is the sum of the partial vapour pressures: P_total = P1° x1 + P2° x2."

Syllabus Focus: Vapour Pressure of Liquid-Liquid Solutions (Raoult's Law)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 93 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Raoult's Law: Ptotal = P°B · χB + P°O · χO

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining Component Mole Fractions

Calculating Component Partial Pressures via Raoult's Law

Computing Total Vapour Pressure

360° Option Elimination Engine Key Feature

NTA Trap Alert: Partial Pressure vs. Total Pressure Trap (Option 2 vs. Option 3)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Solutions

Thermodynamics & Intermolecular Forces of Ideal Solutions

Raoult's Law & Binary Liquid Mixtures Deviations

Classification of Solutions: Solute and Solvent States

Highest Boiling Point of Aqueous Solutions (Effective Molality / Osmolarity)

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