CUET UG Chemistry Booster Test - 3 Ideal Solutions and Colligative Properties -

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CUET UG Chemistry Booster Test - 3 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 - 3 Ideal Solutions and Colligative Properties

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QUESTION 1 OF 20

If two liquids A and B form a nearly ideal solution, what can be fundamentally inferred about their microscopic interactions?

Answer Review

1 If two liquids A and B form a nearly ideal solution, what can be fundamentally inferred about their microscopic interactions?

A Molecules of A and B repel each other completely.

B The cohesive forces in pure liquids A and B are nearly identical in magnitude to the adhesive forces between A and B.

C A and B undergo a strong exothermic chemical reaction.

D The partial pressure of A becomes totally independent of the mole fraction of A.

💡 Short Explanation

�� Ideal solutions have similar molecular interactions. �� A-A, B-B and A-B forces are nearly equal. 1. • This allows Raoult's law behaviour.

✅ Correct Answer Explanation

2. → In a nearly ideal solution, the attractive forces between A-A, B-B and A-B molecules are almost equal. Therefore, mixing does not cause a major change in energy or volume. This is why ideal solutions obey Raoult's law over the entire concentration range.

❓ Why Other Options Are Incorrect

�� Option A → Complete repulsion would cause strong non-ideal behaviour.
�� Option C → Ideal mixing does not involve a strong chemical reaction.
3. • Option D → In Raoult's law, partial pressure depends on mole fraction.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate options showing reaction, repulsion or independence from mole fraction.

Final Logic:

�� Ideal solution means A-B interactions are nearly equal to A-A and B-B interactions.

🧠 Memory Trick

1. → Ideal = Equal Interactions.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Ideal Solution Theory

2 Arrange the following solutions strictly in terms of deviation from ideal behaviour (Closest to ideal → Highest positive deviation → Highest negative deviation):
1. n-hexane and n-heptane
2. Ethanol and acetone
3. Phenol and aniline

A 1, 2, 3

B 3, 2, 1

C 2, 1, 3

D 1, 3, 2

💡 Short Explanation

�� n-hexane and n-heptane are nearly ideal. �� Ethanol and acetone show positive deviation. 1. • Phenol and aniline show negative deviation.

✅ Correct Answer Explanation

2. → n-Hexane and n-heptane have similar structures and intermolecular forces, so they behave nearly ideally. Ethanol and acetone show positive deviation because acetone disrupts ethanol-ethanol hydrogen bonding. Phenol and aniline show negative deviation due to stronger intermolecular hydrogen bonding between unlike molecules.

❓ Why Other Options Are Incorrect

�� Option B → Starts with negative deviation instead of closest ideal behaviour.
�� Option C → Places positive deviation before ideal behaviour.
3. • Option D → Interchanges positive and negative deviation examples.

🎯 Strategy

Used

4. Option Grouping

Application:

�� First identify the ideal pair, then classify positive and negative deviation examples.

Final Logic:

�� n-Hexane + n-heptane = ideal; ethanol + acetone = positive; phenol + aniline = negative.

🧠 Memory Trick

1. → Hexane Ideal, Ethanol Positive, Phenol Negative.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Ideal and Non-Ideal Solutions

3 Consider the stringent thermodynamic parameters defining an ideal solution:
1. Enthalpy of Mixing = 0
2. Volume Change on Mixing = 0
3. Heat is intensely absorbed during the mixing process
4. The total volume of the solution perfectly equals the sum of the volumes of the two components

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Ideal solution has no heat change. �� Ideal solution has no volume change. 1. • Total volume remains additive.

✅ Correct Answer Explanation

2. → For an ideal solution, the enthalpy of mixing is zero, meaning no heat is absorbed or released. The volume change on mixing is also zero, so the final solution volume equals the sum of the volumes of the pure components. Statement 3 is incorrect because intense heat absorption indicates non-ideal behaviour.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 3, which is incorrect.
�� Option C → Includes Statement 3 and omits Statement 1.
3. • Option D → Includes Statement 3 and omits Statement 2.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate all options containing the false heat absorption statement.

Final Logic:

�� Ideal solution requires zero enthalpy change and zero volume change.

🧠 Memory Trick

1. → Ideal Mixing = Zero Heat, Zero Volume Change.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Thermodynamics of Ideal Mixing

4 If exactly 50 mL of pure liquid A and 50 mL of pure liquid B are mixed to form an ideal solution, the expected total volume of the solution is:

A Exactly 100 mL

B Less than 100 mL due to contraction

C Greater than 100 mL due to expansion

D Undeterminable without knowing the temperatures

💡 Short Explanation

�� Ideal solution shows no volume change. �� Volume Change on Mixing = 0. 1. • 50 mL + 50 mL = 100 mL.

✅ Correct Answer Explanation

2. → In an ideal solution, there is neither contraction nor expansion on mixing. Therefore, the total volume is the simple sum of the volumes of the two components. Total Volume = 50 mL + 50 mL = 100 mL

❓ Why Other Options Are Incorrect

�� Option B → Contraction occurs in some negative deviation systems, not ideal solutions.
�� Option C → Expansion occurs in some positive deviation systems, not ideal solutions.
1. • Option D → The ideal condition already determines the volume.

🎯 Strategy

Used

2. Substitution

Application:

�� Add the volumes because ideal mixing has zero volume change.

Final Logic:

�� Ideal solution volume = sum of component volumes.

🧠 Memory Trick

1. → Ideal Volume Adds Exactly.

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

5 Match List-I with List-II regarding thermodynamic parameters and mixtures:

List 1 (Thermodynamic Parameter / Mixture)	List 2 (Description)
1. Positive Deviation	a. Enthalpy of Mixing is greater than zero
2. Negative Deviation	b. Enthalpy of Mixing is less than zero
3. Ideal Solution	c. Enthalpy of Mixing is equal to zero
4. Azeotrope formation	d. Constant boiling mixture

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

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

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

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

💡 Short Explanation

�� Positive deviation is generally endothermic. �� Negative deviation is generally exothermic. 1. • Ideal solution has zero enthalpy change.

✅ Correct Answer Explanation

2. → Positive deviation occurs when A-B interactions are weaker, so heat is absorbed during mixing and enthalpy of mixing is greater than zero. Negative deviation occurs when A-B interactions are stronger, so heat is released and enthalpy of mixing is less than zero. Ideal solutions have zero enthalpy of mixing. Azeotropes are constant boiling mixtures.

❓ Why Other Options Are Incorrect

�� Option B → Positive and negative deviation are interchanged.
�� Option C → Ideal solution and azeotrope are incorrectly matched.
3. • Option D → All major thermodynamic matches are incorrect.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Match deviation type with enthalpy change first, then match ideal solution and azeotrope.

Final Logic:

�� Positive = heat absorbed; Negative = heat released; Ideal = zero; Azeotrope = constant boiling.

🧠 Memory Trick

1. → Positive Heat In, Negative Heat Out.

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

6 In a non-ideal solution exhibiting a profound negative deviation, the actual partial vapour pressure of component A (p₁) is mathematically:

A p₁ = p₁° × x₁

B p₁ > p₁° × x₁

C p₁ < p₁° × x₁

D p₁ = 0 regardless of x₁

💡 Short Explanation

�� Negative deviation lowers vapour pressure. �� Actual vapour pressure is less than Raoult's law value. 1. • Strong A-B attractions reduce escaping tendency.

✅ Correct Answer Explanation

2. → For an ideal solution, Raoult's law predicts: Partial Vapour Pressure of Component 1 = Pure Vapour Pressure of Component 1 × Mole Fraction of Component 1 For negative deviation, the actual partial vapour pressure is lower than this predicted value because A-B interactions are stronger and molecules escape less easily. Therefore: Actual Partial Vapour Pressure of Component 1 < Pure Vapour Pressure of Component 1 × Mole Fraction of Component 1

❓ Why Other Options Are Incorrect

�� Option A → This represents ideal behaviour.
�� Option B → This represents positive deviation.
1. • Option D → Partial pressure does not become zero at all mole fractions.

🎯 Strategy

Used

2. Contextual/Tonal Matching

Application:

�� Connect negative deviation with lower-than-expected vapour pressure.

Final Logic:

�� Negative deviation means actual vapour pressure is less than Raoult's law prediction.

🧠 Memory Trick

1. → Negative Deviation = Pressure Below Prediction.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Negative Deviation from Raoult's Law

📄 Passage

When a solution does not obey Raoult's law over the entire range of concentration, it is called a non-ideal solution. If the vapour pressure of the mixture is higher than expected, the solution exhibits positive deviation. In the case of positive deviation from Raoult's law, the A-B interactions are fundamentally weaker than those between A-A or B-B. Because the intermolecular attractive forces between the solute and solvent molecules are weaker, the molecules of A (or B) will find it significantly easier to escape than they would in their pure states. This physically increases the vapour pressure. Solutions like ethanol and acetone mixtures fall precisely into this category.

From the passage, what directly results at a macroscopic level from the A-B interactions being weaker than A-A or B-B?

A The boiling point of the solution vastly increases.

B The molecules of A or B find it easier to escape into the vapour phase.

C The vapour pressure drops severely.

D The volume of the solution contracts upon mixing.

💡 Short Explanation

�� Weak A-B forces increase escaping tendency. �� More molecules enter vapour phase. 1. • Vapour pressure becomes higher.

✅ Correct Answer Explanation

2. → The passage states that when A-B interactions are weaker than A-A or B-B interactions, molecules are held less strongly in the solution. As a result, they escape more easily into the vapour phase. This produces positive deviation and increases vapour pressure.

❓ Why Other Options Are Incorrect

�� Option A → Positive deviation generally lowers boiling tendency, not vastly increases boiling point.
�� Option C → Vapour pressure increases, not drops.
3. • Option D → Contraction is associated with stronger interactions and negative deviation.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Use the direct explanation from the passage.

Final Logic:

�� Weaker A-B forces cause easier molecular escape.

🧠 Memory Trick

1. → Weak A-B = Easy Escape.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Positive Deviation from Raoult's Law

📄 Passage

Based strictly on the text passage, a mixture of ethanol and acetone will dynamically exhibit:

A A vapour pressure significantly lower than predicted by Raoult's law.

B Stronger hydrogen bonds than found in pure ethanol.

C A vapour pressure observably higher than predicted by Raoult's law.

D Perfectly ideal solution behaviour across all temperatures.

💡 Short Explanation

�� Ethanol-acetone shows positive deviation. �� Positive deviation gives higher vapour pressure. 1. • Molecules escape more easily.

✅ Correct Answer Explanation

2. → The passage explicitly states that ethanol and acetone mixtures show positive deviation. In positive deviation, A-B interactions are weaker, making molecules escape more easily. Therefore, the vapour pressure is higher than predicted by Raoult's law.

❓ Why Other Options Are Incorrect

�� Option A → Lower vapour pressure is a feature of negative deviation.
�� Option B → Stronger hydrogen bonding would cause negative deviation.
3. • Option D → Ethanol-acetone is non-ideal, not perfectly ideal.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Match ethanol-acetone with the passage definition of positive deviation.

Final Logic:

�� Ethanol-acetone → positive deviation → higher vapour pressure.

🧠 Memory Trick

1. → Ethanol-Acetone = Vapour Pressure Up.

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

9 Identify the specific structural interaction process that directly causes a negative deviation in a binary mixture of phenol and aniline:

A Formation of covalent coordinate bonds

B Generation of extensive London dispersion forces

C Formation of strong intermolecular hydrogen bonding

D Induction of ionic lattice bonding

💡 Short Explanation

�� Phenol and aniline form strong hydrogen bonding. �� A-B interactions become stronger. 1. • Vapour pressure decreases.

✅ Correct Answer Explanation

2. → In a phenol-aniline mixture, strong intermolecular hydrogen bonding occurs between the phenolic proton and the lone pair on nitrogen in aniline. These stronger A-B interactions decrease the escaping tendency of molecules and cause negative deviation from Raoult's law.

❓ Why Other Options Are Incorrect

�� Option A → Coordinate covalent bonding is not the NCERT reason.
�� Option B → London dispersion forces alone do not explain strong negative deviation.
3. • Option D → Ionic lattice bonding does not form in this molecular liquid mixture.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Recall the standard NCERT example of phenol-aniline negative deviation.

Final Logic:

�� Phenol + aniline = strong intermolecular hydrogen bonding.

🧠 Memory Trick

1. → Phenol-Aniline = Hydrogen Bond Hold.

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

10 The macroscopic negative deviation observed in a chloroform-acetone mixture fundamentally indicates that the escaping tendency of the constituent molecules:

A Increases symmetrically for both components

B Decreases for both components due to complexation

C Increases for chloroform but strictly decreases for acetone

D Remains completely unchanged from their pure states

💡 Short Explanation

�� Chloroform and acetone interact strongly. �� Hydrogen bonding reduces escaping tendency. 1. • Vapour pressure becomes lower.

✅ Correct Answer Explanation

2. → Chloroform and acetone form strong intermolecular hydrogen bonding. These stronger A-B interactions hold both components more firmly in the liquid phase, decreasing their escaping tendency. Therefore, the solution shows negative deviation and lower vapour pressure than expected.

❓ Why Other Options Are Incorrect

�� Option A → Increased escaping tendency indicates positive deviation.
�� Option C → Negative deviation reduces escaping tendency of both components.
3. • Option D → Interactions change significantly compared to pure states.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate options showing increased or unchanged escaping tendency.

Final Logic:

�� Negative deviation means stronger attraction and reduced escaping tendency.

🧠 Memory Trick

1. → Strong Complex = Less Escape.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Chloroform-Acetone Negative Deviation

11 For an azeotrope successfully reached during continuous boiling, the mole fraction of a component in the liquid phase (x) and its mole fraction in the vapour phase (y) are related as:

A x > y

B x < y

C x = y

D x = 1 - y

💡 Short Explanation

�� Azeotropes have identical liquid and vapour composition. �� They boil at constant composition. 1. • Therefore, x equals y.

✅ Correct Answer Explanation

2. → In an azeotropic mixture, the liquid phase and vapour phase have the same composition at the azeotropic point. Hence, the mole fraction of a component in the liquid phase is equal to its mole fraction in the vapour phase. Readable Expression: Mole Fraction in Liquid Phase = Mole Fraction in Vapour Phase

❓ Why Other Options Are Incorrect

�� Option A → This suggests liquid phase is richer than vapour phase, which is not true for azeotropes.
�� Option B → This suggests vapour phase is richer than liquid phase, which is not true at the azeotropic point.
1. • Option D → x = 1 - y is not the defining relation of an azeotrope.

🎯 Strategy

Used

2. Contextual/Tonal Matching

Application:

�� Use the defining property of azeotropes.

Final Logic:

�� Azeotrope means same liquid and vapour composition.

🧠 Memory Trick

1. → Azeotrope = x equals y.

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

12 Azeotropes physically limit the effectiveness of fractional distillation completely because:

A The entire solution becomes too viscous at the azeotropic boiling point.

B The vapour produced has the identical composition as the boiling liquid, preventing further enrichment of either component.

C The constituent components react chemically at the boiling temperature to form a new substance.

D The boiling point drops below room temperature abruptly.

💡 Short Explanation

�� Fractional distillation needs different liquid and vapour compositions. �� Azeotropes have identical liquid and vapour composition. 1. • Further separation becomes impossible.

✅ Correct Answer Explanation

2. → Fractional distillation works because the vapour phase is usually richer in the more volatile component than the liquid phase. In an azeotrope, the vapour and liquid have the same composition. Therefore, repeated vaporisation and condensation cannot enrich either component further.

❓ Why Other Options Are Incorrect

�� Option A → Viscosity is not the main reason fractional distillation fails.
�� Option C → Azeotrope formation is not due to chemical reaction into a new substance.
3. • Option D → Azeotropes do not necessarily boil below room temperature.

🎯 Strategy

Used

4. Elimination

Application:

�� Remove options unrelated to vapour-liquid composition.

Final Logic:

�� Same vapour and liquid composition prevents separation.

🧠 Memory Trick

1. → Same Vapour, Same Liquid, No Separation.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Fractional Distillation of Azeotropes

13 Identify the classical, industrially relevant minimum boiling azeotrope formed directly by the fermentation of sugars:

A 68% Nitric acid and 32% water

B 95% Ethanol by volume in water

C 100% strictly pure ethanol

D 35% Ethylene glycol by volume in water

💡 Short Explanation

�� Ethanol-water forms a minimum boiling azeotrope. �� It contains about 95% ethanol by volume. 1. • It cannot be separated further by ordinary fractional distillation.

✅ Correct Answer Explanation

2. → Ethanol obtained by fermentation forms an azeotropic mixture with water. This azeotrope contains approximately 95% ethanol by volume and behaves as a minimum boiling azeotrope due to positive deviation from Raoult's law.

❓ Why Other Options Are Incorrect

�� Option A → 68% nitric acid and 32% water forms a maximum boiling azeotrope.
�� Option C → 100% pure ethanol is not the azeotropic composition obtained by ordinary fractional distillation.
3. • Option D → Ethylene glycol in water is used as antifreeze, not the fermentation azeotrope.

🎯 Strategy

Used

4. Odd One Out

Application:

�� Identify the standard NCERT example of minimum boiling azeotrope.

Final Logic:

�� Fermentation ethanol-water azeotrope = 95% ethanol by volume.

🧠 Memory Trick

1. → Ethanol 95 = Minimum Boiling Azeotrope.

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

14 A maximum boiling azeotrope is thermodynamically formed by binary solutions showing:

A Extremely large positive deviations from Raoult's law

B Extremely large negative deviations from Raoult's law

C Zero deviation from Raoult's law across all concentration ranges

D Total immiscibility at high temperatures

💡 Short Explanation

�� Negative deviation lowers vapour pressure. �� Lower vapour pressure raises boiling point. 1. • Large negative deviation forms maximum boiling azeotrope.

✅ Correct Answer Explanation

2. → A maximum boiling azeotrope forms when a solution shows large negative deviation from Raoult's law. In such solutions, A-B interactions are stronger than A-A and B-B interactions, causing lower vapour pressure and a higher boiling point.

❓ Why Other Options Are Incorrect

�� Option A → Large positive deviation forms minimum boiling azeotropes.
�� Option C → Zero deviation represents ideal solution behaviour.
3. • Option D → Immiscibility is not the cause of maximum boiling azeotrope formation.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Relate deviation type to vapour pressure and boiling point.

Final Logic:

�� Negative deviation → lower vapour pressure → maximum boiling azeotrope.

🧠 Memory Trick

1. → Negative Deviation = Maximum Boiling.

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

15 The unit of molality (m), a key concentration term often used because colligative properties depend heavily on particle ratios, is:

A mol L⁻¹

B mol kg⁻¹

C g mol⁻¹

D L mol⁻¹

💡 Short Explanation

�� Molality means moles of solute per kilogram of solvent. �� It uses mass of solvent, not volume of solution. 1. • Its unit is mol kg⁻¹.

✅ Correct Answer Explanation

2. → Molality is defined as the number of moles of solute dissolved per kilogram of solvent. Readable Formula: Molality = Moles of Solute ÷ Mass of Solvent in kg Therefore, the unit of molality is mol kg⁻¹.

❓ Why Other Options Are Incorrect

�� Option A → mol L⁻¹ is the unit of molarity.
�� Option C → g mol⁻¹ is the unit of molar mass.
1. • Option D → L mol⁻¹ is not the unit of molality.

🎯 Strategy

Used

2. Dimensional/Unit Analysis

Application:

�� Use the definition of molality to identify the unit.

Final Logic:

�� Moles per kilogram = mol kg⁻¹.

🧠 Memory Trick

1. → Molality = Mole per kg.

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

16 If 1.0 mol of an arbitrary non-electrolyte non-volatile solute is dissolved in 1 kg of water, the magnitude of the resulting colligative property depends strictly on:

A The exact molar mass of the chosen solute

B The detailed chemical structure and functional groups of the solute

C The total number of particles dissolved in the given mass of solvent

D The solid density of the pure solute prior to dissolution

💡 Short Explanation

�� Colligative properties depend on particle number. �� Non-electrolytes do not dissociate. 1. • Chemical identity is not the deciding factor.

✅ Correct Answer Explanation

2. → Colligative properties depend only on the number of solute particles present relative to the amount of solvent. For a non-electrolyte non-volatile solute, 1.0 mol gives Avogadro's number of particles and does not dissociate into ions. Therefore, the magnitude of the colligative property depends on the total number of dissolved particles, not the molar mass, structure, or density of the solute.

❓ Why Other Options Are Incorrect

�� Option A → Molar mass is not the deciding factor once the number of moles is fixed.
�� Option B → Chemical structure and functional groups do not directly determine colligative property magnitude.
3. • Option D → Density of the pure solid is unrelated to colligative behaviour.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate properties related to solute identity and select particle-number dependence.

Final Logic:

�� Colligative property magnitude depends on number of particles.

🧠 Memory Trick

1. → Colligative = Count, not character.

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

17 In the formal mathematical expression for the relative lowering of vapour pressure, (p₁° - p₁)/p₁° = x₂, what does the specific term p₁° strictly represent?

A The final vapour pressure of the completely mixed solution

B The vapour pressure of the pure unmixed solvent

C The partial pressure of the dissolved solute

D The ambient atmospheric pressure during measurement

💡 Short Explanation

�� p₁° refers to pure solvent vapour pressure. �� It is measured before solute addition. 1. • It is used as the reference pressure.

✅ Correct Answer Explanation

2. → In the relative lowering of vapour pressure expression: Readable Formula: Relative Lowering of Vapour Pressure = (Vapour Pressure of Pure Solvent − Vapour Pressure of Solution) ÷ Vapour Pressure of Pure Solvent Here, p₁° represents the vapour pressure of the pure solvent before adding the non-volatile solute. It acts as the reference value for calculating how much vapour pressure has decreased.

❓ Why Other Options Are Incorrect

�� Option A → This refers to p₁, the vapour pressure of the solution.
�� Option C → Non-volatile solute does not contribute vapour pressure.
1. • Option D → Atmospheric pressure is not represented by p₁°.

🎯 Strategy

Used

2. Contextual/Tonal Matching

Application:

�� Identify the meaning of the superscript zero in vapour pressure notation.

Final Logic:

�� Superscript zero means pure component state.

🧠 Memory Trick

1. → p° = Pure pressure.

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

18 To calculate the exact molar mass of a solute from the relative lowering of vapour pressure without making any dilute-solution approximations, one must accurately use the expression:

A Δp₁/p₁° = n₂ / n₁

B Δp₁/p₁° = n₂ / (n₁ + n₂)

C Δp₁/p₁° = n₁ / n₂

D Δp₁/p₁° = (n₁ + n₂) / n₂

💡 Short Explanation

�� Relative lowering equals solute mole fraction. �� Exact mole fraction uses total moles. 1. • Total moles = moles of solvent + moles of solute.

✅ Correct Answer Explanation

2. → Relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. Readable Formula: Relative Lowering of Vapour Pressure = Moles of Solute ÷ Total Moles of Solution Therefore: Relative Lowering of Vapour Pressure = Moles of Solute ÷ (Moles of Solvent + Moles of Solute) So the exact expression is: Δp₁/p₁° = n₂ / (n₁ + n₂)

❓ Why Other Options Are Incorrect

�� Option A → This is an approximation used only for very dilute solutions.
�� Option C → Inverts the solute and solvent mole relationship.
1. • Option D → Inverts the correct expression.

🎯 Strategy

Used

2. Substitution

Application:

�� Substitute the definition of solute mole fraction without approximation.

Final Logic:

�� Exact solute mole fraction = n₂ ÷ (n₁ + n₂).

🧠 Memory Trick

1. → Exact mole fraction uses total moles.

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

19 Evaluate the following advanced statements regarding boiling point elevation:
1. The complete vapour pressure curve of the solution lies strictly below that of the pure solvent.
2. The solution must be heated to a demonstrably higher temperature to make its newly lowered vapour pressure equal to atmospheric pressure.
3. For dilute solutions, boiling point elevation is directly proportional to molality.
4. The boiling point of any 1 M non-volatile solution is inherently lower than the pure solvent.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Non-volatile solute lowers vapour pressure. �� Higher temperature is required for boiling. 1. • Boiling point elevation is proportional to molality for dilute solutions.

✅ Correct Answer Explanation

2. → When a non-volatile solute is added, the vapour pressure curve of the solution lies below that of the pure solvent. Since boiling occurs when vapour pressure equals atmospheric pressure, the solution must be heated to a higher temperature. For dilute solutions: Readable Formula: Elevation in Boiling Point = Molal Elevation Constant × Molality Thus, boiling point elevation is directly proportional to molality. Statement 4 is incorrect because a non-volatile solute raises, not lowers, the boiling point.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 4, which is incorrect.
�� Option C → Includes Statement 4 and omits Statement 1.
1. • Option D → Includes Statement 4 and omits Statement 2.

🎯 Strategy

Used

2. Elimination

Application:

�� Remove options containing the false statement about boiling point becoming lower.

Final Logic:

�� Statements 1, 2 and 3 correctly describe boiling point elevation.

🧠 Memory Trick

1. → Solute lowers vapour pressure, raises boiling point.

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

20 The molal elevation constant (Kb) can be calculated purely theoretically using the precise enthalpy of vaporisation (ΔvapH) via the thermodynamic relation:

A Kb = (R × M₁ × T_b²) / (1000 × ΔvapH)

B Kb = (1000 × ΔvapH) / (R × M₁ × T_b²)

C Kb = (R × M₁ × T_b) / (1000 × ΔvapH)

D Kb = (R × T_b²) / (1000 × M₁ × ΔvapH)

💡 Short Explanation

�� Kb depends on solvent properties. �� It includes gas constant, molar mass of solvent, boiling point and enthalpy of vaporisation. 1. • The correct relation has Tb squared in the numerator.

✅ Correct Answer Explanation

2. → The theoretical expression for molal elevation constant is: Readable Formula: Molal Elevation Constant = (Gas Constant × Molar Mass of Solvent × Boiling Point of Solvent Squared) ÷ (1000 × Enthalpy of Vaporisation) Symbol Form: Kb = (R × M₁ × T_b²) ÷ (1000 × ΔvapH) Here: R = Gas constant M₁ = Molar mass of solvent T_b = Boiling point of solvent in Kelvin ΔvapH = Enthalpy of vaporisation of solvent

❓ Why Other Options Are Incorrect

�� Option B → Inverts the correct formula.
�� Option C → Uses Tb instead of Tb squared.
1. • Option D → Places molar mass of solvent in the denominator incorrectly.

🎯 Strategy

Used

2. Dimensional/Unit Analysis

Application:

�� Identify the standard NCERT thermodynamic relation for Kb.

Final Logic:

�� Kb contains R, M₁ and Tb² in the numerator, and 1000 × ΔvapH in the denominator.

🧠 Memory Trick

1. → Kb = R M T squared over 1000 ΔH.

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

CUET UG Chemistry Booster Test – 3 Ideal Solutions and Colligative Properties

CUET UG Chemistry Booster Test - 3 Ideal Solutions and Colligative Properties

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