Q.64 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 64 Solution

Topic: Solutions · Official NTA Booklet Code M1

Question

The following solutions were prepared by dissolving 10 g of glucose (C6H12O6) in 250 ml of water (P1), 10 g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is:

(1)P2 > P1 > P3
(2)P1 > P2 > P3
(3)P2 > P3 > P1
(4)P3 > P1 > P2

✓

Verified Against NTA Final Answer Key

Option (1) : P2 > P1 > P3

📚 Solutions Easy

📋 Given Parameters & Problem Setup

Solute 1 10 g glucose (C6H12O6, M = 180 g/mol) in 250 mL (P1)

Solute 2 10 g urea (CH4N2O, M = 60 g/mol) in 250 mL (P2)

Solute 3 10 g sucrose (C12H22O11, M = 342 g/mol) in 250 mL (P3)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Equal Mass & Volume: Π ∝ 1/Molar Mass!

⏱️ Save ~20 seconds

Osmotic pressure $\Pi = \frac{w}{M V} R T$ . With identical dissolved mass ( $10\text{ g}$ ) and solution volume ( $250\text{ mL}$ ), osmotic pressure is strictly inversely proportional to molar mass: $\Pi \propto \frac{1}{M}$ ! Urea ( $M=60$ ) has lowest molar mass $\implies$ highest pressure ( $P_2$ ). Glucose ( $M=180$ ) is intermediate ( $P_1$ ). Sucrose ( $M=342$ ) is highest molar mass $\implies$ lowest pressure ( $P_3$ ). Decreasing order: $P_2 > P_1 > P_3$ ! Match Option (1)!

1 Formula:

\Pi = C R T = \frac{n}{V} R T = \left(\frac{w}{M}\right) \frac{R T}{V}

2 Since

w = 10\text{ g}

V = 250\text{ mL}

, and

T

are identical for all solutions,

\Pi \propto \frac{1}{M}

3 Compare molar masses: Urea (

M = 60\text{ g/mol}

) < Glucose (

M = 180\text{ g/mol}

) < Sucrose (

M = 342\text{ g/mol}

4 Invert order for osmotic pressure:

P_{\text{urea}} (P_2) > P_{\text{glucose}} (P_1) > P_{\text{sucrose}} (P_3)

5 Match Option (1) P2 > P1 > P3.

🛡️

Why You Can 100% Trust This Shortcut: Canonical van 't Hoff colligative property definition for non-electrolytes.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulating Osmotic Pressure for Non-Electrolytes

For dilute non-electrolyte solutions that neither associate nor dissociate (van 't Hoff factor $i = 1$ ), osmotic pressure $\Pi$ is given by:

\Pi = C R T = \left(\frac{n}{V}\right) R T = \left(\frac{w}{M \cdot V}\right) R T

Step 2

Calculating Number of Moles for Each Solution

Each solution contains $w = 10\text{ g}$ of solute dissolved in $V = 250\text{ mL} = 0.25\text{ L}$ of water: - Solution 1 (Glucose, $C_6H_{12}O_6$ ): $M_1 = 180\text{ g/mol}$ $n_1 = \frac{10}{180} = 0.0556\text{ mol} \implies C_1 = \frac{0.0556}{0.25} = 0.222\text{ M}$ - Solution 2 (Urea, $NH_2CONH_2$ ): $M_2 = 60\text{ g/mol}$ $n_2 = \frac{10}{60} = 0.1667\text{ mol} \implies C_2 = \frac{0.1667}{0.25} = 0.667\text{ M}$ - Solution 3 (Sucrose, $C_{12}H_{22}O_{11}$ ): $M_3 = 342\text{ g/mol}$ $n_3 = \frac{10}{342} = 0.0292\text{ mol} \implies C_3 = \frac{0.0292}{0.25} = 0.117\text{ M}$

Step 3

Ranking the Osmotic Pressures in Decreasing Order

Because osmotic pressure is directly proportional to molar concentration $C$ :

C_2 > C_1 > C_3 \implies P_2 > P_1 > P_3

⭐

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

✓ 100% Correct

Verdict & Scientific Reason:

Exact decreasing order of osmotic pressure: Urea ( $P_2$ ) > Glucose ( $P_1$ ) > Sucrose ( $P_3$ ).

💡

Why this is airtight: None; rigorous colligative property computation.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Places glucose above urea, which violates $n_2 (10/60) > n_1 (10/180)$ .

⚠️

Student Fallacy (Why candidates pick this trap): Assuming larger molecule exerts higher osmotic pressure.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Places sucrose above glucose ( $10/342 > 10/180$ ), which is false.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting glucose and sucrose molar masses.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Directly proportional to molar mass (complete inversion of colligative property).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming colligative property scales with molecular weight rather than mole count.

⚠️

NTA Trap Alert: Molecular Size vs. Particle Number Trap

Students frequently think "sucrose is the largest molecule, so it pushes harder on the semipermeable membrane and gives the highest pressure". Colligative properties depend STRICTLY on the NUMBER of particles, NOT their size or molecular weight. Smaller molar mass $\implies$ MORE moles for equal mass $\implies$ HIGHER pressure.

🛡️ How to avoid the -1 mark: For equal masses: Smallest molar mass (

M=60

) has the MOST moles

\implies

highest

\Pi

📖

Core NEET Syllabus Standard

NCERT Chemistry · Solutions Standard Benchmark

Core Scientific Principle Tested:

"Osmotic pressure is a colligative property proportional to the number of solute particles per unit volume: $\Pi = \frac{w R T}{M V}$ ."

Syllabus Focus: Colligative Properties and Determination of Molar Mass (Osmotic Pressure)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 64 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Equal Mass & Volume: Π ∝ 1/Molar Mass!

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulating Osmotic Pressure for Non-Electrolytes

Calculating Number of Moles for Each Solution

Ranking the Osmotic Pressures in Decreasing Order

360° Option Elimination Engine Key Feature

NTA Trap Alert: Molecular Size vs. Particle Number Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Solutions

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

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