CUET UG Chemistry Booster Test - 2 Concentration Measures and Solubility -

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CUET UG Chemistry Booster Test - 2 Concentration Measures and Solubility

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Concentration Measures and Solubility

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QUESTION 1 OF 20

What is the unit of mole fraction, xi?

Answer Review

1 What is the unit of mole fraction, xi?

A mol L⁻¹

B mol kg⁻¹

C Dimensionless (no unit)

D g mol⁻¹

💡 Short Explanation

�� Mole fraction is a ratio. �� It compares moles of a component with total moles. 1. • Ratios have no unit.

✅ Correct Answer Explanation

2. → Mole fraction is defined as moles of a component divided by total moles of all components. Since both numerator and denominator have the same unit, the units cancel out. Therefore, mole fraction is dimensionless.

❓ Why Other Options Are Incorrect

�� Option A → mol L⁻¹ is the unit of molarity.
�� Option B → mol kg⁻¹ is the unit of molality.
3. • Option D → g mol⁻¹ is the unit of molar mass.

🎯 Strategy

Used

4. Dimensional/Unit Analysis

Application:

�� Compare the units in the numerator and denominator of mole fraction.

Final Logic:

�� mole/mole cancels, so mole fraction has no unit.

🧠 Memory Trick

1. → Fraction has no unit.

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

2 In a ternary mixture containing components A, B, and C, if xA = 0.3 and xB = 0.4, what is the value of xC?

A 0.7

B 0.3

C 1.0

D 0.1

💡 Short Explanation

�� Sum of mole fractions is 1. �� xA + xB + xC = 1. 1. • xC = 1 − 0.7 = 0.3.

✅ Correct Answer Explanation

2. → In any solution, the sum of mole fractions of all components is unity. Here, xA = 0.3 and xB = 0.4. Therefore, xC = 1 − (0.3 + 0.4) = 0.3.

❓ Why Other Options Are Incorrect

�� Option A → 0.7 is the sum of xA and xB, not xC.
�� Option C → 1.0 is the total sum of all mole fractions.
3. • Option D → 0.1 is not obtained from the mole fraction relation.

🎯 Strategy

Used

4. Substitution

Application:

�� Substitute given values into xA + xB + xC = 1.

Final Logic:

�� xC = 1 − 0.3 − 0.4 = 0.3.

🧠 Memory Trick

1. → Mole fractions make one whole.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Mole Fraction in Multicomponent Solutions

3 Arrange the following solutions in decreasing order of molarity:
1. 0.5 mol of NaOH in 1.0 L solution
2. 2.0 mol of KCl in 2.0 L solution
3. 0.1 mol of Glucose in 0.5 L solution
4. 1.5 mol of NaCl in 0.5 L solution

A 4, 2, 1, 3

B 1, 2, 3, 4

C 2, 1, 3, 4

D 4, 2, 3, 1

💡 Short Explanation

�� Molarity = moles/litre. �� Calculate each molarity first. 1. • Arrange from highest to lowest.

✅ Correct Answer Explanation

1 = 0.5/1.0 = 0.5 M 2 = 2.0/2.0 = 1.0 M 3 = 0.1/0.5 = 0.2 M 2. 4 = 1.5/0.5 = 3.0 M Decreasing order: 4 > 2 > 1 > 3.

❓ Why Other Options Are Incorrect

�� Option B → Places 0.5 M before 3.0 M and 1.0 M.
�� Option C → Places 1.0 M before 3.0 M.
1. • Option D → Places 0.2 M before 0.5 M.

🎯 Strategy

Used

2. Dimensional/Unit Analysis

Application:

�� Convert all given values into mol L⁻¹ before comparing.

Final Logic:

�� 3.0 M > 1.0 M > 0.5 M > 0.2 M.

🧠 Memory Trick

1. → Calculate before arranging.

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

4 If an aqueous solution is heated from 25°C to 80°C, how will its molarity behave, assuming no solvent evaporates?

A Molarity will increase due to a decrease in volume.

B Molarity will decrease because the volume of the solution expands with temperature.

C Molarity will remain constant as mass is conserved.

D Molarity will decrease because the mass of the solvent increases.

💡 Short Explanation

�� Molarity depends on volume. �� Heating expands the solution. 1. • Moles remain constant, so molarity decreases.

✅ Correct Answer Explanation

2. → Molarity is moles of solute per litre of solution. On heating, the volume of the solution generally increases due to thermal expansion. Since the number of moles remains unchanged, molarity decreases.

❓ Why Other Options Are Incorrect

�� Option A → Heating usually increases volume, not decreases it.
�� Option C → Mass conservation does not make molarity constant because molarity depends on volume.
3. • Option D → Mass of solvent does not increase on heating.

🎯 Strategy

Used

4. Dimensional/Unit Analysis

Application:

�� Apply molarity = moles/volume and analyze the effect of increased volume.

Final Logic:

�� Volume increases, denominator increases, molarity decreases.

🧠 Memory Trick

1. → Heat ↑, Volume ↑, Molarity ↓.

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

5 Match the solute amounts with their corresponding molality in 1 kg of water.

List 1 (Solute Amount)	List 2 (Corresponding Molality)
1. 40 g NaOH	a. 1.0 m
2. 60 g Urea	b. 1.0 m
3. 90 g Glucose	c. 0.5 m
4. 20 g NaOH	d. 0.5 m

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

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

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

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

💡 Short Explanation

�� 40 g NaOH = 1 mole. �� 60 g urea = 1 mole. 1. • 90 g glucose = 0.5 mole.

✅ Correct Answer Explanation

Molality is moles of solute per kg of solvent. Since the solvent is 1 kg water, molality equals the number of moles of solute. 40 g NaOH ÷ 40 = 1.0 m 60 g urea ÷ 60 = 1.0 m 90 g glucose ÷ 180 = 0.5 m 2. 20 g NaOH ÷ 40 = 0.5 m

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly matches 40 g NaOH with 0.5 m.
�� Option C → Incorrectly matches 60 g urea and 20 g NaOH.
3. • Option D → Incorrectly matches all calculated molality values.

🎯 Strategy

Used

4. Dimensional/Unit Analysis

Application:

�� Convert grams into moles using molar mass, then divide by 1 kg solvent.

Final Logic:

�� In 1 kg water, molality = moles of solute.

🧠 Memory Trick

1. → 1 kg solvent makes m = moles.

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

6 Why is molality an independent concentration measure in relation to thermal variation?
1. It relates to the volume of the solution directly.
2. Mass is not a function of temperature.
3. Volume is not a function of temperature.
4. It uses moles per litre instead of kilograms.

A 1, 3

B 2 only

C 2, 3

D 3, 4

💡 Short Explanation

�� Molality uses mass of solvent. �� Mass does not change with temperature. 1. • Therefore, molality is temperature independent.

✅ Correct Answer Explanation

2. → Molality is defined as moles of solute per kilogram of solvent. Since it depends on mass and not volume, it remains independent of temperature. Mass is not a function of temperature, but volume is affected by temperature.

❓ Why Other Options Are Incorrect

�� Option A → Statements 1 and 3 are incorrect.
�� Option C → Statement 3 is incorrect because volume changes with temperature.
3. • Option D → Both Statements 3 and 4 are incorrect.

🎯 Strategy

Used

4. Elimination

Application:

�� Remove statements involving volume independence or moles per litre.

Final Logic:

�� Molality is temperature independent because mass is temperature independent.

🧠 Memory Trick

1. → Molality = mass-based.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Temperature Independence of Molality

7 Identify the specific term used to refer to the concentration of a solute in a saturated solution at a dynamic equilibrium.

A Molality

B Molarity

C Solubility

D Mole fraction

💡 Short Explanation

�� Saturated solution contains maximum dissolved solute. �� Dynamic equilibrium exists. 1. • This concentration is called solubility.

✅ Correct Answer Explanation

2. → Solubility is the concentration of solute in a saturated solution at a given temperature and pressure. At saturation, dissolution and crystallisation occur at equal rates, establishing dynamic equilibrium.

❓ Why Other Options Are Incorrect

�� Option A → Molality is a concentration unit, not the specific saturated concentration term.
�� Option B → Molarity is also a concentration unit, not the term for saturated concentration.
3. • Option D → Mole fraction is a way of expressing composition, not specifically solubility.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Link saturated solution concentration with the term used in NCERT.

Final Logic:

�� Concentration of saturated solution = solubility.

🧠 Memory Trick

1. → Saturated concentration = solubility.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solubility

8 Which combination is most likely to result in a successful dissolution based on the nature of substances?

A Sodium chloride in benzene

B Anthracene in water

C Sugar in benzene

D Naphthalene in benzene

💡 Short Explanation

�� Benzene is non-polar. �� Naphthalene is non-polar. 1. • Like dissolves like.

✅ Correct Answer Explanation

2. → A solute dissolves best in a solvent having similar intermolecular interactions. Naphthalene is non-polar and benzene is also non-polar, so naphthalene dissolves readily in benzene.

❓ Why Other Options Are Incorrect

�� Option A → Sodium chloride is ionic and does not dissolve well in non-polar benzene.
�� Option B → Anthracene is non-polar and does not dissolve well in polar water.
3. • Option C → Sugar is polar and does not dissolve well in non-polar benzene.

🎯 Strategy

Used

4. Odd One Out

Application:

�� Identify the pair with similar polarity and intermolecular interactions.

Final Logic:

�� Non-polar naphthalene dissolves in non-polar benzene.

🧠 Memory Trick

1. → Like dissolves like.

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

9 Which principle or rule governs the dissolution of polar solutes in polar solvents and non-polar solutes in non-polar solvents?

A Henry's Law

B Le Chatelier's Principle

C Like dissolves like

D Raoult's Law

💡 Short Explanation

�� Polar dissolves polar. �� Non-polar dissolves non-polar. 1. • This is called like dissolves like.

✅ Correct Answer Explanation

2. → The principle "like dissolves like" explains that substances with similar intermolecular interactions dissolve in each other. Polar solutes dissolve in polar solvents, while non-polar solutes dissolve in non-polar solvents.

❓ Why Other Options Are Incorrect

�� Option A → Henry's law deals with solubility of gases in liquids.
�� Option B → Le Chatelier's principle explains equilibrium shifts.
3. • Option D → Raoult's law relates vapour pressure to mole fraction.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Match the phrase about polar-polar and non-polar-non-polar dissolution.

Final Logic:

�� Similar nature supports dissolution.

🧠 Memory Trick

1. → Like likes like.

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

10 Identify the process type regarding energetic intermolecular dynamics: For a solute to dissolve in a solvent, the interactions must be:

A Exclusively ionic.

B Similar in the two components (solute and solvent).

C Vastly stronger in solute-solute than solute-solvent.

D Completely repulsive.

💡 Short Explanation

�� Similar intermolecular interactions favour dissolution. �� Solute-solvent attraction must be effective. 1. • This supports solution formation.

✅ Correct Answer Explanation

2. → For dissolution to occur, solute-solvent interactions must be similar or comparable to the interactions present among solute-solute and solvent-solvent particles. This is the basis of the "like dissolves like" principle.

❓ Why Other Options Are Incorrect

�� Option A → Dissolution is not exclusively ionic; covalent and non-polar substances may also dissolve.
�� Option C → If solute-solute interactions dominate strongly, dissolution becomes difficult.
3. • Option D → Repulsive interactions prevent dissolution.

🎯 Strategy

Used

4. Elimination

Application:

�� Remove extreme or scientifically incorrect conditions.

Final Logic:

�� Similar interactions between solute and solvent favour dissolution.

🧠 Memory Trick

1. → Similar forces mix.

📚 Questions 11–15

11 During the initial stages of adding a solid solute to a pure solvent, which process dominates the system?

A Crystallisation

B Dissolution

C Precipitation

D Sublimation

💡 Short Explanation

Initially, only solvent is present. Solute particles begin entering the solvent. Dissolution rate is greater than crystallisation.

✅ Correct Answer Explanation

When a solid solute is first added to a pure solvent, there are almost no dissolved solute particles available to undergo crystallisation. Therefore, dissolution dominates the system, increasing the solute concentration until saturation is approached. Only after sufficient solute dissolves does crystallisation begin to compete with dissolution.

❓ Why Other Options Are Incorrect

Option A → Crystallisation requires dissolved solute particles and does not dominate initially.
Option C → Precipitation is the formation of an insoluble solid from a reaction and is not the process occurring here.
Option D → Sublimation is the direct conversion of a solid into a gas and is unrelated to solution formation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Focus on the phrase "initial stages", where dissolution naturally exceeds crystallisation.

Final Logic:

�� Pure solvent + added solute = dissolution dominates.

🧠 Memory Trick

Start = Dissolve First

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

12 In a saturated solution, if radioactively tagged solute crystals are added, the radioactivity will soon be detected in the liquid phase as well. This proves that:

A Saturated solutions are actually unsaturated.

B Crystallisation has permanently ceased.

C Dissolution and crystallisation are ongoing in a dynamic equilibrium.

D Radioactivity alters solubility limits.

💡 Short Explanation

Solute particles continuously exchange. Both processes occur simultaneously. Dynamic equilibrium exists.

✅ Correct Answer Explanation

The movement of radioactive particles from the crystal into the solution demonstrates that solute particles continuously dissolve while an equal number simultaneously crystallise. Although there is no net change in concentration, both processes continue at equal rates, proving the existence of dynamic equilibrium.

❓ Why Other Options Are Incorrect

Option A → The solution remains saturated despite particle exchange.
Option B → Crystallisation continues at the same rate as dissolution.
Option D → Radioactivity acts only as a tracer and does not change solubility.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Interpret the tracer experiment using the concept of dynamic equilibrium.

Final Logic:

�� Particle exchange without concentration change indicates dynamic equilibrium.

🧠 Memory Trick

Dynamic = Both Directions Continue

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

13 A saturated solution's concentration limit is specifically defined by its:

A Molality limit

B Saturation index

C Solubility

D Osmotic point

💡 Short Explanation

Saturated solution contains maximum dissolved solute. Maximum concentration is called solubility. It depends on temperature.

✅ Correct Answer Explanation

Solubility is defined as the maximum amount of solute that can dissolve in a specified amount of solvent at a given temperature and pressure. Therefore, the concentration limit of a saturated solution is its solubility.

❓ Why Other Options Are Incorrect

Option A → Molality is only a concentration unit and does not define the saturation limit.
Option B → Saturation index is not the NCERT definition of concentration limit.
Option D → Osmotic point is unrelated to saturation concentration.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the NCERT term specifically defining the saturation concentration.

Final Logic:

�� Saturated concentration = Solubility.

🧠 Memory Trick

Saturated = Solubility Limit

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

14 If a solution is unsaturated, the rate of dissolution compared to the rate of crystallisation is:

A Slower

B Exactly equal

C Faster

D Zero

💡 Short Explanation

More solute can still dissolve. Dissolution exceeds crystallisation. Concentration continues increasing.

✅ Correct Answer Explanation

In an unsaturated solution, the solvent can still accommodate additional solute particles. Therefore, the rate of dissolution is greater than the rate of crystallisation, resulting in a net increase in dissolved solute concentration until saturation is reached.

❓ Why Other Options Are Incorrect

Option A → Dissolution is not slower than crystallisation in an unsaturated solution.
Option B → Equal rates occur only at dynamic equilibrium in a saturated solution.
Option D → Crystallisation is not necessarily zero; however, dissolution predominates.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate options describing saturated equilibrium or impossible conditions.

Final Logic:

�� Unsaturated solution → Dissolution > Crystallisation.

🧠 Memory Trick

Unsaturated = Dissolve More

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

📄 Passage

The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium represented by equation 1.10. This, being dynamic equilibrium, must follow Le Chatelier's Principle. In general, if in a nearly saturated solution, the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in temperature and if it is exothermic (Δsol H < 0) the solubility should decrease. These trends are also observed experimentally. Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not.

If a salt dissolves in water and noticeably cools the surrounding solution (endothermic), how can its solubility be increased?

A By decreasing the temperature

B By increasing the pressure

C By increasing the temperature

D By decreasing the pressure

💡 Short Explanation

Endothermic dissolution absorbs heat. Heating supplies additional heat. Solubility increases.

✅ Correct Answer Explanation

An endothermic dissolution process absorbs heat from the surroundings. According to Le Chatelier's Principle, increasing the temperature supplies additional heat and shifts the equilibrium toward dissolution, thereby increasing the solubility of the solid.

❓ Why Other Options Are Incorrect

Option A → Lowering temperature shifts equilibrium away from dissolution.
Option B → Pressure has negligible effect on the solubility of solids in liquids.
Option D → Decreasing pressure does not significantly affect solid-liquid solubility.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the passage statement "endothermic dissolution increases with rise in temperature."

Final Logic:

�� Endothermic process + Heat added = Greater solubility.

🧠 Memory Trick

Endothermic Loves Heat

📚 Questions 16–20

📄 Passage

For an exothermic dissolution process (Δsol H < 0), what happens to the equilibrium if the solution is externally heated?

A It shifts to dissolve more solute.

B Solubility decreases.

C More solvent is produced.

D Solubility becomes infinite.

💡 Short Explanation

Exothermic dissolution releases heat. Heating opposes the dissolution process. Solubility decreases.

✅ Correct Answer Explanation

According to Le Chatelier's Principle, heat acts as a product in an exothermic dissolution process. When the temperature is increased, the equilibrium shifts in the reverse direction to consume the added heat, resulting in less solute dissolving and hence a decrease in solubility.

❓ Why Other Options Are Incorrect

Option A → Heating favors dissolution only in endothermic processes.
Option C → Heating does not produce additional solvent.
Option D → Solubility cannot become infinite.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Apply Le Chatelier's Principle to an exothermic dissolution equilibrium.

Final Logic:

�� Heat added + exothermic process = decreased solubility.

🧠 Memory Trick

Exothermic Hates Heat

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

17 The primary reason pressure has minimal effect on the solubility of solids in liquids is because:

A Solids and liquids undergo rapid expansion under pressure.

B Solids and liquids are highly compressible.

C Solids and liquids are highly incompressible.

D Liquids instantly vaporize at high pressure.

💡 Short Explanation

Solids and liquids show negligible compression. Pressure causes very little volume change. Solubility remains almost unchanged.

✅ Correct Answer Explanation

Pressure significantly affects gases because gases are highly compressible. Solids and liquids, however, are highly incompressible, so increasing pressure produces only a negligible change in their volume. Consequently, pressure has almost no effect on the solubility of solids in liquids.

❓ Why Other Options Are Incorrect

Option A → Solids and liquids do not undergo rapid expansion under pressure.
Option B → They are incompressible, not highly compressible.
Option D → Liquids do not instantly vaporize under high pressure.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate statements contradicting the physical properties of solids and liquids.

Final Logic:

�� Incompressibility explains the negligible pressure effect.

🧠 Memory Trick

Solid + Liquid = Pressure Little

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

18 Which equilibrium system is LEAST affected by a change in pressure?

A Gas dissolved in liquid

B Solid dissolved in liquid

C Gas reacting with gas

D Liquid evaporating into gas

💡 Short Explanation

Solids and liquids are nearly incompressible. Pressure has negligible influence. Solid-liquid equilibrium remains almost unchanged.

✅ Correct Answer Explanation

Pressure mainly affects systems involving gases because gases undergo significant volume changes. A solid dissolved in a liquid involves two incompressible phases, so pressure produces almost no change in equilibrium or solubility.

❓ Why Other Options Are Incorrect

Option A → Gas solubility in liquids increases with pressure (Henry's Law).
Option C → Gas-phase equilibria are strongly influenced by pressure.
Option D → Liquid-vapour equilibrium is also pressure dependent.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the equilibrium system without a gaseous phase.

Final Logic:

�� No gaseous phase = least pressure effect.

🧠 Memory Trick

No Gas = No Pressure Effect

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

19 Which of the following gases is greatly affected by pressure and temperature when dissolved in liquids?

A Only temperature

B Only pressure

C Neither pressure nor temperature

D Both pressure and temperature

💡 Short Explanation

Gas solubility changes with pressure. Gas solubility also changes with temperature. Both factors are important.

✅ Correct Answer Explanation

Unlike solids, the solubility of gases in liquids depends significantly on both pressure and temperature. According to Henry's Law, increasing pressure generally increases gas solubility, while increasing temperature usually decreases the solubility of gases in liquids.

❓ Why Other Options Are Incorrect

Option A → Pressure also significantly affects gas solubility.
Option B → Temperature also significantly affects gas solubility.
Option C → Gas solubility is influenced by both variables.

🎯 Strategy

Used

Elimination

Application:

�� Remove options considering only one variable or none.

Final Logic:

�� Gas solubility depends on both pressure and temperature.

🧠 Memory Trick

Gas = Pressure + Temperature

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

20 The dissolved oxygen in aquatic environments sustains life, despite its solubility being:

A Exceedingly high

B A small extent

C Reactive with water to form acid

D Unaffected by pressure

💡 Short Explanation

Oxygen is only sparingly soluble. Even small dissolved amounts support aquatic respiration. Aquatic ecosystems depend on this dissolved oxygen.

✅ Correct Answer Explanation

Oxygen dissolves in water only to a small extent, yet this dissolved oxygen is sufficient to sustain fish and other aquatic organisms through respiration. Its concentration is limited but biologically essential.

❓ Why Other Options Are Incorrect

Option A → Oxygen is not highly soluble in water.
Option C → Oxygen does not react with water to form an acid under normal conditions.
Option D → The solubility of gases, including oxygen, is affected by pressure.

🎯 Strategy

Used

Odd One Out

Application:

�� Compare the known physical properties of dissolved oxygen with the given statements.

Final Logic:

�� Aquatic life survives on oxygen that is dissolved only to a small extent.

🧠 Memory Trick

Small O₂, Big Life

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

CUET UG Chemistry Booster Test – 2 Concentration Measures and Solubility

CUET UG Chemistry Booster Test - 2 Concentration Measures and Solubility

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