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

🚀

CUET UG Chemistry Booster Test - 2 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 - 2 Henry's Law and Raoult's Law

1 / 20

QUESTION 1 OF 20

If a gas over a solution is compressed to a smaller volume, the number of gaseous particles per unit volume will:

Answer Review

1 If a gas over a solution is compressed to a smaller volume, the number of gaseous particles per unit volume will:

A Decrease

B Remain the same

C Increase

D Become zero

💡 Short Explanation

Compression reduces volume. Same gas particles occupy smaller space. Particles per unit volume increase.

✅ Correct Answer Explanation

When a gas is compressed, its volume decreases while the number of gas particles remains the same. Therefore, the number of gaseous particles per unit volume increases. This increases the frequency of gas particles striking the liquid surface, promoting gas solubility.

❓ Why Other Options Are Incorrect

Option A → Compression increases, not decreases, particles per unit volume.
Option B → The value changes because volume decreases.
Option D → Gas particles do not become zero.

🎯 Strategy

Used

Substitution

Application:

�� Apply the relation: particles per unit volume = number of particles / volume.

Final Logic:

�� Smaller volume with same particles means higher particle density.

🧠 Memory Trick

Compress = Concentrate Gas

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

2 Arrange the following states of an oxygen-water system in decreasing order of oxygen solubility (assume constant T):
1. Under 5 atm pressure
2. Under 1 atm pressure
3. Under 10 atm pressure
4. Under 0.5 atm pressure

A 3, 1, 2, 4

B 1, 2, 3, 4

C 4, 2, 1, 3

D 2, 3, 4, 1

💡 Short Explanation

Gas solubility increases with pressure. Highest pressure gives highest solubility. Lowest pressure gives lowest solubility.

✅ Correct Answer Explanation

According to Henry's Law, at constant temperature, the solubility of a gas in a liquid increases with pressure. Therefore, the decreasing order of oxygen solubility is 10 atm > 5 atm > 1 atm > 0.5 atm.

❓ Why Other Options Are Incorrect

Option B → Does not place 10 atm first.
Option C → Gives increasing order instead of decreasing order.
Option D → Incorrectly places 1 atm before 10 atm.

🎯 Strategy

Used

Option Grouping

Application:

�� Rank the given pressures from highest to lowest.

Final Logic:

�� Greater pressure = greater gas solubility.

🧠 Memory Trick

Pressure Up, Gas In

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law and Pressure Effect

3 Henry's Law properties:
1. Mole fraction in solution is proportional to partial pressure
2. Used to calculate the quantitative solubility of a gas
3. Dalton independently gave a similar conclusion
4. Valid only at varying temperature

A 1, 2, 3

B 1, 3

C 1, 2, 3, 4

D 2, 4

💡 Short Explanation

Henry's Law relates pressure and mole fraction. It gives quantitative gas solubility. It is applied at constant temperature.

✅ Correct Answer Explanation

Henry's Law states that the solubility of a gas in a liquid is proportional to its partial pressure above the liquid. It is used to calculate quantitative solubility of gases. The relation is valid at constant temperature, so Statement 4 is incorrect.

❓ Why Other Options Are Incorrect

Option B → Omits Statement 2, which is correct.
Option C → Includes Statement 4, which is incorrect.
Option D → Includes Statement 4 and omits Statements 1 and 3.

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the false statement about varying temperature.

Final Logic:

�� Henry's Law applies at constant temperature and relates pressure to gas solubility.

🧠 Memory Trick

Henry = Gas Pressure Law

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law Fundamentals

4 Match List-I (Gas at 293 K) with List-II (KH value in kbar):

List 1 (Gas at 293 K)	List 2 (KH value in kbar)
1. He	a. 144.97
2. H₂	b. 69.16
3. N₂	c. 76.48
4. O₂	d. 34.86

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

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

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

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

💡 Short Explanation

He has the highest KH value. O₂ has the lowest KH among the listed gases. KH depends on the nature of gas.

✅ Correct Answer Explanation

At 293 K, the NCERT values of Henry's Law constant are approximately: He = 144.97 kbar, H₂ = 69.16 kbar, N₂ = 76.48 kbar, and O₂ = 34.86 kbar. Hence, the correct matching is 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

Option B → Incorrectly matches He with O₂ value and O₂ with He value.
Option C → Incorrectly matches He and H₂ values.
Option D → Multiple gas-KH pairings are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

�� Match known NCERT Henry's Law constant values at 293 K.

Final Logic:

�� He-144.97, H₂-69.16, N₂-76.48, O₂-34.86.

🧠 Memory Trick

He Highest, O₂ Lowest

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

5 The unit of KH matches the unit of which physical quantity in the equation p = KH x?

A Mole fraction

B Temperature

C Partial pressure

D Volume

💡 Short Explanation

Henry's Law: p = KHx. Mole fraction has no unit. KH has same unit as pressure.

✅ Correct Answer Explanation

In Henry's Law, p = KHx, where x is mole fraction and is dimensionless. Therefore, KH must have the same unit as p, which is partial pressure.

❓ Why Other Options Are Incorrect

Option A → Mole fraction is dimensionless.
Option B → Temperature is not the unit of KH.
Option D → Volume is not involved as the unit of KH.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Compare units on both sides of p = KHx.

Final Logic:

�� Since x has no unit, KH has the unit of pressure.

🧠 Memory Trick

KH Carries p Unit

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

6 According to the equation p = KH x, if KH is higher at a given pressure, the solubility of the gas in the liquid will be:

A Higher

B Lower

C Unchanged

D Equal to KH

💡 Short Explanation

Henry's Law: p = KHx. At fixed pressure, x = p/KH. Higher KH means lower mole fraction.

✅ Correct Answer Explanation

From Henry's Law: [x = {p}/{Kh}] At a given pressure, the mole fraction of gas in solution is inversely proportional to KH. Therefore, a higher KH value means lower solubility of the gas.

❓ Why Other Options Are Incorrect

Option A → Higher KH indicates lower solubility, not higher.
Option C → Solubility changes with KH.
Option D → Solubility is not equal to KH.

🎯 Strategy

Used

Substitution

Application:

�� Rearrange Henry's Law equation to x = p/KH.

Final Logic:

�� KH ↑ ⇒ x ↓ ⇒ solubility ↓.

🧠 Memory Trick

High KH = Hard to Dissolve

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law Constant and Solubility

7 Identify the process type that is considered similar to the dissolution of a gas in a liquid phase:

A Condensation

B Sublimation

C Vaporization

D Fusion

💡 Short Explanation

Gas enters liquid phase. This resembles gas changing to condensed phase. Hence it is like condensation.

✅ Correct Answer Explanation

Dissolution of a gas in a liquid resembles condensation because gas molecules move from a gaseous state into a more condensed liquid environment. This process is generally exothermic.

❓ Why Other Options Are Incorrect

Option B → Sublimation is solid to gas.
Option C → Vaporization is liquid to gas, opposite in direction.
Option D → Fusion is solid to liquid.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the phase change most similar to gas entering liquid.

Final Logic:

�� Gas into liquid resembles condensation.

🧠 Memory Trick

Gas In Liquid = Condensation-like

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

8 Aquatic life finds it more comfortable in cold water than warm water because:

A Oxygen's partial pressure is zero in warm water

B The solubility of oxygen is higher at lower temperatures

C Warm water triggers endothermic condensation

D KH of oxygen is zero at lower temperatures

💡 Short Explanation

Gas solubility decreases with temperature. Cold water holds more dissolved oxygen. Aquatic organisms need dissolved oxygen.

✅ Correct Answer Explanation

Dissolution of gases in liquids is generally exothermic. Therefore, increasing temperature decreases gas solubility, while lower temperature increases it. Cold water contains more dissolved oxygen, making it more suitable for aquatic life.

❓ Why Other Options Are Incorrect

Option A → Oxygen's partial pressure is not zero in warm water.
Option C → Gas dissolution is exothermic, not endothermic condensation.
Option D → KH of oxygen is not zero at lower temperatures.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect aquatic life with dissolved oxygen and temperature effect.

Final Logic:

�� Cold water holds more O₂.

🧠 Memory Trick

Cold Water Holds Oxygen

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solubility of Gases and Temperature

📄 Passage

Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium.

Why does nitrogen form bubbles in the blood as a diver ascends?

A Temperature of the water drops rapidly

B Decreased pressure reduces nitrogen's solubility, forcing it out of solution

C Nitrogen reacts with helium in the tank

D The mole fraction of oxygen drops to zero

💡 Short Explanation

Pressure decreases during ascent. Gas solubility decreases with pressure. Nitrogen comes out as bubbles.

✅ Correct Answer Explanation

Underwater, high pressure increases the solubility of gases in blood. As the diver ascends, pressure decreases, reducing nitrogen solubility. Nitrogen then escapes from the blood as bubbles, blocking capillaries and causing bends.

❓ Why Other Options Are Incorrect

Option A → The key cause is pressure decrease, not temperature drop.
Option C → Nitrogen does not react with helium.
Option D → Oxygen mole fraction does not become zero.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the passage explanation of pressure decrease and nitrogen bubble formation.

Final Logic:

�� Pressure ↓ ⇒ Nitrogen solubility ↓ ⇒ Bubbles form.

🧠 Memory Trick

Ascent Releases Nitrogen

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Henry's Law and Scuba Diving

📄 Passage

Based on the passage, helium is added to the scuba diver's tank primarily to:

A Make the diver float easier

B Increase the total pressure inside the lungs

C Avoid bends and toxic effects of high nitrogen concentrations

D React chemically with oxygen

💡 Short Explanation

Helium dilutes nitrogen in breathing gas. It reduces nitrogen toxicity. It helps prevent bends.

✅ Correct Answer Explanation

Scuba tanks are filled with air diluted with helium to reduce the amount of nitrogen breathed under high pressure. This helps avoid excessive nitrogen dissolving in blood, thereby reducing the risk of bends and nitrogen toxicity.

❓ Why Other Options Are Incorrect

Option A → Helium is not added to make the diver float.
Option B → The purpose is not to increase lung pressure.
Option D → Helium is chemically inert and does not react with oxygen.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the exact purpose stated in the passage.

Final Logic:

�� Helium dilution reduces nitrogen-related dangers.

🧠 Memory Trick

Helium Helps Divers

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

11 Identify the name of the pressure exerted by the vapours of a liquid over the liquid phase when the rates of evaporation and condensation are equal:

A Partial Vapour Pressure

B Equilibrium Vapour Pressure

C Atmospheric Pressure

D Osmotic Pressure

💡 Short Explanation

Evaporation and condensation occur simultaneously. Equal rates establish dynamic equilibrium. The resulting pressure is equilibrium vapour pressure.

✅ Correct Answer Explanation

When a volatile liquid is kept in a closed container, molecules continuously evaporate and condense. A stage is reached where the rate of evaporation becomes equal to the rate of condensation. The pressure exerted by the vapour at this dynamic equilibrium is called the equilibrium vapour pressure.

❓ Why Other Options Are Incorrect

Option A → Partial vapour pressure refers to the pressure contributed by one component in a mixture.
Option C → Atmospheric pressure is the pressure exerted by the atmosphere.
Option D → Osmotic pressure is associated with osmosis through a semipermeable membrane.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the term directly associated with dynamic equilibrium between liquid and vapour.

Final Logic:

�� Equal evaporation and condensation rates define equilibrium vapour pressure.

🧠 Memory Trick

Equal Rates = Equilibrium Vapour Pressure

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

12 In a binary solution of two volatile liquids, the total vapour pressure over the solution:

A Derives exclusively from the solvent

B Remains independent of the mole fractions

C Is the sum of the partial pressures of both components

D Decreases to zero at equilibrium

💡 Short Explanation

Both volatile liquids contribute vapour. Each exerts its own partial pressure. Total pressure is their sum.

✅ Correct Answer Explanation

According to Dalton's Law of Partial Pressures, the total vapour pressure over a binary solution of volatile liquids equals the sum of the partial vapour pressures of both components. [P_{total}=P_1+P_2] Thus, both liquids contribute to the total vapour pressure.

❓ Why Other Options Are Incorrect

Option A → Both components contribute, not only the solvent.
Option B → Vapour pressure depends on mole fractions through Raoult's Law.
Option D → Vapour pressure does not become zero at equilibrium.

🎯 Strategy

Used

Substitution

Application:

�� Recall Dalton's Law equation.

Final Logic:

�� Total pressure equals sum of partial pressures.

🧠 Memory Trick

Dalton Adds Pressures

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

13 For an ideal solution of volatile liquids, the plot of partial vapour pressure (p₁) versus mole fraction (x₁) is:

A A linear line passing through the origin

B A downward curving parabola

C A horizontal straight line

D A hyperbola

💡 Short Explanation

Raoult's Law: (p_1 = p_1^0x_1) Direct proportionality exists. Graph is a straight line through the origin.

✅ Correct Answer Explanation

Raoult's Law states: [p_1 = p_1^0 x_1] Since partial vapour pressure is directly proportional to mole fraction, a graph of (p_1) versus (x_1) is a straight line passing through the origin with slope (p_1^0).

❓ Why Other Options Are Incorrect

Option B → No quadratic relationship exists.
Option C → Pressure changes with mole fraction.
Option D → Hyperbolic variation is not predicted by Raoult's Law.

🎯 Strategy

Used

Substitution

Application:

�� Use the Raoult's Law equation to determine graph shape.

Final Logic:

�� Direct proportionality produces a straight line through the origin.

🧠 Memory Trick

Raoult = Straight Line

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Raoult's Law Graph

14 Binary liquid equations properties based on
[P_{total}=P_1^0+(P_2^0-P_1^0)x_2]
1. Total vapour pressure varies linearly with mole fraction of component 2
2. Total vapour pressure can be related to the mole fraction of any one component
3. Dependent entirely on atmospheric pressure
4. The equation proves Raoult's Law fails for ideal gases

A 1, 2

B 2, 3

C 1, 3, 4

D 3, 4

💡 Short Explanation

The equation is linear in (x_2). Vapour pressure can be expressed using either mole fraction. Atmospheric pressure is not involved.

✅ Correct Answer Explanation

The equation [P_{total}=P_1^0+(P_2^0-P_1^0)x_2] shows that total vapour pressure changes linearly with mole fraction (x_2). Similar expressions can also be written using (x_1). Therefore, Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

Option B → Statement 3 is incorrect because atmospheric pressure does not appear in the equation.
Option C → Statements 3 and 4 are false.
Option D → Both statements are incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing statements unrelated to Raoult's Law.

Final Logic:

�� Only Statements 1 and 2 follow directly from the equation.

🧠 Memory Trick

Linear Equation → Linear Graph

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Total Vapour Pressure of Binary Solutions

15 The composition of the vapour phase in equilibrium with the solution can be calculated using Dalton's law. What is the equation for the mole fraction of component 1 in the vapour phase (y₁)?

A (y_1 = \frac{p_1}{p_{total}})

B (y_1 = \frac{p_{total}}{p_1})

C (y_1 = x_1 \times p_2^0)

D (y_1 = p_1 \times p_{total})

💡 Short Explanation

Dalton's law relates vapour composition to partial pressure. Vapour mole fraction equals pressure fraction. Formula uses partial pressure divided by total pressure.

✅ Correct Answer Explanation

According to Dalton's Law of Partial Pressures, the mole fraction of a component in the vapour phase is: [ y_1=\frac{p_1}{p_{total}} ] where (p_1) is the partial pressure of component 1 and (p_{total}) is the total vapour pressure.

❓ Why Other Options Are Incorrect

Option B → Inverts the correct relationship.
Option C → Does not represent vapour-phase mole fraction.
Option D → Multiplication of pressures has no physical meaning here.

🎯 Strategy

Used

Substitution

Application:

�� Directly apply Dalton's Law expression.

Final Logic:

�� Vapour mole fraction = Partial pressure / Total pressure.

🧠 Memory Trick

y = Part / Total

📚 Questions 16–20

16 If p₁⁰ = 200 mm Hg, p₂⁰ = 415 mm Hg, and x₂ = 0.688, the total vapour pressure is calculated as 200 + (415 - 200) × 0.688. The total pressure is approximately:

A 285.5 mm Hg

B 347.9 mm Hg

C 615.0 mm Hg

D 147.9 mm Hg

💡 Short Explanation

Use the given expression directly. Difference in pure vapour pressures = 215 mm Hg. Total pressure ≈ 347.9 mm Hg.

✅ Correct Answer Explanation

Given: [ p_{total}=200+(415-200)\times0.688 ] [ =200+215\times0.688 ] [ =200+147.92 ] [ =347.92 \text{ mm Hg} ] So, the approximate total vapour pressure is 347.9 mm Hg.

❓ Why Other Options Are Incorrect

Option A → 285.5 mm Hg is obtained from incorrect calculation.
Option C → 615.0 mm Hg incorrectly adds the pure vapour pressures directly.
Option D → 147.9 mm Hg represents only the added term, not the total pressure.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Substitute values carefully into the vapour pressure equation and retain the unit mm Hg.

Final Logic:

�� (200 + 147.92 = 347.92), so the answer is 347.9 mm Hg.

🧠 Memory Trick

Calculate bracket first, then add base pressure.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Vapour Pressure of Binary Liquid Solutions

17 When Raoult's law is considered a special case of Henry's law, which parameter aligns perfectly in both mathematical forms representing the proportionality constant?

A The vapour pressure of pure component (p₁⁰) and KH

B The total pressure and mole fraction

C Partial pressure and temperature

D Volume and atmospheric pressure

💡 Short Explanation

Henry's Law: (p = K_Hx) Raoult's Law: (p_1 = p_1^0x_1) (p_1^0) corresponds to (K_H)

✅ Correct Answer Explanation

Henry's Law is expressed as: [p = K_Hx] Raoult's Law is expressed as: [p_1 = p_1^0x_1] Both equations show pressure directly proportional to mole fraction. In Raoult's Law, the proportionality constant is the vapour pressure of the pure component, (p_1^0), which corresponds to (K_H) in Henry's Law.

❓ Why Other Options Are Incorrect

Option B → Total pressure and mole fraction are not corresponding proportionality constants.
Option C → Partial pressure is the dependent variable, not the proportionality constant.
Option D → Volume and atmospheric pressure are not part of the proportionality comparison.

🎯 Strategy

Used

Substitution

Application:

�� Compare both equations term by term.

Final Logic:

�� (K_H) in Henry's Law aligns with (p_1^0) in Raoult's Law.

🧠 Memory Trick

Henry has KH, Raoult has p⁰.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Raoult's Law as a Special Case of Henry's Law

18 Raoult's Law becomes exactly Henry's Law for a volatile gas in a liquid when:

A The pure vapour pressure of the gas (p₁⁰) equals the Henry's Law constant (KH)

B The mole fraction of the gas approaches 1

C The solvent has zero vapour pressure

D KH is infinitely large

💡 Short Explanation

Henry's Law uses (K_H). Raoult's Law uses (p_1^0). They match when (K_H = p_1^0).

✅ Correct Answer Explanation

Henry's Law is: [ p = K_Hx ] Raoult's Law is: [ p_1 = p_1^0x_1 ] For Raoult's Law to become exactly equivalent to Henry's Law, the proportionality constants must be equal. Therefore, (p_1^0 = K_H).

❓ Why Other Options Are Incorrect

Option B → Mole fraction approaching 1 does not make both laws exactly identical.
Option C → Solvent vapour pressure becoming zero is unrelated to the equality of the two laws.
Option D → Infinite (K_H) would not make the laws equivalent.

🎯 Strategy

Used

Substitution

Application:

�� Compare the mathematical constants in both equations.

Final Logic:

�� Same form becomes exact only when (p_1^0 = K_H).

🧠 Memory Trick

Same law when constants same.

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

19 The decrease in vapour pressure of water by adding 1.0 mol of sucrose compared to adding 1.0 mol of urea to 1 kg of water is:

A Much higher for sucrose

B Much higher for urea

C Nearly similar for both

D Zero in both cases

💡 Short Explanation

Vapour pressure lowering is a colligative property. It depends mainly on number of solute particles. 1 mol sucrose and 1 mol urea give similar particle amounts.

✅ Correct Answer Explanation

Lowering of vapour pressure is a colligative property. It depends on the number of solute particles present, not on the chemical identity of the solute. Since sucrose and urea are both non-electrolytes and 1.0 mol of each gives approximately the same number of solute particles, the decrease in vapour pressure is nearly similar for both.

❓ Why Other Options Are Incorrect

Option A → Sucrose does not cause much higher lowering because particle number is the same.
Option B → Urea also does not cause much higher lowering for the same reason.
Option D → Vapour pressure lowering is not zero when a non-volatile solute is added.

🎯 Strategy

Used

Option Grouping

Application:

�� Identify the property as colligative and compare number of solute particles.

Final Logic:

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

🧠 Memory Trick

Colligative counts particles, not names.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Lowering of Vapour Pressure as a Colligative Property

20 The reduction in vapour pressure upon adding a non-volatile solute depends strictly on:

A The chemical identity of the solute

B The quantity of non-volatile solute present

C The colour of the solution

D The volume of the vapour phase

💡 Short Explanation

Non-volatile solute lowers vapour pressure. The effect depends on amount of solute particles. It is independent of colour or identity.

✅ Correct Answer Explanation

When a non-volatile solute is added to a solvent, fewer solvent molecules occupy the surface and fewer escape into the vapour phase. The extent of vapour pressure lowering depends on the quantity or number of non-volatile solute particles present, not on their colour or chemical identity.

❓ Why Other Options Are Incorrect

Option A → Vapour pressure lowering is a colligative property and does not depend strictly on chemical identity.
Option C → Colour has no role in vapour pressure lowering.
Option D → Vapour pressure reduction is not determined by vapour phase volume.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate properties unrelated to colligative behaviour.

Final Logic:

�� More non-volatile solute particles cause greater vapour pressure lowering.

🧠 Memory Trick

More Solute, Less Vapour.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Vapour Pressure Lowering by Non-volatile Solutes

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

CUET UG Chemistry Booster Test - 2 Henry's Law and Raoult's Law

Test Complete!

Answer Review