CUET UG Chemistry Booster Test - 2 Freezing, Osmosis, and van't Hoff Factor -

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CUET UG Chemistry Booster Test - 3 Freezing, Osmosis, and van\'t Hoff Factor

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 3 Freezing, Osmosis, and van\'t Hoff Factor

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QUESTION 1 OF 20

Statements regarding the freezing point of substances:
1. At the freezing point, the solid phase is in dynamic equilibrium with the liquid phase.
2. Vapour pressure of the liquid phase is higher than the solid phase at the freezing point.
3. Addition of a non-volatile solute decreases the solvent's vapour pressure.
4. A solution freezes when its vapour pressure is greater than the pure solid solvent.

Answer Review

1 Statements regarding the freezing point of substances:
1. At the freezing point, the solid phase is in dynamic equilibrium with the liquid phase.
2. Vapour pressure of the liquid phase is higher than the solid phase at the freezing point.
3. Addition of a non-volatile solute decreases the solvent's vapour pressure.
4. A solution freezes when its vapour pressure is greater than the pure solid solvent.

A 1, 2

B 1, 3

C 3, 4

D 1, 2, 3

💡 Short Explanation

�� Freezing point involves solid-liquid equilibrium. �� At freezing point, vapour pressures are equal. 1. • Non-volatile solute lowers solvent vapour pressure.

✅ Correct Answer Explanation

2. → At the freezing point, solid and liquid phases coexist in dynamic equilibrium. Also, addition of a non-volatile solute lowers the vapour pressure of the solvent. Statement 2 is incorrect because vapour pressures are equal at freezing point. Statement 4 is incorrect because freezing occurs when solution vapour pressure equals that of the pure solid solvent.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 2, which is incorrect.
�� Option C → Includes Statement 4, which is incorrect.
3. • Option D → Includes Statement 2, which is incorrect.

🎯 Strategy

Used

4. Elimination

Application:

�� Remove options containing incorrect vapour pressure statements.

Final Logic:

�� Only Statements 1 and 3 are correct.

🧠 Memory Trick

1. → Freezing = Equal Vapour Pressure.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Depression in Freezing Point

2 Arrange the following states in decreasing order of temperature at 1 atm pressure:
1. Boiling point of aqueous sucrose solution
2. Boiling point of pure water
3. Freezing point of pure water
4. Freezing point of aqueous sucrose solution

A 1, 2, 3, 4

B 2, 1, 3, 4

C 1, 2, 4, 3

D 4, 3, 2, 1

💡 Short Explanation

�� Solute elevates boiling point. �� Solute depresses freezing point. 1. • Therefore, solution BP > pure water BP > pure water FP > solution FP.

✅ Correct Answer Explanation

2. → Adding a non-volatile solute like sucrose increases the boiling point and decreases the freezing point of water. Therefore, in decreasing order of temperature: Boiling point of sucrose solution > Boiling point of pure water > Freezing point of pure water > Freezing point of sucrose solution.

❓ Why Other Options Are Incorrect

�� Option B → Places pure water boiling point above solution boiling point.
�� Option C → Places solution freezing point above pure water freezing point.
1. • Option D → Gives reverse-type ordering.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Apply boiling point elevation and freezing point depression.

Final Logic:

�� Non-volatile solute raises BP and lowers FP.

🧠 Memory Trick

1. → Solute: Boiling Up, Freezing Down.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Colligative Properties

3 Match List-I (Solvent) with List-II (Kf value in K kg mol⁻¹):

List 1 (Solvent)	List 2 (Kf Value)
1. Water	a. 1.86
2. Benzene	b. 5.12
3. Cyclohexane	c. 20.00
4. Carbon tetrachloride	d. 31.8

A 1-a, 2-b, 3-c, 4-d

B 1-d, 2-b, 3-a, 4-c

C 1-c, 2-a, 3-b, 4-d

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

�� Water has Kf = 1.86. �� Benzene has Kf = 5.12. �� Cyclohexane has Kf = 20.00. 1. • Carbon tetrachloride has Kf = 31.8.

✅ Correct Answer Explanation

2. → The molal depression constant is solvent-specific. The correct NCERT values are: Water = 1.86 K kg mol⁻¹ Benzene = 5.12 K kg mol⁻¹ Cyclohexane = 20.00 K kg mol⁻¹ Carbon tetrachloride = 31.8 K kg mol⁻¹

❓ Why Other Options Are Incorrect

�� Option B → Water and carbon tetrachloride values are incorrectly matched.
�� Option C → Water and benzene values are interchanged.
1. • Option D → Multiple Kf values are incorrectly matched.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Match standard Kf values with their solvents.

Final Logic:

�� Water-1.86, Benzene-5.12, Cyclohexane-20.00, Carbon tetrachloride-31.8.

🧠 Memory Trick

1. → Water lowest, CCl₄ highest.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molal Depression Constant

4 What formal term is used for the constant 'Kf' when calculating the molar mass of a solute from freezing point data?

A Ebullioscopic Constant

B Cryoscopic Constant

C Osmotic Constant

D Henry's Constant

💡 Short Explanation

�� Kf is used in freezing point depression. �� It is called molal depression constant. 1. • It is also called cryoscopic constant.

✅ Correct Answer Explanation

2. → The constant Kf used in freezing point depression calculations is known as the molal depression constant or cryoscopic constant. Readable Formula: Depression in Freezing Point = Kf × Molality

❓ Why Other Options Are Incorrect

�� Option A → Ebullioscopic constant refers to boiling point elevation constant Kb.
�� Option C → Osmotic constant is not the term used for Kf.
1. • Option D → Henry's constant is used in gas solubility.

🎯 Strategy

Used

2. Odd One Out

Application:

�� Connect "cryo" with freezing.

Final Logic:

�� Freezing point depression constant = cryoscopic constant.

🧠 Memory Trick

1. → Cryo means freezing.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Cryoscopic Constant

📄 Passage

Membranes having a network of submicroscopic holes or pores are known as semipermeable membranes (SPM). Small solvent molecules, like water, can pass through these holes but the passage of bigger molecules like solute is hindered. If this membrane is placed between the solvent and solution, the solvent molecules will flow through the membrane from pure solvent to the solution. This process of flow of the solvent is called osmosis. The flow will continue till the equilibrium is attained.

Which of the following can freely pass through the submicroscopic holes of a semipermeable membrane?

A Macromolecules

B Solute molecules only

C Small solvent molecules

D Both solute and solvent molecules

💡 Short Explanation

�� SPM allows selective passage. �� Small solvent molecules pass through. 1. • Larger solute particles are blocked.

✅ Correct Answer Explanation

2. → According to the passage, semipermeable membranes contain submicroscopic pores that allow small solvent molecules like water to pass through. Larger solute molecules cannot pass freely.

❓ Why Other Options Are Incorrect

�� Option A → Macromolecules are too large to pass through.
�� Option B → Solute molecules are hindered.
3. • Option D → Both solute and solvent cannot freely pass through SPM.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Use the direct statement from the passage.

Final Logic:

�� SPM permits small solvent molecules only.

🧠 Memory Trick

1. → SPM = Solvent Passes Membrane.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Semipermeable Membrane

📄 Passage

During osmosis, the natural flow of solvent molecules across the SPM occurs from:

A Solution to pure solvent

B Higher concentration of solution to lower concentration of solution

C Lower concentration of solution to higher concentration of solution

D Both directions equally at all times

💡 Short Explanation

�� Osmosis is solvent movement. �� Solvent moves from dilute side to concentrated side. 1. • Pure solvent moves toward solution.

✅ Correct Answer Explanation

2. → During osmosis, solvent molecules pass through a semipermeable membrane from pure solvent or dilute solution toward the more concentrated solution. This continues until equilibrium is reached.

❓ Why Other Options Are Incorrect

�� Option A → Natural osmosis occurs from pure solvent to solution, not generally solution to pure solvent.
�� Option B → This reverses the correct direction.
3. • Option D → Net flow is not equal in both directions before equilibrium.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Use the passage definition of solvent flow.

Final Logic:

�� Osmosis = solvent flow from lower concentration to higher concentration.

🧠 Memory Trick

1. → Solvent moves to stronger solution.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Osmosis

7 The excess pressure that just stops the flow of solvent across a semipermeable membrane is defined as:

A Atmospheric pressure

B Vapour pressure

C Osmotic pressure

D Hydrostatic pressure

💡 Short Explanation

�� Osmotic pressure stops osmosis. �� It is applied on the solution side. 1. • It prevents net solvent flow.

✅ Correct Answer Explanation

2. → Osmotic pressure is the minimum excess pressure that must be applied to a solution to stop the flow of solvent through a semipermeable membrane.

❓ Why Other Options Are Incorrect

�� Option A → Atmospheric pressure is external air pressure.
�� Option B → Vapour pressure is pressure exerted by vapour.
3. • Option D → Hydrostatic pressure is pressure due to liquid column, not the defined colligative pressure.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Match the phrase "just stops solvent flow" with osmotic pressure.

Final Logic:

�� Pressure required to stop osmosis = osmotic pressure.

🧠 Memory Trick

1. → Osmotic Pressure Stops Osmosis.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Osmotic Pressure

8 In the osmotic pressure equation P = CRT, if C is expressed in mol L⁻¹ and T in Kelvin, what is the unit of the gas constant R typically used when P is measured in bar?

A J K⁻¹ mol⁻¹

B L atm K⁻¹ mol⁻¹

C L bar mol⁻¹ K⁻¹

D Pa m³ K⁻¹ mol⁻¹

💡 Short Explanation

�� Osmotic pressure is measured in bar. �� Concentration is mol per litre. 1. • R must match litre and bar units.

✅ Correct Answer Explanation

2. → In the equation: Readable Formula: Osmotic Pressure = Molarity × Gas Constant × Temperature If osmotic pressure is in bar, concentration is in mol L⁻¹ and temperature is in Kelvin, then R should have the unit: L bar mol⁻¹ K⁻¹ This ensures the final pressure unit becomes bar.

❓ Why Other Options Are Incorrect

�� Option A → Used when energy is expressed in joules.
�� Option B → Used when pressure is measured in atm.
1. • Option D → Used when pressure is in pascal and volume in cubic metre.

🎯 Strategy

Used

2. Dimensional/Unit Analysis

Application:

�� Match the units of pressure, concentration and temperature in the equation.

Final Logic:

�� For pressure in bar, R must contain L bar mol⁻¹ K⁻¹.

🧠 Memory Trick

1. → Bar pressure needs R in L bar.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Osmotic Pressure Equation

9 When blood cells are placed in a 0.9% (mass/volume) sodium chloride solution, what type of net osmotic process occurs?

A Plasmolysis

B Hemolysis

C Reverse osmosis

D No net osmosis

💡 Short Explanation

�� 0.9% NaCl is normal saline. �� It is isotonic with blood cells. 1. • No net water movement occurs.

✅ Correct Answer Explanation

2. → A 0.9% mass/volume NaCl solution is isotonic with the fluid inside blood cells. Since the osmotic pressure inside and outside the cells is balanced, there is no net movement of water across the cell membrane.

❓ Why Other Options Are Incorrect

�� Option A → Plasmolysis occurs in hypertonic solution when cells lose water.
�� Option B → Hemolysis occurs in hypotonic solution when cells swell and burst.
3. • Option C → Reverse osmosis requires applied pressure greater than osmotic pressure.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Connect 0.9% NaCl with isotonic solution.

Final Logic:

�� Isotonic solution causes no net osmosis.

🧠 Memory Trick

1. → 0.9% = No Net Flow.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Isotonic Solutions

10 If blood cells are placed in a solution containing more than 0.9% (mass/volume) sodium chloride, what will happen?

A Water flows into the cells, causing them to swell.

B Water flows out of the cells, causing them to shrink.

C No flow of water occurs.

D The cells undergo reverse osmosis.

💡 Short Explanation

�� More than 0.9% NaCl is hypertonic. �� Water leaves the cells. 1. • Cells shrink.

✅ Correct Answer Explanation

2. → A solution containing more than 0.9% NaCl is hypertonic compared to blood cells. Water moves out of the cells toward the more concentrated external solution by osmosis. As a result, the cells shrink.

❓ Why Other Options Are Incorrect

�� Option A → This happens in hypotonic solution.
�� Option C → No net flow occurs only in isotonic solution.
3. • Option D → Reverse osmosis requires external pressure, not normal cell osmosis.

🎯 Strategy

Used

4. Elimination

Application:

�� Identify hypertonic solution and remove options describing hypotonic or isotonic conditions.

Final Logic:

�� Hypertonic solution pulls water out of cells.

🧠 Memory Trick

1. → Hypertonic = Water exits.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Osmosis in Blood Cells

11 Statements regarding reverse osmosis in desalination:
1. Pure water is squeezed out of the sea water through the membrane.
2. Pressure less than osmotic pressure must be applied.
3. The direction of osmosis is reversed.
4. The pressure required is quite low.

A 1, 3

B 2, 3, 4

C 1, 4

D 1, 2

💡 Short Explanation

�� Reverse osmosis reverses natural osmosis. �� Pressure greater than osmotic pressure is applied. 1. • Pure water passes through the membrane.

✅ Correct Answer Explanation

2. → In reverse osmosis, pressure greater than the osmotic pressure is applied to the solution side. This forces pure water through the semipermeable membrane, reversing the natural direction of osmosis. Hence Statements 1 and 3 are correct. Statements 2 and 4 are incorrect because high pressure, not low pressure, is required.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Statement 4 is incorrect.
3. • Option D → Statement 2 is incorrect.

🎯 Strategy

Used

4. Elimination

Application:

�� Remove statements contradicting the definition of reverse osmosis.

Final Logic:

�� Reverse osmosis requires pressure greater than osmotic pressure and reverses solvent flow.

🧠 Memory Trick

1. → RO = Reverse + High Pressure.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Reverse Osmosis

12 Match List-I with List-II concerning membrane types and processes:

List 1 (Membrane / Process)	List 2 (Description)
1. Cellulose acetate	a. Synthetic membrane for RO
2. Pig's bladder	b. Natural semipermeable membrane
3. Reverse Osmosis	c. Pressure larger than osmotic pressure applied
4. Desalination	d. Used to meet potable water requirements

A 1-a, 2-b, 3-c, 4-d

B 1-b, 2-d, 3-c, 4-a

C 1-a, 2-c, 3-d, 4-b

D 1-d, 2-c, 3-b, 4-a

💡 Short Explanation

�� Cellulose acetate is a synthetic RO membrane. �� Pig's bladder is a natural semipermeable membrane. �� Reverse osmosis requires pressure greater than osmotic pressure. 1. • Desalination provides potable water.

✅ Correct Answer Explanation

2. → The correct matching is: Cellulose acetate → Synthetic membrane for RO Pig's bladder → Natural semipermeable membrane Reverse Osmosis → Pressure larger than osmotic pressure applied Desalination → Used to meet potable water requirements This is the standard NCERT description of reverse osmosis and membrane applications.

❓ Why Other Options Are Incorrect

�� Option B → Multiple membrane-process mismatches.
�� Option C → Pig's bladder and reverse osmosis incorrectly matched.
1. • Option D → Cellulose acetate and desalination incorrectly matched.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Match each membrane/process with its defining characteristic.

Final Logic:

�� Cellulose acetate–RO membrane, Pig's bladder–natural membrane, RO–high pressure, Desalination–drinking water.

🧠 Memory Trick

1. → Cellulose–RO, Pig–Natural.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Reverse Osmosis and Semipermeable Membranes

13 When 1 mole of KCl dissolves in water, assuming complete dissociation, the experimentally determined molar mass is lower than the true value. What is this resultant calculated molar mass called?

A Normal molar mass

B Abnormal molar mass

C Theoretical molar mass

D Associated molar mass

💡 Short Explanation

�� KCl dissociates into ions. �� Colligative properties become larger than expected. 1. • Calculated molar mass becomes lower than true molar mass.

✅ Correct Answer Explanation

2. → KCl dissociates into K⁺ and Cl⁻ ions, increasing the number of particles in solution. Since colligative properties depend on particle number, the experimentally calculated molar mass becomes smaller than the actual molar mass. Such a value is called an abnormal molar mass.

❓ Why Other Options Are Incorrect

�� Option A → Normal molar mass is the actual molar mass.
�� Option C → Theoretical molar mass is not the NCERT term.
3. • Option D → Associated molar mass is related to association, not dissociation.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Relate dissociation with abnormal colligative behavior.

Final Logic:

�� Dissociation causes abnormal molar mass values.

🧠 Memory Trick

1. → More particles → Lower calculated molar mass.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Abnormal Molar Masses

14 Arrange the following terms based on the extent of particles present in 1 mole of substance after dissolution (from highest to lowest):
1. Complete dissociation of K₂SO₄
2. Complete dissociation of NaCl
3. Glucose in water
4. Complete dimerisation of ethanoic acid in benzene

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 1, 3, 4

D 1, 3, 2, 4

💡 Short Explanation

�� K₂SO₄ gives 3 particles. �� NaCl gives 2 particles. �� Glucose gives 1 particle. 1. • Dimerisation reduces particles to 0.5 mole equivalents.

✅ Correct Answer Explanation

2. → Number of particles produced: K₂SO₄ → 2K⁺ + SO₄²⁻ = 3 particles NaCl → Na⁺ + Cl⁻ = 2 particles Glucose → no dissociation = 1 particle Ethanoic acid dimerisation → particle number decreases to about half Therefore: K₂SO₄ > NaCl > Glucose > Ethanoic acid (dimerised)

❓ Why Other Options Are Incorrect

�� Option B → Completely reverses the correct trend.
�� Option C → Places NaCl above K₂SO₄ incorrectly.
1. • Option D → Places glucose above NaCl incorrectly.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Count particles after dissociation or association.

Final Logic:

�� More particles correspond to higher position in the order.

🧠 Memory Trick

1. → 3 > 2 > 1 > 0.5

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Association, Dissociation and van't Hoff Factor

15 What is the symbol used to denote the ratio of normal molar mass to abnormal molar mass?

A Kf

B i

C P

D R

💡 Short Explanation

�� i denotes van't Hoff factor. �� It explains abnormal colligative behavior. 1. • It relates normal and abnormal molar masses.

✅ Correct Answer Explanation

2. → The van't Hoff factor (i) is used to account for association and dissociation of solutes. Readable Expression: i = Normal Molar Mass / Abnormal Molar Mass It also equals: i = Observed Colligative Property / Calculated Colligative Property

❓ Why Other Options Are Incorrect

�� Option A → Kf is cryoscopic constant.
�� Option C → P commonly denotes pressure.
1. • Option D → R is the gas constant.

🎯 Strategy

Used

2. Direct Recall

Application:

�� Identify the symbol associated with abnormal molar mass.

Final Logic:

�� van't Hoff factor is represented by i.

🧠 Memory Trick

1. → i = irregularity factor.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: van't Hoff Factor (i)

16 Identify the mathematical relationship representing the van't Hoff factor (i):

A Abnormal molar mass / Normal molar mass

B Calculated colligative property / Observed colligative property

C Observed colligative property / Calculated colligative property

D Moles of particles before association / Moles of particles after

💡 Short Explanation

�� van't Hoff factor corrects abnormal colligative values. �� It compares observed and calculated colligative properties. 1. • It accounts for association or dissociation.

✅ Correct Answer Explanation

2. → The van't Hoff factor is represented as: Readable Expression: van't Hoff factor = Observed Colligative Property ÷ Calculated Colligative Property It can also be written as: van't Hoff factor = Normal Molar Mass ÷ Abnormal Molar Mass Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → This is the inverse of the correct molar mass relationship.
�� Option B → This is the inverse of the correct colligative property relationship.
1. • Option D → This is not the standard NCERT definition.

🎯 Strategy

Used

2. Elimination

Application:

�� Compare each option with the standard definition of van't Hoff factor.

Final Logic:

�� i = Observed value ÷ Calculated value.

🧠 Memory Trick

1. → i = Observed over Calculated.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: van't Hoff Factor

17 Statements regarding modified colligative equations:
1. Elevation of boiling point is ΔTb = i Kb m
2. Osmotic pressure is P = i n₂ R T / V
3. Relative lowering of vapour pressure incorporates 'i'
4. 'i' is only used for non-electrolytes

A 1, 2, 3

B 1, 2

C 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Modified equations include van't Hoff factor. �� i corrects values for association or dissociation. 1. • i is not only for non-electrolytes.

✅ Correct Answer Explanation

2. → Modified colligative equations include the van't Hoff factor to correct for association or dissociation. Readable Formulae: Elevation in Boiling Point = i × Kb × Molality Osmotic Pressure = i × Number of Moles of Solute × Gas Constant × Temperature ÷ Volume Relative Lowering of Vapour Pressure also incorporates i when abnormal association or dissociation affects particle count. Statement 4 is incorrect because i is mainly used when solutes associate or dissociate, especially electrolytes.

❓ Why Other Options Are Incorrect

�� Option B → Omits Statement 3, which is correct.
�� Option C → Includes Statement 4, which is incorrect.
1. • Option D → Includes Statement 4, which is incorrect.

🎯 Strategy

Used

2. Elimination

Application:

�� Remove options containing the false statement that i is only used for non-electrolytes.

Final Logic:

�� Statements 1, 2 and 3 correctly use van't Hoff factor.

🧠 Memory Trick

1. → Add i when particles change.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Modified Colligative Equations

18 For ethanoic acid in benzene undergoing complete association (dimerisation), the value of the van't Hoff factor 'i' incorporated in equations is nearly:

A 0.5

B 1.0

C 2.0

D 3.0

💡 Short Explanation

�� Two ethanoic acid molecules associate into one dimer. �� Particle number becomes half. 1. • Therefore, i becomes nearly 0.5.

✅ Correct Answer Explanation

2. → In benzene, ethanoic acid molecules associate due to hydrogen bonding. Readable Expression: 2 Ethanoic Acid Molecules combine to form 1 Dimer So, the number of solute particles becomes half of the expected number. Therefore, the van't Hoff factor becomes nearly 0.5.

❓ Why Other Options Are Incorrect

�� Option B → Represents no association or dissociation.
�� Option C → Represents dissociation into two particles.
1. • Option D → Represents dissociation into three particles.

🎯 Strategy

Used

2. Substitution

Application:

�� Convert two associated molecules into one particle and compare particle count.

Final Logic:

�� Complete dimerisation halves particle number, so i = 0.5.

🧠 Memory Trick

1. → Dimer means two become one.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Association and van't Hoff Factor

19 Match List-I (Electrolyte) with List-II (Limiting van't Hoff factor 'i' at complete dissociation):

List 1 (Electrolyte / Solute)	List 2 (Limiting van't Hoff Factor)
1. K₂SO₄	a. i approaches 3
2. NaCl	b. i approaches 2
3. Ethanoic acid in benzene	c. i approaches 0.5
4. Glucose	d. i is 1

A 1-a, 2-b, 3-c, 4-d

B 1-a, 2-c, 3-d, 4-b

C 1-b, 2-d, 3-c, 4-a

D 1-c, 2-d, 3-a, 4-b

💡 Short Explanation

�� K₂SO₄ gives 3 ions. �� NaCl gives 2 ions. �� Ethanoic acid associates, so i is nearly 0.5. 1. • Glucose does not dissociate, so i = 1.

✅ Correct Answer Explanation

2. → The correct matching is: K₂SO₄ → 2K⁺ + SO₄²⁻ → i approaches 3 NaCl → Na⁺ + Cl⁻ → i approaches 2 Ethanoic acid in benzene → dimerisation → i approaches 0.5 Glucose → no association or dissociation → i is 1

❓ Why Other Options Are Incorrect

�� Option B → NaCl, ethanoic acid and glucose are incorrectly matched.
�� Option C → K₂SO₄ and glucose are incorrectly matched.
1. • Option D → Multiple solute-i relationships are incorrect.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Match each solute based on particle number after dissociation or association.

Final Logic:

�� i depends on the number of particles formed.

🧠 Memory Trick

1. → K₂SO₄ = 3, NaCl = 2, Acid dimer = 0.5, Glucose = 1.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Electrolyte Behaviour and van't Hoff Factor

20 In very dilute solutions (e.g., 0.001 m), the experimental van't Hoff factor (i) for MgSO₄ approaches:

A 1.21

B 1.82

C 2.00

D 3.00

💡 Short Explanation

�� MgSO₄ dissociates into two ions. �� Very dilute solution supports near-complete dissociation. 1. • Therefore, i approaches 2.

✅ Correct Answer Explanation

2. → Magnesium sulfate dissociates as: Readable Expression: MgSO₄ dissociates into Mg²⁺ and SO₄²⁻ Total particles formed = 2 In very dilute solutions, dissociation becomes nearly complete. Hence, the van't Hoff factor approaches 2.00.

❓ Why Other Options Are Incorrect

�� Option A → Too low for near-complete dissociation.
�� Option B → Possible for partial dissociation, not the limiting dilute value.
1. • Option D → Would require formation of three particles.

🎯 Strategy

Used

2. Substitution

Application:

�� Count ions produced by complete dissociation of MgSO₄.

Final Logic:

�� MgSO₄ produces two ions, so i approaches 2.

🧠 Memory Trick

1. → MgSO₄ = 2 ions.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: van't Hoff Factor of Electrolytes

CUET UG Chemistry Booster Test – 2 Freezing, Osmosis, and van’t Hoff Factor

CUET UG Chemistry Booster Test - 3 Freezing, Osmosis, and van\'t Hoff Factor

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