CUET UG Chemistry Booster Test - 3 Henry's Law and Raoult's Law -

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CUET UG Chemistry Booster Test - 3 Henry's Law and Raoult's Law

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Henry's Law and Raoult's Law

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QUESTION 1 OF 20

📄 Passage

For solution of gases in a solvent, consider a system where the lower part is solution and the upper part is a gaseous system at pressure p and temperature T. Assume this system to be in a state of dynamic equilibrium; under these conditions rate of gaseous particles entering and leaving the solution phase is the same. Now increase the pressure over the solution phase by compressing the gas to a smaller volume. This will increase the number of gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of solution to enter it. The solubility of the gas will increase until a new equilibrium is reached.

Before the new equilibrium is reached upon compressing the gas, what dynamic change occurs?

Answer Review

📄 Passage

Before the new equilibrium is reached upon compressing the gas, what dynamic change occurs?

A The rate of evaporation instantly drops to zero

B The rate of gaseous particles striking the surface to enter the solution increases

C The solvent molecules start freezing

D Gaseous particles stop leaving the solution entirely

💡 Short Explanation

�� Compression increases gas particles per unit volume. �� More particles strike the solution surface. 1. • Gas solubility increases until new equilibrium is reached.

✅ Correct Answer Explanation

2. → When the gas is compressed, the same number of gaseous particles occupy a smaller volume. This increases the number of gaseous particles per unit volume above the solution. As a result, more gas particles strike the surface of the solution and enter the liquid phase, increasing solubility until a new dynamic equilibrium is established.

❓ Why Other Options Are Incorrect

�� Option A → Evaporation does not instantly become zero.
�� Option C → Solvent freezing is unrelated to pressure increase in this context.
3. • Option D → Gas particles do not completely stop leaving the solution.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Use the direct cause-effect explanation given in the passage.

Final Logic:

�� Compression increases gas particle collisions with the solution surface.

🧠 Memory Trick

1. → Compress Gas, More Particles Hit.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Pressure Effect on Gas Solubility

2 Based on the passage, the newly established dynamic equilibrium after compression is defined by:

A A lower partial pressure of the gas

B An increased solubility and higher pressure of the gas above the solution

C A complete halt in particle movement between phases

D A volume expansion of the liquid

💡 Short Explanation

�� Compression increases gas pressure. �� Higher pressure increases gas solubility. 1. • A new dynamic equilibrium is reached.

✅ Correct Answer Explanation

2. → After compression, pressure over the solution increases. This increases the number of gas particles striking and entering the liquid phase. Therefore, more gas dissolves until a new equilibrium is reached at higher gas pressure and increased solubility.

❓ Why Other Options Are Incorrect

�� Option A → Compression increases pressure, not lowers it.
�� Option C → Dynamic equilibrium does not mean particle movement stops.
3. • Option D → The liquid does not expand significantly due to compression of gas.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate options contradicting pressure-solubility relation and dynamic equilibrium.

Final Logic:

�� Higher pressure produces higher solubility before new equilibrium.

🧠 Memory Trick

1. → New Pressure, New Solubility.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Gas-Liquid Dynamic Equilibrium

3 Henry's Law properties:
1. High KH values imply lower solubility at a constant given pressure
2. Mole fraction of gas in solution is inversely proportional to partial pressure
3. Only applicable for solids dissolving in liquids
4. Provides a qualitative rather than quantitative relationship

A 1

B 1, 2

C 3, 4

D 1, 3, 4

💡 Short Explanation

�� Henry's Law: p = KHx. �� At constant pressure, x = p/KH. 1. • Higher KH means lower solubility.

✅ Correct Answer Explanation

2. → Henry's Law is expressed as p = KHx. At a given pressure, x = p/KH. Therefore, if KH is high, the mole fraction of gas dissolved in the solution is low. Statement 2 is incorrect because mole fraction is directly proportional to pressure, not inversely proportional. Statement 3 is incorrect because Henry's Law applies to gases in liquids. Statement 4 is incorrect because Henry's Law gives a quantitative relationship.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

4. Elimination

Application:

�� Use Henry's Law equation to check each statement.

Final Logic:

�� Only Statement 1 correctly follows from p = KHx.

🧠 Memory Trick

1. → High KH = Hard to Dissolve.

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

4 If N₂ gas exerts a partial pressure of 0.987 bar and KH for N₂ at 293 K is 76,480 bar, the mole fraction of N₂ in the aqueous solution is:

A 0.987 / 76.48

B 76,480 / 0.987

C 0.987 / 76,480

D 0.987 × 76,480

💡 Short Explanation

�� Henry's Law: p = KHx. �� Rearranged form: x = p/KH. 1. • Substitute p = 0.987 bar and KH = 76,480 bar.

✅ Correct Answer Explanation

2. → According to Henry's Law: p = KHx Therefore, x = p / KH Substituting the given values: x = 0.987 / 76,480 Thus, the mole fraction of N₂ in the aqueous solution is represented by Option C.

❓ Why Other Options Are Incorrect

�� Option A → Uses 76.48 instead of 76,480.
�� Option B → Inverts the correct formula.
1. • Option D → Multiplies p and KH instead of dividing.

🎯 Strategy

Used

2. Substitution

Application:

�� Rearrange Henry's Law and substitute the given values.

Final Logic:

�� x = p/KH = 0.987/76,480.

🧠 Memory Trick

1. → Henry: x = Pressure ÷ KH.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law Calculation

5 Arrange the following gases in decreasing order of their KH values in water at 293 K:
1. He
2. H₂
3. N₂
4. O₂

A 1, 3, 2, 4

B 1, 2, 3, 4

C 4, 3, 2, 1

D 3, 1, 4, 2

💡 Short Explanation

�� He has KH = 144.97 kbar. �� N₂ has KH = 76.48 kbar. 1. • H₂ has KH = 69.16 kbar and O₂ has KH = 34.86 kbar.

✅ Correct Answer Explanation

2. → Given KH values: He = 144.97 kbar H₂ = 69.16 kbar N₂ = 76.48 kbar O₂ = 34.86 kbar Decreasing order: He > N₂ > H₂ > O₂ Therefore, the correct order is 1, 3, 2, 4.

❓ Why Other Options Are Incorrect

�� Option B → Places H₂ before N₂, which is incorrect.
�� Option C → Gives reverse-type order starting with the lowest KH.
1. • Option D → Places N₂ before He, which is incorrect.

🎯 Strategy

Used

2. Option Grouping

Application:

�� Compare numerical KH values from highest to lowest.

Final Logic:

�� 144.97 > 76.48 > 69.16 > 34.86.

🧠 Memory Trick

1. → He Highest, O₂ Lowest.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law Constant Values

6 The unit of KH is expressed as kbar in some tables. The conversion to standard bar units means 1 kbar is equal to:

A 100 bar

B 10 bar

C 1000 bar

D 0.001 bar

💡 Short Explanation

�� k means kilo. �� 1 kilo = 1000. 1. • Therefore, 1 kbar = 1000 bar.

✅ Correct Answer Explanation

2. → The prefix "kilo" represents 1000. Hence, 1 kilobar equals 1000 bar. Therefore, when KH is expressed in kbar, it can be converted to bar by multiplying by 1000.

❓ Why Other Options Are Incorrect

�� Option A → 100 bar is too small.
�� Option B → 10 bar is too small.
3. • Option D → 0.001 bar is the inverse conversion idea, not 1 kbar.

🎯 Strategy

Used

4. Dimensional/Unit Analysis

Application:

�� Apply metric prefix conversion.

Final Logic:

�� kbar = kilobar = 1000 bar.

🧠 Memory Trick

1. → kilo = 1000.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law Constant Units

7 Based on Le Chatelier's Principle, because dissolution of a gas in a liquid releases heat (ΔsolH < 0), what happens when the ambient temperature is elevated?

A The equilibrium shifts backward, decreasing gas solubility

B The equilibrium shifts forward, maximizing dissolution

C The process becomes endothermic spontaneously

D The KH value drops to zero

💡 Short Explanation

�� Gas dissolution is exothermic. �� Heating opposes the exothermic direction. 1. • Gas solubility decreases.

✅ Correct Answer Explanation

2. → Dissolution of gases in liquids is generally exothermic. According to Le Chatelier's Principle, increasing temperature shifts the equilibrium in the direction that absorbs heat. This favours the reverse process, causing gas to escape from the solution and decreasing gas solubility.

❓ Why Other Options Are Incorrect

�� Option B → Heating does not favour exothermic dissolution.
�� Option C → The process does not spontaneously change its thermodynamic nature.
3. • Option D → KH does not become zero.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Connect exothermic dissolution with the effect of heating.

Final Logic:

�� Heat added shifts equilibrium backward, reducing gas solubility.

🧠 Memory Trick

1. → Heat Pushes Gas Out.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Temperature Effect on Gas Solubility

8 Identify the reaction type for the phase change when a gas molecule enters the liquid phase and releases heat:

A Exothermic condensation process

B Endothermic vaporization process

C Isochoric heating

D Adiabatic expansion

💡 Short Explanation

�� Gas enters liquid phase. �� Heat is released. 1. • This resembles condensation.

✅ Correct Answer Explanation

2. → When a gas molecule dissolves in a liquid, it moves from the gaseous phase into a more condensed liquid environment. This resembles condensation and releases heat, so it is treated as an exothermic process.

❓ Why Other Options Are Incorrect

�� Option B → Vaporization is liquid to gas and usually endothermic.
�� Option C → Isochoric heating refers to heating at constant volume.
3. • Option D → Adiabatic expansion involves expansion without heat exchange.

🎯 Strategy

Used

4. Odd One Out

Application:

�� Identify the process most similar to gas entering a liquid phase.

Final Logic:

�� Gas to condensed phase with heat release = exothermic condensation.

🧠 Memory Trick

1. → Gas In, Heat Out.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Dissolution of Gases in Liquids

9 Identify the name of the medical condition characterized by climbers becoming weak and unable to think clearly due to low blood oxygen at high altitudes:

A The Bends

B Anoxia

C Decompression Sickness

D Edema

💡 Short Explanation

�� High altitude has low oxygen partial pressure. �� Blood oxygen concentration decreases. 1. • This condition is called anoxia.

✅ Correct Answer Explanation

2. → At high altitudes, the partial pressure of oxygen is low. This reduces the concentration of oxygen in blood and body tissues. As a result, climbers may become weak and unable to think clearly. This condition is called anoxia.

❓ Why Other Options Are Incorrect

�� Option A → The bends occur in scuba divers due to nitrogen bubbles.
�� Option C → Decompression sickness is another name for bends.
3. • Option D → Edema refers to fluid accumulation in tissues.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Link high altitude oxygen deficiency with the correct medical term.

Final Logic:

�� Low oxygen supply at high altitude = anoxia.

🧠 Memory Trick

1. → Anoxia = No Oxygen.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Applications of Henry's Law

10 Match the phenomenological application (List-I) with its underlying reasoning (List-II):

List 1 (Phenomenological Application)	List 2 (Underlying Reasoning)
1. Soda bottles sealed under high pressure	a. Increases the solubility of CO₂ in the beverage
2. Climbers experience Anoxia	b. Low atmospheric partial pressure lowers blood O₂ concentration
3. Scuba tanks contain 11.7% Helium	c. Dilution minimizes toxic effects of N₂
4. Scuba divers suffer from Bends	d. Ascending decreases pressure, releasing dissolved N₂ bubbles

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

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

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

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

💡 Short Explanation

�� Soda bottles use high pressure to dissolve CO₂. �� Climbers face low oxygen partial pressure. 1. • Helium reduces nitrogen toxicity in scuba tanks.

✅ Correct Answer Explanation

2. → Soda bottles are sealed under high pressure to increase the solubility of CO₂ in the beverage. Climbers experience anoxia because low atmospheric partial pressure lowers oxygen concentration in blood. Scuba tanks contain helium to dilute nitrogen and reduce toxic effects. Bends occur when ascending divers face lower pressure, causing dissolved nitrogen to form bubbles.

❓ Why Other Options Are Incorrect

�� Option B → Soda and climber applications are incorrectly matched.
�� Option C → Scuba and soda applications are incorrectly paired.
3. • Option D → All major applications are mismatched.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Match each real-life phenomenon with the appropriate Henry's Law application.

Final Logic:

�� Soda-CO₂ pressure; Climber-anoxia; Helium-N₂ dilution; Bends-N₂ bubbles.

🧠 Memory Trick

1. → Soda CO₂, Climber O₂, Diver N₂.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Applications of Henry's Law

11 At equilibrium, if component 1 has p₁⁰ = 200 mm Hg and component 2 has p₂⁰ = 415 mm Hg, the vapour phase will always be richer in:

A Component 1, because it is less volatile

B Component 2, because it is more volatile

C Neither, they evaporate equally

D The component with the larger molar mass

💡 Short Explanation

�� Higher vapour pressure indicates greater volatility. �� More volatile components contribute more to the vapour phase. 1. • Component 2 has a higher vapour pressure than component 1.

✅ Correct Answer Explanation

2. → Vapour pressure is a measure of a liquid's tendency to escape into the vapour phase. Since component 2 has a vapour pressure of 415 mm Hg compared to 200 mm Hg for component 1, it is more volatile. Therefore, the vapour phase above the solution will contain a greater proportion of component 2 and will be richer in that component.

❓ Why Other Options Are Incorrect

�� Option A → Less volatile components contribute less to the vapour phase.
�� Option C → Components do not evaporate equally unless they have identical vapour pressures.
3. • Option D → Vapour composition depends primarily on volatility, not molar mass.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Compare vapour pressures to identify the more volatile component.

Final Logic:

�� Higher vapour pressure means greater vapour-phase concentration.

🧠 Memory Trick

1. → Higher Vapour Pressure = Higher Vapour Presence.

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

12 Vapour pressure of binary liquid solution properties:
(A) The partial vapour pressure of each component varies linearly with its mole fraction
(B) The minimum value of p_total corresponds to the vapour pressure of the less volatile pure component
(C) The plots for partial pressures pass through points where x = 1 and x = 0
(D) The total pressure over the solution is the product of partial pressures

A A, B, C

B A, C

C B, D

D A, B, C, D

💡 Short Explanation

�� Raoult's law predicts linear variation. �� Total pressure lies between the vapour pressures of pure components. 1. • Total pressure is the sum, not the product, of partial pressures.

✅ Correct Answer Explanation

2. → For an ideal binary solution, Raoult's law gives: p₁ = x₁p₁⁰ p₂ = x₂p₂⁰ Thus, partial vapour pressures vary linearly with mole fraction. The graph passes through x = 0 and x = 1. The minimum total vapour pressure corresponds to the vapour pressure of the less volatile component. Total vapour pressure is given by: p_total = p₁ + p₂ not their product.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement B, which is correct.
�� Option C → Statement D is incorrect.
1. • Option D → Total pressure is not the product of partial pressures.

🎯 Strategy

Used

2. Elimination

Application:

�� Check each statement using Raoult's law and Dalton's law.

Final Logic:

�� A, B and C are correct; D is incorrect.

🧠 Memory Trick

1. → Raoult = Linear, Dalton = Sum.

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

13 According to Raoult's Law, the equation p_total = p₁⁰ + (p₂⁰ - p₁⁰) x₂ mathematically proves that:

A Total vapour pressure decreases non-linearly with x₂

B Total vapour pressure can be expressed purely in terms of the mole fraction of one component

C Total vapour pressure is independent of the pure vapour pressures

D Ideal solutions cannot contain two volatile components

💡 Short Explanation

�� The equation contains only x₂. �� Total pressure changes linearly with mole fraction. 1. • Pure vapour pressures remain important constants.

✅ Correct Answer Explanation

2. → Since x₁ + x₂ = 1, Raoult's law can be rearranged into: p_total = p₁⁰ + (p₂⁰ − p₁⁰)x₂ This expression shows that total vapour pressure can be represented entirely using the mole fraction of a single component (x₂), along with the pure vapour pressures.

❓ Why Other Options Are Incorrect

�� Option A → The relationship is linear, not non-linear.
�� Option C → Pure vapour pressures appear directly in the equation.
1. • Option D → Ideal solutions commonly contain two volatile liquids.

🎯 Strategy

Used

2. Substitution

Application:

�� Use x₁ + x₂ = 1 to derive the equation.

Final Logic:

�� Total pressure can be written solely in terms of x₂.

🧠 Memory Trick

1. → One Mole Fraction Can Describe the Whole System.

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

14 In Raoult's formula p₁ = p₁⁰ × x₁, the identity of p₁⁰ denotes:

A The partial pressure of the component in the mixture

B The absolute atmospheric pressure at boiling point

C The vapour pressure of the pure component at that exact temperature

D The Henry's law constant for the solvent

💡 Short Explanation

�� p₁⁰ refers to the pure liquid. �� It is measured at the same temperature. 1. • It acts as the proportionality constant in Raoult's law.

✅ Correct Answer Explanation

2. → In Raoult's law: p₁ = x₁p₁⁰ p₁⁰ represents the vapour pressure exerted by the pure liquid component at the given temperature. It is a characteristic property of that liquid and serves as the proportionality constant.

❓ Why Other Options Are Incorrect

�� Option A → p₁ is the partial pressure in the mixture.
�� Option B → Atmospheric pressure is unrelated.
1. • Option D → Henry's constant is represented by KH.

🎯 Strategy

Used

2. Contextual/Tonal Matching

Application:

�� Identify the meaning of the superscript zero.

Final Logic:

�� Superscript zero indicates the pure component state.

🧠 Memory Trick

1. → p⁰ = Pure Pressure.

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

15 Binary liquid mole fraction calculations (vapour phase) properties:
(A) The mole fraction of a component in vapour phase is yi = pi / p_total
(B) Based directly on Dalton's law of partial pressures
(C) Always richer in the more volatile component
(D) Depends on the osmotic pressure of the solution

A A, B, C

B A, B

C C, D

D B, C, D

💡 Short Explanation

�� Vapour composition is calculated using Dalton's law. �� More volatile components dominate the vapour phase. 1. • Osmotic pressure is unrelated.

✅ Correct Answer Explanation

2. → Dalton's law gives: yi = pi / p_total Therefore statement A is correct. The equation is derived directly from Dalton's law, making statement B correct. Since the more volatile component contributes more vapour, the vapour phase becomes richer in that component, making statement C correct. Osmotic pressure has no role in vapour-phase composition.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement C.
�� Option C → Statement D is incorrect.
1. • Option D → Statement D is not related to vapour-phase composition.

🎯 Strategy

Used

2. Elimination

Application:

�� Verify each statement using Dalton's law and volatility concepts.

Final Logic:

�� A, B and C are correct; D is incorrect.

🧠 Memory Trick

1. → Vapour Loves the More Volatile Component.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Dalton's Law and Vapour Phase Composition

16 For a binary solution where x₂ = 0.688, x₁ = 0.312, and p_total = 347.9 mm Hg, if p₂ = 285.5 mm Hg, what is the mole fraction y₂ of component 2 in the vapour phase?

A 285.5 / 415

B 285.5 / 347.9

C 0.688

D 347.9 / 285.5

💡 Short Explanation

�� Vapour-phase mole fraction is calculated using Dalton's law. �� y₂ = p₂ / p_total. 1. • Substitute p₂ = 285.5 and p_total = 347.9.

✅ Correct Answer Explanation

2. → According to Dalton's law of partial pressures, the mole fraction of a component in the vapour phase is given by: y₂ = p₂ / p_total Here, p₂ = 285.5 mm Hg p_total = 347.9 mm Hg Therefore, y₂ = 285.5 / 347.9 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uses p₂⁰ instead of p_total in the denominator.
�� Option C → Represents liquid-phase mole fraction x₂, not vapour-phase mole fraction y₂.
1. • Option D → Inverts the correct formula.

🎯 Strategy

Used

2. Dimensional/Unit Analysis

Application:

�� Use Dalton's law to compare partial pressure with total pressure.

Final Logic:

�� Vapour mole fraction = partial pressure / total pressure.

🧠 Memory Trick

1. → y = vapour pressure part / total pressure.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Vapour Phase Composition using Dalton's Law

17 Comparing p₁ = x₁p₁⁰ and p = KHx, Raoult's Law behaves essentially as a specialized instance of Henry's Law. What fundamentally distinguishes p₁⁰ from KH in typical solutions?

A p₁⁰ is experimentally measurable for a pure liquid, whereas KH describes a gas solute in a specific solvent.

B KH is unitless while p₁⁰ is measured in Pascals.

C KH only applies to non-ideal solutions.

D p₁⁰ varies with pressure but KH does not.

💡 Short Explanation

�� p₁⁰ is pure component vapour pressure. �� KH is Henry's law constant for gas solubility. 1. • Both act as proportionality constants.

✅ Correct Answer Explanation

2. → In Raoult's Law, p₁⁰ represents the vapour pressure of the pure liquid component at a given temperature. In Henry's Law, KH represents the proportionality constant for the solubility of a gas in a particular solvent at a given temperature. Thus, both are constants in pressure–mole fraction relations, but they refer to different physical systems.

❓ Why Other Options Are Incorrect

�� Option B → KH is not unitless; it has pressure units.
�� Option C → Henry's Law is not limited only to non-ideal solutions.
3. • Option D → p₁⁰ mainly depends on temperature, not external pressure in this context.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Compare the physical meaning of each proportionality constant.

Final Logic:

�� p₁⁰ belongs to pure liquid vapour pressure; KH belongs to gas-solvent solubility.

🧠 Memory Trick

1. → p⁰ = Pure liquid; KH = Gas Henry constant.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Relationship between Raoult's Law and Henry's Law

18 In comparing Henry's Law and Raoult's Law, the partial pressure of the volatile component is:

A Directly proportional to its mole fraction in both laws

B Inversely proportional to its mole fraction in both laws

C Proportional to volume in Raoult's law

D Independent of mole fraction in Henry's law

💡 Short Explanation

�� Henry's Law: p = KHx. �� Raoult's Law: p₁ = p₁⁰x₁. 1. • Both show direct proportionality.

✅ Correct Answer Explanation

2. → Henry's Law states that the partial pressure of a gas is directly proportional to its mole fraction in solution. Raoult's Law also states that the partial vapour pressure of a volatile component is directly proportional to its mole fraction in the liquid phase. Therefore, both laws use mole fraction as the proportional variable.

❓ Why Other Options Are Incorrect

�� Option B → Both laws show direct, not inverse, proportionality.
�� Option C → Raoult's law does not relate partial pressure to volume.
3. • Option D → Henry's law directly depends on mole fraction.

🎯 Strategy

Used

4. Substitution

Application:

�� Compare the two equations term by term.

Final Logic:

�� In both laws, p ∝ x.

🧠 Memory Trick

1. → Pressure follows mole fraction.

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

19 The fraction of the surface covered by solvent molecules is reduced when a non-volatile solute is added. This physical mechanism directly results in:

A An increase in boiling point due to reduced vapour pressure

B A sudden spike in the liquid's partial pressure

C Complete immiscibility of the solute

D The solvent molecules turning into a solid state

💡 Short Explanation

�� Non-volatile solute reduces solvent surface coverage. �� Fewer solvent molecules escape. 1. • Vapour pressure decreases, causing boiling point elevation.

✅ Correct Answer Explanation

2. → When a non-volatile solute is added, part of the liquid surface is occupied by solute particles. This reduces the number of solvent molecules escaping into the vapour phase and lowers vapour pressure. Since a lower vapour pressure requires a higher temperature to equal atmospheric pressure, the boiling point increases.

❓ Why Other Options Are Incorrect

�� Option B → Vapour pressure decreases instead of spiking.
�� Option C → Solubility or miscibility is not the direct result described.
3. • Option D → Solvent molecules do not turn into a solid because of vapour pressure lowering.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Connect surface reduction with vapour pressure lowering and boiling point elevation.

Final Logic:

�� Reduced solvent escape → lower vapour pressure → higher boiling point.

🧠 Memory Trick

1. → Less vapour, more heat to boil.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Vapour Pressure Lowering and Boiling Point Elevation

20 If 1.0 mol of urea and 1.0 mol of sucrose are added separately to 1 kg samples of water at the same temperature, Raoult's law predicts that the vapour pressure lowering in both containers will be:

A Dependent on the molar mass of urea vs sucrose

B Exactly identical because the number of solute particles is identical

C Higher for sucrose due to its larger molecular size

D Lower for urea due to hydrogen bonding

💡 Short Explanation

�� Vapour pressure lowering is colligative. �� It depends on number of solute particles. 1. • Urea and sucrose are non-electrolytes.

✅ Correct Answer Explanation

2. → Vapour pressure lowering is a colligative property. It depends on the number of solute particles present, not on their chemical identity, molar mass, or molecular size. Since 1.0 mol urea and 1.0 mol sucrose contain the same number of molecules and both behave as non-electrolytes, the vapour pressure lowering will be identical.

❓ Why Other Options Are Incorrect

�� Option A → Colligative properties do not depend on molar mass directly.
�� Option C → Larger molecular size does not increase vapour pressure lowering when particle number is the same.
3. • Option D → Hydrogen bonding is not the deciding factor in Raoult's law prediction here.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Identify the property as colligative and compare particle number.

Final Logic:

�� Same moles of non-electrolyte solute = same vapour pressure lowering.

🧠 Memory Trick

1. → Colligative counts particles, not identity.

📚 2. → Class XII – Chemistry Part I – Chapter: Solutions – Topic: Colligative Property and Vapour Pressure Lowering

CUET UG Chemistry Booster Test – 3 Henry’s Law and Raoult’s Law

CUET UG Chemistry Booster Test - 3 Henry's Law and Raoult's Law

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