CUET UG Chemistry Booster Test - 3 Concentration Measures and Solubility -

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CUET UG Chemistry Booster Test - 3 Concentration Measures and Solubility

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Chemistry Booster Test - 3 Concentration Measures and Solubility

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QUESTION 1 OF 20

In a multi-component solution, if the moles of components are n₁, n₂, ... nᵢ, the mole fraction x₁ mathematically approaches 1 only when:

Answer Review

1 In a multi-component solution, if the moles of components are n₁, n₂, ... nᵢ, the mole fraction x₁ mathematically approaches 1 only when:

A The solvent evaporates completely.

B The solution is infinitely dilute.

C The number of moles of all other components approaches zero.

D The temperature reaches the boiling point.

💡 Short Explanation

Mole fraction is a ratio. Total moles appear in the denominator. x₁ approaches 1 when other components become negligible.

✅ Correct Answer Explanation

Mole fraction is defined as [x_1=\frac{n_1}{n_1+n_2+n_3+….] If the moles of all other components approach zero, the denominator becomes approximately equal to n₁, making x₁ approach 1. This represents an almost pure component.

❓ Why Other Options Are Incorrect

Option A → Solvent evaporation alone does not necessarily make x₁ equal to 1.
Option B → An infinitely dilute solution makes the solute mole fraction approach zero, not one.
Option D → Boiling point has no direct mathematical relation to mole fraction.

🎯 Strategy

Used

Substitution

Application:

�� Substitute limiting values into the mole fraction formula.

Final Logic:

�� Other components → 0 ⇒ x₁ → 1.

🧠 Memory Trick

Only One Component → x = 1

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Mole Fraction

2 Consider a binary mixture of components A and B:
1. xA + xB = 1
2. xA = nA / (nA + nB)
3. Mole fraction strongly depends on the volume of the solution.
4. Mole fraction is useful for relating vapour pressure with concentration.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 4

💡 Short Explanation

Binary mole fractions add to one. Mole fraction uses moles only. It is used in Raoult's law.

✅ Correct Answer Explanation

In a binary solution, [ x_A+x_B=1 ] and [ x_A=\frac{n_A}{n_A+n_B} ] Mole fraction is independent of volume and temperature and is widely used in vapour pressure calculations through Raoult's law.

❓ Why Other Options Are Incorrect

Option A → Statement 3 is incorrect because mole fraction does not depend on volume.
Option C → Includes incorrect Statement 3.
Option D → Omits correct Statement 2.

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the false statement about volume dependence.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Mole Fraction = Moles Only

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Mole Fraction Properties

3 Molarity is calculated as moles of solute divided by the volume of solution. If volume is expressed in cubic decimetres (dm³), the unit becomes:

A mol dm⁻³

B mol dm³

C mol kg⁻¹

D ppm

💡 Short Explanation

Molarity = moles/volume. Volume is measured in dm³. Unit becomes mol dm⁻³.

✅ Correct Answer Explanation

Since molarity is moles of solute per cubic decimetre (litre) of solution, [ \text{Molarity}=\frac{\text{mol}}{\text{dm}^3} ] its unit is mol dm⁻³.

❓ Why Other Options Are Incorrect

Option B → Missing the denominator.
Option C → Unit of molality.
Option D → Unit for trace concentration.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Apply the definition of molarity directly.

Final Logic:

�� mol/dm³ = mol dm⁻³.

🧠 Memory Trick

Molarity = Mole per dm³

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molarity

4 Arrange the following conditions based on the predicted molarity of a fixed mass of solute in water (assuming the density of water decreases as temperature increases above 4°C), from highest molarity to lowest:
1. Solution at 20°C
2. Solution at 80°C
3. Solution at 50°C

A 1, 3, 2

B 2, 3, 1

C 3, 1, 2

D 1, 2, 3

💡 Short Explanation

Molarity depends on volume. Higher temperature increases volume. Larger volume means lower molarity.

✅ Correct Answer Explanation

The number of moles remains constant, but solution volume increases with increasing temperature. Therefore, molarity decreases as temperature rises. Thus, 20°C > 50°C > 80°C.

❓ Why Other Options Are Incorrect

Option B → Places highest temperature first.
Option C → Incorrectly ranks 50°C above 20°C.
Option D → Places 80°C above 50°C.

🎯 Strategy

Used

Option Grouping

Application:

�� Compare volume changes due to temperature.

Final Logic:

�� Temperature ↑ ⇒ Volume ↑ ⇒ Molarity ↓.

🧠 Memory Trick

Heat Up, Molarity Down

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Temperature Dependence of Molarity

5 Match the solution parameters to determine molality (assume 100 g of water as the solvent in all cases):

List 1 (Solution Parameter)	List 2 (Molality)
1. 0.1 mol NaCl	a. 1.0 m
2. 0.5 mol Glucose	b. 5.0 m
3. 0.2 mol Urea	c. 2.0 m
4. 0.05 mol KCl	d. 0.5 m

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

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

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

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

💡 Short Explanation

100 g water = 0.1 kg. Molality = moles/kg solvent. Divide moles by 0.1.

✅ Correct Answer Explanation

0.1 mol / 0.1 kg = 1.0 m 0.5 mol / 0.1 kg = 5.0 m 0.2 mol / 0.1 kg = 2.0 m 0.05 mol / 0.1 kg = 0.5 m Hence the correct matching is Option A.

❓ Why Other Options Are Incorrect

Option B → Incorrectly assigns 0.1 mol NaCl as 5.0 m.
Option C → Multiple incorrect molality values.
Option D → Does not satisfy molality calculations.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Calculate molality using moles/kg solvent.

Final Logic:

�� m = moles/0.1 kg.

🧠 Memory Trick

100 g = Divide by 0.1

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molality

6 Identify the concentration term that utilizes the "Mass of solvent in kg" as its fundamental denominator, rendering it independent of thermal expansion.

A Molarity

B Molality

C Volume percentage

D Parts per million

💡 Short Explanation

Uses kg of solvent. Depends on mass. Independent of temperature.

✅ Correct Answer Explanation

Molality is defined as moles of solute per kilogram of solvent. Since mass does not change with temperature, molality remains independent of thermal expansion.

❓ Why Other Options Are Incorrect

Option A → Depends on volume.
Option C → Depends on volume.
Option D → ppm is not fundamentally defined using kg of solvent.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Identify the concentration expression using mass.

Final Logic:

�� Mass-based concentration = Molality.

🧠 Memory Trick

Molality = Kilogram

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Molality

7 When the maximum dissolved amount is reached at a specified temperature, what type of chemical state is established between the solid solute and the solution?

A Static Equilibrium

B Dynamic Equilibrium

C Thermal Runaway

D Supersaturation

💡 Short Explanation

Dissolution continues. Crystallisation continues. Both occur at equal rates.

✅ Correct Answer Explanation

In a saturated solution, dissolution and crystallisation occur simultaneously at equal rates, establishing dynamic equilibrium while the concentration remains constant.

❓ Why Other Options Are Incorrect

Option A → The system is not static.
Option C → Thermal runaway is unrelated.
Option D → Supersaturation is a different condition.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the equilibrium described in NCERT.

Final Logic:

�� Equal forward and reverse processes = Dynamic equilibrium.

🧠 Memory Trick

Dynamic = Both Continue

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Dynamic Equilibrium

8 Why is anthracene highly soluble in benzene but largely insoluble in water?

A Anthracene is ionic, matching benzene's high dielectric constant.

B Anthracene is non-polar, and similar intermolecular forces exist between it and the non-polar benzene.

C Water degrades anthracene upon contact.

D Benzene is highly polar.

💡 Short Explanation

Anthracene is non-polar. Benzene is non-polar. Like dissolves like.

✅ Correct Answer Explanation

Similar intermolecular interactions allow anthracene to dissolve readily in benzene, whereas highly polar water cannot effectively replace anthracene-anthracene interactions.

❓ Why Other Options Are Incorrect

Option A → Anthracene is not ionic.
Option C → Water does not chemically degrade anthracene.
Option D → Benzene is non-polar.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Apply "like dissolves like."

Final Logic:

�� Non-polar dissolves in non-polar.

🧠 Memory Trick

Anthracene Likes Benzene

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Like Dissolves Like

9 If a newly discovered synthetic solute dissolves perfectly in carbon tetrachloride (a non-polar solvent) but forms a separate immiscible layer in water, the solute is most likely:

A A strong electrolyte

B Highly polar

C Non-polar

D A hydrated salt

💡 Short Explanation

Carbon tetrachloride is non-polar. Water is polar. Solute matches carbon tetrachloride.

✅ Correct Answer Explanation

A substance that dissolves in non-polar carbon tetrachloride but remains immiscible with polar water is most likely non-polar, following the "like dissolves like" principle.

❓ Why Other Options Are Incorrect

Option A → Electrolytes dissolve well in polar solvents.
Option B → Polar substances generally dissolve in water.
Option D → Hydrated salts are ionic and polar.

🎯 Strategy

Used

Odd One Out

Application:

�� Compare solvent polarity with solute behavior.

Final Logic:

�� Solubility pattern identifies a non-polar solute.

🧠 Memory Trick

CCl₄ Loves Non-polar

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Nature of Solute and Solvent

10 The "like dissolves like" principle is rooted in the energetics of solution formation. For a solid to dissolve effectively, its intermolecular interactions must be:

A Radically different from the solvent to create a high enthalpy of mixing.

B Overcome by the kinetic energy of the solvent molecules alone without forming new bonds.

C Similar to the solvent, allowing solvent-solute interactions to smoothly replace solute-solute and solvent-solvent interactions.

D Completely ionic to conduct electricity.

💡 Short Explanation

Similar interactions favor dissolution. New solvent-solute interactions compensate for broken bonds. Solution formation becomes energetically favorable.

✅ Correct Answer Explanation

A substance dissolves effectively when solvent-solute interactions are comparable to solute-solute and solvent-solvent interactions. This allows smooth replacement of existing interactions and stabilizes the solution.

❓ Why Other Options Are Incorrect

Option A → Radically different interactions hinder dissolution.
Option B → Solvent-solute interactions are essential.
Option D → Dissolution is not restricted to ionic substances.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Apply the energetic basis of "like dissolves like."

Final Logic:

�� Similar intermolecular forces produce stable solutions.

🧠 Memory Trick

Similar Forces Mix Best

📚 Questions 11–15

11 When assessing a dynamic equilibrium in a saturated solution, the rate of solute particles going into the solution must be:

A Zero

B Exactly equal to the rate of solute particles separating out.

C Greater than the rate of separating out.

D Continuously increasing.

💡 Short Explanation

Saturated solution shows dynamic equilibrium. Dissolution and crystallisation continue. Their rates become equal.

✅ Correct Answer Explanation

In a saturated solution, solute particles continue to dissolve and crystallise at the same time. At dynamic equilibrium, the rate of particles entering the solution is exactly equal to the rate of particles separating out. Therefore, the concentration of solute remains constant.

❓ Why Other Options Are Incorrect

Option A → Dissolution does not stop at dynamic equilibrium.
Option C → If dissolution is greater, the solution is not at equilibrium.
Option D → The rate does not continuously increase at equilibrium.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect the phrase "dynamic equilibrium" with equal forward and reverse rates.

Final Logic:

�� Dynamic equilibrium means equal dissolution and crystallisation rates.

🧠 Memory Trick

Dynamic = Equal Rates

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Dynamic Equilibrium in Saturated Solution

12 If a saturated solution is in a state of dynamic equilibrium with undissolved solute, and the temperature and pressure are maintained constant, the concentration of the solute in the solution will:

A Fluctuate randomly

B Steadily decrease

C Remain constant

D Exponentially increase

💡 Short Explanation

Dissolution rate equals crystallisation rate. No net change occurs. Solute concentration remains constant.

✅ Correct Answer Explanation

In a saturated solution at dynamic equilibrium, solute particles dissolve and separate out at equal rates. Since temperature and pressure are constant, the solubility limit does not change. Hence, the concentration of solute in the solution remains constant.

❓ Why Other Options Are Incorrect

Option A → Equilibrium does not cause random concentration fluctuation.
Option B → Concentration does not steadily decrease because dissolution continues.
Option D → Concentration does not exponentially increase at equilibrium.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate options that show net concentration change.

Final Logic:

�� Equal opposing rates keep concentration constant.

🧠 Memory Trick

Equal Rates = Constant Concentration

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Saturated Solution and Dynamic Equilibrium

13 If a solution contains the maximum amount of solute dissolved in a given amount of solvent at a specific temperature, it is termed:

A An unsaturated solution

B A saturated solution

C A dilute solution

D A supersaturated solution

💡 Short Explanation

Maximum solute has dissolved. Same temperature is specified. This defines a saturated solution.

✅ Correct Answer Explanation

A saturated solution contains the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature. Any additional solute will not dissolve under the same conditions.

❓ Why Other Options Are Incorrect

Option A → An unsaturated solution can still dissolve more solute.
Option C → Dilute means relatively small solute quantity.
Option D → Supersaturated solution contains more solute than normally possible under stable conditions.

🎯 Strategy

Used

Extreme Word Filter

Application:

�� Focus on the word "maximum," which directly indicates saturation.

Final Logic:

�� Maximum dissolved solute = saturated solution.

🧠 Memory Trick

Saturated = Full

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Saturated Solution

14 Which condition proves a liquid solution is strictly unsaturated?

A It is in dynamic equilibrium with undissolved solute.

B More solute can be dissolved in it at the same temperature.

C Crystals begin to form upon agitation.

D It contains suspended solid particles.

💡 Short Explanation

Unsaturated solution has not reached maximum solubility. It can dissolve additional solute. Same temperature condition is important.

✅ Correct Answer Explanation

An unsaturated solution is one in which more solute can still dissolve at the same temperature. It has not reached the solubility limit, so additional solute can enter the solution.

❓ Why Other Options Are Incorrect

Option A → Dynamic equilibrium with undissolved solute indicates a saturated solution.
Option C → Crystal formation may indicate saturation or supersaturation.
Option D → Suspended particles indicate a heterogeneous mixture, not proof of unsaturation.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate options describing saturation, supersaturation, or heterogeneity.

Final Logic:

�� Ability to dissolve more solute proves unsaturation.

🧠 Memory Trick

Unsaturated = More Can Dissolve

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Unsaturated Solution

15 For an endothermic dissolution, what is the thermodynamic sign of Δsol H, and how does Le Chatelier's principle predict its solubility shift when heated?

A Δsol H < 0; solubility increases

B Δsol H > 0; solubility increases

C Δsol H > 0; solubility decreases

D Δsol H < 0; solubility decreases

💡 Short Explanation

Endothermic process absorbs heat. Δsol H is positive. Heating favours dissolution.

✅ Correct Answer Explanation

In an endothermic dissolution process, heat is absorbed, so Δsol H > 0. According to Le Chatelier's Principle, increasing temperature favours the heat-absorbing forward process. Therefore, solubility increases when the solution is heated.

❓ Why Other Options Are Incorrect

Option A → Δsol H < 0 indicates exothermic dissolution, not endothermic.
Option C → Heating increases solubility for endothermic dissolution.
Option D → This represents exothermic dissolution with heating, not endothermic behavior.

🎯 Strategy

Used

Option Grouping

Application:

�� First identify the sign of endothermic dissolution, then link heating to solubility increase.

Final Logic:

�� Endothermic = Δsol H > 0; heating increases solubility.

🧠 Memory Trick

Endothermic = Positive Heat

📚 Questions 16–20

16 If a solute forms a solution and the mixing beaker becomes noticeably hot (exothermic), placing the beaker in an ice bath will logically:

A Decrease its solubility

B Increase its solubility

C Force immediate crystallisation

D Have no effect

💡 Short Explanation

Exothermic dissolution releases heat. Cooling favors dissolution. Solubility increases.

✅ Correct Answer Explanation

In an exothermic dissolution process (Δsol H < 0), heat behaves as a product. According to Le Chatelier's Principle, lowering the temperature by placing the beaker in an ice bath shifts the equilibrium toward the dissolution side to produce more heat. As a result, more solute dissolves and the solubility increases.

❓ Why Other Options Are Incorrect

Option A → Cooling does not decrease solubility for an exothermic dissolution.
Option C → Crystallisation is not immediately forced; equilibrium simply shifts toward greater dissolution.
Option D → Temperature significantly affects exothermic dissolution equilibria.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recognize that cooling favors an exothermic process according to Le Chatelier's Principle.

Final Logic:

�� Exothermic + Cooling = Increased solubility.

🧠 Memory Trick

Exothermic + Cold = Dissolve More

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

17 Why does the volume of a solid or liquid phase remain virtually constant under varying pressure, leading to an independent solubility characteristic?

A Particles have exceptionally high kinetic energy.

B The intermolecular distances are already minimal, making them highly incompressible.

C They perfectly obey Henry's law.

D Pressure only acts on gaseous bonds.

💡 Short Explanation

Solids and liquids are highly incompressible. Particle spacing changes very little. Pressure has negligible effect on solubility.

✅ Correct Answer Explanation

In solids and liquids, particles are already packed closely together with very small intermolecular distances. Because of this, applying external pressure produces only negligible volume changes. Consequently, pressure has almost no influence on the solubility of solids in liquids.

❓ Why Other Options Are Incorrect

Option A → High kinetic energy is not responsible for incompressibility.
Option C → Henry's law applies to gases dissolved in liquids, not solids.
Option D → Pressure acts on all phases, although gases respond much more significantly.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate statements inconsistent with the physical properties of solids and liquids.

Final Logic:

�� Closely packed particles ⇒ High incompressibility ⇒ Negligible pressure effect.

🧠 Memory Trick

Close Packing = No Compression

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

18 In industrial chemical applications, if the concentration of a solid dissolved in a liquid needs to be altered, which parameter is practically useless to adjust?

A Mass of the solvent

B Temperature of the solution

C Nature of the solute

D External pressure on the liquid surface

💡 Short Explanation

Solids and liquids are incompressible. Pressure produces negligible solubility change. Temperature is much more effective.

✅ Correct Answer Explanation

The solubility of solids in liquids is practically independent of pressure because both phases are highly incompressible. Therefore, changing external pressure is ineffective for altering concentration, whereas changing temperature or solvent composition can significantly affect solubility.

❓ Why Other Options Are Incorrect

Option A → Changing solvent amount directly changes concentration.
Option B → Temperature significantly affects the solubility of many solids.
Option C → The nature of the solute strongly influences its solubility.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the parameter known to have negligible influence on solid-liquid solubility.

Final Logic:

�� Pressure has almost no practical effect on solid-liquid solutions.

🧠 Memory Trick

Solid + Liquid = Ignore Pressure

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

📄 Passage

Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure. Many gases dissolve in water. Oxygen dissolves only to a small extent in water. It is this dissolved oxygen which sustains all aquatic life. On the other hand, hydrogen chloride gas (HCl) is highly soluble in water. Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of gases increase with increase of pressure.

Unlike solid-liquid solutions, gas-liquid solutions are heavily influenced by pressure. According to the text, what happens to the gas solubility when the pressure is increased?

A It decreases due to gas compression.

B It remains unaffected.

C It increases.

D It rapidly fluctuates.

💡 Short Explanation

Gas solubility depends on pressure. Higher pressure forces more gas into solution. Solubility increases.

✅ Correct Answer Explanation

As stated in the passage and explained by Henry's Law, increasing the pressure of a gas above a liquid increases the amount of gas that dissolves in the liquid. Therefore, gas solubility increases with increasing pressure.

❓ Why Other Options Are Incorrect

Option A → Increased pressure promotes dissolution rather than decreasing it.
Option B → Gas solubility is significantly affected by pressure.
Option D → Solubility changes predictably rather than fluctuating randomly.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the direct statement given in the passage.

Final Logic:

�� Pressure ↑ ⇒ Gas solubility ↑.

🧠 Memory Trick

More Pressure = More Gas Dissolves

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solubility of Gases in Liquids

📄 Passage

Which crucial biological phenomenon relies on the specific behavior of gas solubility limits in liquids?

A Photosynthesis in terrestrial plants.

B The survival of aquatic life through dissolved oxygen.

C The formation of coral reefs from dissolved calcium.

D The incompressibility of ocean water protecting the sea floor.

💡 Short Explanation

Oxygen dissolves only to a small extent. Dissolved oxygen supports respiration. Aquatic organisms depend on it.

✅ Correct Answer Explanation

Although oxygen is only sparingly soluble in water, the dissolved oxygen present is essential for the respiration and survival of fish and other aquatic organisms. This is one of the most important biological applications of gas solubility in liquids.

❓ Why Other Options Are Incorrect

Option A → Terrestrial photosynthesis does not primarily depend on dissolved oxygen in water.
Option C → Coral reef formation mainly involves calcium carbonate deposition rather than oxygen solubility.
Option D → Ocean water incompressibility is unrelated to dissolved oxygen supporting life.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the biological application explicitly mentioned in the passage.

Final Logic:

�� Dissolved oxygen ⇒ Aquatic life survives.

🧠 Memory Trick

Small O₂ → Big Life

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solubility of Gases in Liquids and Dissolved Oxygen

CUET UG Chemistry Booster Test – 3 Concentration Measures and Solubility

CUET UG Chemistry Booster Test - 3 Concentration Measures and Solubility

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