CUET UG Chemistry Booster Test - 2 Ideal Solutions and Colligative Properties -

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CUET UG Chemistry Booster Test - 2 Ideal Solutions and Colligative Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Ideal Solutions and Colligative Properties

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QUESTION 1 OF 20

If a binary solution of A and B obeys Raoult's law perfectly, the plot of total vapour pressure vs mole fraction (x₂) will be:

Answer Review

1 If a binary solution of A and B obeys Raoult's law perfectly, the plot of total vapour pressure vs mole fraction (x₂) will be:

A A curve showing a distinct maximum

B A curve showing a distinct minimum

C A linear straight line

D A perfectly horizontal line

💡 Short Explanation

�� Ideal solutions obey Raoult's law. �� Total vapour pressure changes linearly with composition. �� The graph is a straight line.

✅ Correct Answer Explanation

For an ideal binary solution: Total Vapour Pressure = p₁° + (p₂° − p₁°) × x₂ This equation is of the form y = mx + c, representing a straight line. Therefore, the graph of total vapour pressure versus mole fraction is linear.

❓ Why Other Options Are Incorrect

�� Option A → Maximum occurs in positive deviation systems.
�� Option B → Minimum occurs in negative deviation systems.
�� Option D → Vapour pressure changes with composition.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Recognize the linear equation form.

Final Logic:

�� Linear equation produces a straight-line graph.

🧠 Memory Trick

Ideal Solution = Straight Line.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Vapour Pressure of Ideal Solutions

2 Consider the intermolecular forces in a mixture of n-hexane and n-heptane:
1. It forms a nearly ideal solution.
2. Intermolecular forces between hexane and heptane are nearly equal to those in pure hexane and pure heptane.
3. The mixing process is highly exothermic.
4. The solution obeys Raoult's law over the entire range of concentrations.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Hexane and heptane are chemically similar. �� Their intermolecular forces are nearly identical. �� They form nearly ideal solutions.

✅ Correct Answer Explanation

n-Hexane and n-heptane possess similar molecular structures and intermolecular forces. Therefore: • They form an ideal or nearly ideal solution. • A-B interactions ≈ A-A interactions ≈ B-B interactions. • Raoult's law is obeyed over the entire composition range. Statement 3 is incorrect because ideal mixing has: Enthalpy of Mixing (ΔmixH) = 0 not highly exothermic.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the incorrect statement.

Final Logic:

�� Similar molecules produce nearly ideal behaviour.

🧠 Memory Trick

Hexane + Heptane = Ideal Partners.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Ideal Solutions

3 Why is the enthalpy of mixing (ΔmixH) precisely zero for an ideal solution?

A Heat is rapidly lost to the surroundings before it can be measured.

B The intermolecular attractive forces between A-A and B-B are nearly equal to those between A-B, so no net energy is absorbed or released.

C The chemical bonds of the solvent are completely severed.

D The kinetic energy of the molecules drops abruptly to zero.

💡 Short Explanation

�� Similar intermolecular forces exist before and after mixing. �� Energy required equals energy released. �� Net enthalpy change becomes zero.

✅ Correct Answer Explanation

In an ideal solution: A-B interactions ≈ A-A interactions ≈ B-B interactions The energy needed to separate molecules is approximately equal to the energy released when new interactions form. Hence: Enthalpy of Mixing = 0

❓ Why Other Options Are Incorrect

�� Option A → Not the reason for ΔmixH = 0.
�� Option C → Chemical bonds are not broken.
�� Option D → Molecular kinetic energy never drops to zero.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the NCERT condition defining ideal solutions.

Final Logic:

�� Equal intermolecular forces lead to zero enthalpy change.

🧠 Memory Trick

Equal Forces = Zero Heat.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Enthalpy of Mixing

4 Arrange the following processes in terms of their volume of mixing (ΔmixV) from highest to lowest:
1. Solution showing large positive deviation
2. Ideal solution
3. Solution showing large negative deviation

A 1, 2, 3

B 3, 2, 1

C 2, 1, 3

D 1, 3, 2

💡 Short Explanation

�� Positive deviation often causes expansion. �� Ideal solutions show zero volume change. �� Negative deviation often causes contraction.

✅ Correct Answer Explanation

Positive deviation occurs when A-B attractions are weaker, causing expansion: ΔmixV > 0 Ideal solutions have: ΔmixV = 0 Negative deviation occurs when A-B attractions are stronger: ΔmixV < 0 Therefore: Positive Deviation > Ideal Solution > Negative Deviation

❓ Why Other Options Are Incorrect

�� Option B → Reverses the correct trend.
�� Option C → Places ideal solution first.
�� Option D → Places negative deviation ahead of ideal.

🎯 Strategy

Used

Option Grouping

Application:

�� Compare expansion and contraction behaviour.

Final Logic:

�� Expansion > Zero Change > Contraction.

🧠 Memory Trick

Positive Expands, Negative Contracts.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Volume of Mixing

5 Identify the specific deviation type when the total vapour pressure is significantly lower than expected from Raoult's law calculations:

A Positive deviation

B Negative deviation

C Ideal deviation

D Neutral deviation

💡 Short Explanation

�� Stronger A-B interactions reduce escaping tendency. �� Vapour pressure becomes lower than predicted. �� This is negative deviation.

✅ Correct Answer Explanation

In negative deviation systems: A-B interactions > A-A and B-B interactions Stronger attractions hold molecules in the liquid phase more effectively, reducing vapour pressure below Raoult's law predictions.

❓ Why Other Options Are Incorrect

�� Option A → Produces higher vapour pressure.
�� Option C → Not an NCERT classification.
�� Option D → Not a recognized term.

🎯 Strategy

Used

Elimination

Application:

�� Link lower vapour pressure with stronger intermolecular forces.

Final Logic:

�� Lower-than-expected vapour pressure indicates negative deviation.

🧠 Memory Trick

Negative = Lower Vapour Pressure.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Negative Deviation

6 The core molecular cause for a positive deviation from Raoult's law is:

A Solute molecules associating strongly into dimers.

B The escaping tendency of A and B molecules increasing due to weaker A-B forces compared to A-A and B-B.

C Strong hydrogen bonding between A and B blocking evaporation.

D Increase in the solution's overall density.

💡 Short Explanation

�� Weaker A-B attractions allow molecules to escape easily. �� Vapour pressure becomes higher. �� Positive deviation results.

✅ Correct Answer Explanation

Positive deviation occurs when: A-B interactions < A-A and B-B interactions Because of weaker attractions, molecules escape more readily into the vapour phase, increasing vapour pressure beyond Raoult's law predictions.

❓ Why Other Options Are Incorrect

�� Option A → Association usually promotes negative deviation.
�� Option C → Strong hydrogen bonding causes negative deviation.
�� Option D → Density is not the determining factor.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Relate vapour pressure increase to intermolecular force weakening.

Final Logic:

�� Weak A-B forces increase escaping tendency.

🧠 Memory Trick

Weak Bonds = More Escape.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Positive Deviation

7 Identify the exact chemical mixture that explicitly represents weaker A-B interactions compared to A-A and B-B interactions:

A Ethanol and acetone

B Nitric acid and water

C Chloroform and acetone

D Phenol and aniline

💡 Short Explanation

�� Ethanol-acetone shows positive deviation. �� Hydrogen bonding in ethanol is disrupted. �� Vapour pressure increases.

✅ Correct Answer Explanation

Acetone disrupts the hydrogen-bonding network present in pure ethanol. The resulting ethanol-acetone interactions are weaker than the original interactions, producing positive deviation from Raoult's law.

❓ Why Other Options Are Incorrect

�� Option B → Shows negative deviation.
�� Option C → Shows negative deviation.
�� Option D → Shows negative deviation.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the classic NCERT positive deviation example.

Final Logic:

�� Ethanol + Acetone = Positive Deviation.

🧠 Memory Trick

Ethanol-Acetone Escapes More.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Positive Deviation

8 In a mixture of ethanol and acetone, the positive deviation from Raoult's law occurs strictly because:

A Acetone molecules interact and break the hydrogen bonds present between pure ethanol molecules.

B Ethanol molecules break the covalent bonds inside acetone.

C Acetone forms much stronger hydrogen bonds with ethanol than with itself.

D Both molecules associate together to form an inseparable complex.

💡 Short Explanation

�� Ethanol contains strong hydrogen bonds. �� Acetone disrupts these interactions. �� Weaker A-B interactions increase vapour pressure.

✅ Correct Answer Explanation

Pure ethanol molecules are strongly associated through hydrogen bonding. When acetone is added, these hydrogen bonds are disrupted. The resulting ethanol-acetone attractions are weaker, leading to increased escaping tendency and positive deviation.

❓ Why Other Options Are Incorrect

�� Option B → Covalent bonds are not broken.
�� Option C → Stronger hydrogen bonding would cause negative deviation.
�� Option D → No inseparable complex is formed.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the NCERT explanation for ethanol-acetone behaviour.

Final Logic:

�� Breaking ethanol hydrogen bonds causes positive deviation.

🧠 Memory Trick

Acetone Breaks Ethanol H-Bonds.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Positive Deviation

📄 Passage

In the case of negative deviations from Raoult's law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B. This stronger A-B interaction decreases the escaping tendency of molecules for each component, resulting in a decrease in vapour pressure. An example of this type is a mixture of phenol and aniline, where intermolecular hydrogen bonding between the phenolic proton and the lone pair on the nitrogen atom of aniline is stronger than in the pure components. Similarly, a mixture of chloroform and acetone forms a solution with a negative deviation.

According to the passage, what specific phenomenon causes the vapour pressure to decrease, leading to negative deviations?

A Complete dissociation of the solvent molecules

B The escaping tendency of molecules increases drastically

C Stronger A-B intermolecular attractive forces than A-A and B-B

D Weaker A-B interactions than A-A and B-B

💡 Short Explanation

�� Stronger A-B attractions hold molecules in solution. �� Escaping tendency decreases. �� Vapour pressure falls.

✅ Correct Answer Explanation

The passage explicitly states that stronger A-B attractions decrease the tendency of molecules to escape into the vapour phase. This reduction lowers vapour pressure and causes negative deviation from Raoult's law.

❓ Why Other Options Are Incorrect

�� Option A → Dissociation is unrelated.
�� Option B → Increased escaping tendency causes positive deviation.
�� Option D → Weaker A-B interactions cause positive deviation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Extract the key concept directly from the passage.

Final Logic:

�� Strong A-B attractions lower vapour pressure.

🧠 Memory Trick

Strong Bonds = Low Vapour.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Negative Deviation

📄 Passage

In a mixture of chloroform and acetone, the negative deviation is attributed to which strong intermolecular force?

A Hydrogen bonding

B Weak London dispersion forces

C Ionic bonding

D Covalent coordinate bonding

💡 Short Explanation

�� Chloroform and acetone form hydrogen bonds. �� Strong intermolecular attraction develops. �� Vapour pressure decreases.

✅ Correct Answer Explanation

Chloroform contains a hydrogen atom capable of participating in hydrogen bonding with the oxygen atom of acetone. This strong intermolecular attraction lowers the escaping tendency of molecules and produces negative deviation from Raoult's law.

❓ Why Other Options Are Incorrect

�� Option B → Weak dispersion forces cannot explain the strong negative deviation.
�� Option C → Ionic bonding is not formed.
�� Option D → Coordinate bonding is not responsible.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the standard NCERT example of chloroform-acetone interaction.

Final Logic:

�� Hydrogen bonding causes negative deviation.

🧠 Memory Trick

Chloroform + Acetone = H-Bond.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Negative Deviation and Hydrogen Bonding

11 Match the following mixtures with their boiling characteristics:

List 1 (Mixture Type)	List 2 (Boiling Characteristic)
1. Ideal solution	a. Obeys Raoult's law entirely
2. Minimum boiling azeotrope	b. Shows large positive deviation from Raoult's law
3. Maximum boiling azeotrope	c. Shows large negative deviation from Raoult's law

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

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

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

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

💡 Short Explanation

�� Ideal solutions obey Raoult's law. �� Minimum boiling azeotropes show positive deviation. �� Maximum boiling azeotropes show negative deviation.

✅ Correct Answer Explanation

Ideal solutions obey Raoult's law over the entire range of concentration. Minimum boiling azeotropes arise due to large positive deviation because vapour pressure becomes higher and boiling point becomes lower. Maximum boiling azeotropes arise due to large negative deviation because vapour pressure becomes lower and boiling point becomes higher.

❓ Why Other Options Are Incorrect

�� Option B → Ideal solution and minimum boiling azeotrope are incorrectly matched.
�� Option C → Ideal solution and minimum boiling azeotrope are incorrectly matched.
�� Option D → Ideal solution and maximum boiling azeotrope are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

�� First match ideal solution with Raoult's law, then match azeotrope types using deviation.

Final Logic:

�� Ideal = Raoult; Positive deviation = Minimum boiling; Negative deviation = Maximum boiling.

🧠 Memory Trick

Positive = Minimum Boiling, Negative = Maximum Boiling.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Azeotropes and Deviations from Raoult's Law

12 Why is it chemically impossible to separate an azeotropic mixture completely by standard fractional distillation?

A Because the boiling point changes continuously throughout the process.

B Because the liquid and vapour phases have the exact same composition at the azeotropic point.

C Because one component completely evaporates before the other.

D Because the solution crystallizes upon boiling.

💡 Short Explanation

�� Azeotropes boil at constant composition. �� Liquid and vapour compositions are identical. �� Fractional distillation cannot enrich either component.

✅ Correct Answer Explanation

Fractional distillation separates liquids when vapour composition differs from liquid composition. In an azeotrope, the liquid and vapour phases have the same composition at the azeotropic point. Therefore, repeated vaporisation and condensation cannot separate the components completely.

❓ Why Other Options Are Incorrect

�� Option A → Azeotropes boil at constant temperature, not continuously changing temperature.
�� Option C → Both components distil together at azeotropic composition.
�� Option D → Crystallisation is unrelated to azeotropic distillation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Focus on the defining feature of azeotropes.

Final Logic:

�� Same liquid-vapour composition prevents complete separation.

🧠 Memory Trick

Same Vapour, Same Liquid, No Separation.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Azeotropic Mixtures

13 Consider the properties of an ethanol-water minimum boiling azeotrope:
1. It contains approximately 95% ethanol by volume.
2. It forms due to large positive deviations from Raoult's law.
3. Further separation by fractional distillation is easily possible beyond 95%.
4. Liquid and vapour have the identical composition at this point.

A 1, 2, 4

B 1, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Ethanol-water forms a minimum boiling azeotrope. �� It contains about 95% ethanol by volume. �� It cannot be separated further by fractional distillation.

✅ Correct Answer Explanation

Ethanol and water form a minimum boiling azeotrope at approximately 95% ethanol by volume. This occurs due to large positive deviation from Raoult's law. At the azeotropic composition, liquid and vapour phases have identical composition, so further separation by ordinary fractional distillation is not possible.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 3, which is incorrect, and omits correct Statements 2 and 4.
�� Option C → Includes Statement 3, which is incorrect.
�� Option D → Includes Statement 3, which is false.

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the false statement about easy further separation.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Ethanol 95 = Azeotrope Stop.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Minimum Boiling Azeotrope

14 The concentration unit explicitly used to express the approximate composition of the nitric acid maximum boiling azeotrope in the text is:

A % by mass

B % by volume

C Molarity (M)

D Parts per million (ppm)

💡 Short Explanation

�� Nitric acid-water azeotrope is expressed by mass. �� NCERT mentions about 68% nitric acid by mass. �� It is a maximum boiling azeotrope.

✅ Correct Answer Explanation

The nitric acid-water maximum boiling azeotrope is commonly described as containing approximately 68% nitric acid and 32% water by mass. Therefore, the unit used is percentage by mass.

❓ Why Other Options Are Incorrect

�� Option B → Ethanol-water azeotrope is commonly expressed by volume, not nitric acid-water.
�� Option C → Molarity is not used for this azeotrope composition.
�� Option D → ppm is used for trace quantities, not azeotropic composition.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the NCERT azeotrope example and its composition unit.

Final Logic:

�� Nitric acid-water azeotrope is expressed as % by mass.

🧠 Memory Trick

Nitric Acid Azeotrope = Mass Percent.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Maximum Boiling Azeotrope

15 Identify the specific property type that depends solely on the number of particles rather than their chemical identity:

A Additive property

B Colligative property

C Constitutive property

D Chemical property

💡 Short Explanation

�� Colligative properties depend on particle number. �� Chemical identity is not important. �� Examples include vapour pressure lowering and boiling point elevation.

✅ Correct Answer Explanation

Colligative properties are properties of solutions that depend only on the number of solute particles relative to the solvent particles and not on the chemical nature of the solute. Examples include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

❓ Why Other Options Are Incorrect

�� Option A → Additive properties depend on the sum of individual contributions.
�� Option C → Constitutive properties depend on molecular arrangement or structure.
�� Option D → Chemical properties depend on chemical identity and reactivity.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the term linked specifically with particle count.

Final Logic:

�� Particle-count-dependent property = Colligative property.

🧠 Memory Trick

Colligative = Counting Particles.

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

16 The independence of colligative properties from the specific nature of the solute practically implies that:

A 1 mol of urea and 1 mol of sucrose in 1 kg of water lower the vapour pressure by vastly different amounts.

B 1 mol of urea and 1 mol of sucrose in 1 kg of water lower the vapour pressure by nearly the exact same amount.

C The solvent's nature does not dictate the outcome at all.

D Dissociation of ionic solutes has absolutely no effect on the values.

💡 Short Explanation

�� Colligative properties depend on the number of solute particles. �� Urea and sucrose are non-electrolytes. �� Equal moles give nearly equal particle numbers.

✅ Correct Answer Explanation

Colligative properties depend mainly on the number of solute particles present, not on the chemical identity of the solute. Since urea and sucrose are both non-electrolytes, 1 mol of each gives approximately the same number of particles in solution. Therefore, both lower the vapour pressure of water by nearly the same amount.

❓ Why Other Options Are Incorrect

�� Option A → This contradicts the particle-count basis of colligative properties.
�� Option C → The solvent's nature matters because solvent constants and pure solvent properties affect the value.
�� Option D → Ionic solute dissociation increases the number of particles and affects colligative values.

🎯 Strategy

Used

Option Grouping

Application:

�� Compare the number of dissolved particles produced by both solutes.

Final Logic:

�� Same number of non-electrolyte particles gives nearly same vapour pressure lowering.

🧠 Memory Trick

Colligative = Count particles, not names.

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

17 For a dilute solution containing a non-volatile solute, the mathematical term for relative lowering of vapour pressure, (p₁° - p₁)/p₁°, is directly equal to:

A The mass of the solute

B The mole fraction of the pure solvent

C The mole fraction of the solute

D The total pressure of the solution

💡 Short Explanation

�� Relative lowering of vapour pressure is a colligative property. �� It depends on solute particle concentration. �� It equals the mole fraction of the non-volatile solute.

✅ Correct Answer Explanation

For a dilute solution containing a non-volatile solute: Relative Lowering of Vapour Pressure = (Vapour Pressure of Pure Solvent − Vapour Pressure of Solution) ÷ Vapour Pressure of Pure Solvent This value is equal to the mole fraction of the solute.

❓ Why Other Options Are Incorrect

�� Option A → Mass alone does not define relative lowering of vapour pressure.
�� Option B → The relation is equal to solute mole fraction, not solvent mole fraction.
�� Option D → Total pressure is not used in this relation.

🎯 Strategy

Used

Substitution

Application:

�� Apply the NCERT relation for relative lowering of vapour pressure.

Final Logic:

�� Relative lowering of vapour pressure = mole fraction of solute.

🧠 Memory Trick

RLVP = Solute mole fraction.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Relative Lowering of Vapour Pressure

18 When determining the molar mass of a solute from the relative lowering of vapour pressure, the assumption often made for very dilute solutions is:

A Moles of solute >> Moles of solvent

B Moles of solute << Moles of solvent

C Moles of solute exactly equal Moles of solvent

D The solution completely disobeys Raoult's Law

💡 Short Explanation

�� Dilute solutions contain very little solute. �� Solvent is present in much larger amount. �� This simplifies mole fraction calculations.

✅ Correct Answer Explanation

In very dilute solutions, the number of moles of solute is much smaller than the number of moles of solvent. Therefore: Moles of solute are much less than moles of solvent This assumption allows the total moles to be approximated mainly by the moles of solvent.

❓ Why Other Options Are Incorrect

�� Option A → This represents a concentrated solute condition, not a dilute solution.
�� Option C → Dilute solutions do not have equal solute and solvent moles.
�� Option D → Raoult's law is used for this calculation, not disobeyed.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement consistent with a dilute solution.

Final Logic:

�� Very dilute solution = solute moles much smaller than solvent moles.

🧠 Memory Trick

Dilute = Solute tiny, solvent huge.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molar Mass from Relative Lowering of Vapour Pressure

19 How does the addition of a non-volatile solute physically elevate the boiling point of the solution?

A It increases the vapour pressure, driving the boiling point down.

B It lowers the vapour pressure, requiring a higher temperature to reach atmospheric pressure.

C It forms new covalent bonds that require more heat energy to break.

D It makes the vapour pressure equal to atmospheric pressure precisely at room temperature.

💡 Short Explanation

�� Non-volatile solute lowers vapour pressure. �� Boiling occurs when vapour pressure equals atmospheric pressure. �� Higher temperature is needed to boil.

✅ Correct Answer Explanation

Addition of a non-volatile solute reduces the number of solvent molecules escaping into vapour phase. This lowers the vapour pressure of the solution. Since boiling occurs when vapour pressure becomes equal to atmospheric pressure, the solution must be heated to a higher temperature to boil.

❓ Why Other Options Are Incorrect

�� Option A → Vapour pressure decreases, not increases.
�� Option C → Boiling point elevation is not due to formation of new covalent bonds.
�� Option D → Vapour pressure does not become equal to atmospheric pressure at room temperature.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect vapour pressure lowering with the boiling point condition.

Final Logic:

�� Lower vapour pressure means more heat is required to boil.

🧠 Memory Trick

Lower vapour pressure = higher boiling point.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Elevation of Boiling Point

20 The Molal Elevation Constant (Kb) depends entirely on:

A The chemical identity of the solute

B The chemical nature of the solvent

C The concentration of the solution prepared

D The atmospheric pressure on that given day

💡 Short Explanation

�� Kb is a solvent constant. �� It depends on the nature of the solvent. �� It is independent of solute identity.

✅ Correct Answer Explanation

The molal elevation constant, Kb, is a characteristic property of the solvent. It depends on the chemical nature of the solvent and its physical properties. In boiling point elevation: Elevation in Boiling Point = Kb × Molality For a given solvent, Kb remains constant.

❓ Why Other Options Are Incorrect

�� Option A → Kb does not depend on solute identity.
�� Option C → Concentration affects boiling point elevation, not the value of Kb.
�� Option D → Atmospheric pressure does not define the solvent constant Kb.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify which factor defines a solvent-specific constant.

Final Logic:

�� Kb belongs to the solvent.

🧠 Memory Trick

Kb = Boiling constant of solvent.

📚 → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molal Elevation Constant

CUET UG Chemistry Booster Test – 2 Ideal Solutions and Colligative Properties

CUET UG Chemistry Booster Test - 2 Ideal Solutions and Colligative Properties

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