-CUET UG Chemistry Booster Test - 2 Fundamentals of Solutions -

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CUET UG Chemistry Booster Test - 2 Fundamentals of Solutions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Fundamentals of Solutions

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QUESTION 1 OF 20

Arrange the following mixtures in decreasing order of macroscopic uniformity (most homogeneous to most heterogeneous):
1. A perfectly uniform alloy of brass
2. An aqueous solution containing 0.1 M NaCl
3. A partially settled suspension of chalk in water
4. Freshly mixed sand and water

Answer Review

1 Arrange the following mixtures in decreasing order of macroscopic uniformity (most homogeneous to most heterogeneous):
1. A perfectly uniform alloy of brass
2. An aqueous solution containing 0.1 M NaCl
3. A partially settled suspension of chalk in water
4. Freshly mixed sand and water

A 1, 2, 3, 4

B 2, 1, 4, 3

C 1, 2, 4, 3

D 2, 3, 1, 4

💡 Short Explanation

�� Brass alloy and NaCl solution are homogeneous. �� Chalk suspension is less uniform. 1. • Sand and water is clearly heterogeneous.

✅ Correct Answer Explanation

2. → A uniform brass alloy is highly homogeneous at the macroscopic level. A 0.1 M NaCl solution is also homogeneous because salt is uniformly dissolved in water. A partially settled chalk suspension shows non-uniform distribution, while freshly mixed sand and water is visibly heterogeneous.

❓ Why Other Options Are Incorrect

�� Option B → Places sand-water before chalk suspension, which is less appropriate.
�� Option C → Places sand-water before partially settled chalk suspension.
3. • Option D → Places suspension before homogeneous alloy, which is incorrect.

🎯 Strategy

Used

4. Option Grouping

Application:

�� Group homogeneous mixtures first, then arrange visibly non-uniform mixtures later.

Final Logic:

�� Homogeneous solutions/alloys come before suspensions and coarse mixtures.

🧠 Memory Trick

1. → Uniform first, visible particles last.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Homogeneous and Heterogeneous Mixtures

2 In a syrup containing 60 g of sugar and 40 g of water, what is the analytical reason for classifying water as the solvent despite its lower mass?

A Water is the universal solvent.

B The physical state of the resulting syrup matches the liquid state of water.

C Sugar is always a solute regardless of concentration.

D The mole fraction of water is still higher than that of sugar due to their molar masses.

💡 Short Explanation

�� Syrup remains liquid. �� Water determines the final physical state. 1. • Therefore, water acts as solvent.

✅ Correct Answer Explanation

2. → The solvent generally determines the physical state of the solution. Even though sugar is present in greater mass, the final syrup remains liquid because of water. Hence, water is considered the solvent.

❓ Why Other Options Are Incorrect

�� Option A → Water being a universal solvent is not the analytical reason here.
�� Option C → Sugar is not always a solute regardless of context.
3. • Option D → Mole fraction is not the main basis used here.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Focus on the clue that solvent determines the physical state of the solution.

Final Logic:

�� Final state is liquid, so liquid water is the solvent.

🧠 Memory Trick

1. → Solvent sets state.

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

3 Identify the specific classification of a solution containing exactly one solvent and one solute, such as an aqueous solution of glucose.

A Ternary solution

B Aqueous colloidal suspension

C Binary solution

D Supersaturated mixture

💡 Short Explanation

�� Binary means two components. �� One solvent and one solute make two components. 1. • Glucose in water is a binary solution.

✅ Correct Answer Explanation

2. → A binary solution contains exactly two components: one solute and one solvent. An aqueous solution of glucose contains glucose as solute and water as solvent, so it is a binary solution.

❓ Why Other Options Are Incorrect

�� Option A → Ternary solution contains three components.
�� Option B → Glucose solution is not a colloidal suspension.
3. • Option D → Supersaturated refers to solute amount beyond normal saturation, not component number.

🎯 Strategy

Used

4. Substitution

Application:

�� Substitute the number of components into the definition.

Final Logic:

�� One solute + one solvent = binary solution.

🧠 Memory Trick

1. → Binary = Two.

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

4 If a binary solution is in a solid state at room temperature, which combination of physical states for the isolated solute and solvent prior to mixing is analytically impossible based on the solvent's role?

A Solid solute, solid solvent

B Gas solute, solid solvent

C Liquid solute, solid solvent

D Solid solute, liquid solvent

💡 Short Explanation

�� Solvent determines final physical state. �� Solid solution requires solid solvent. 1. • Liquid solvent would give liquid solution.

✅ Correct Answer Explanation

2. → The physical state of a solution is generally determined by the solvent. If the final binary solution is solid, the solvent must be solid. Therefore, a solid solute with a liquid solvent is analytically inconsistent with a final solid solution.

❓ Why Other Options Are Incorrect

�� Option A → Solid solute and solid solvent can form solid-in-solid solution.
�� Option B → Gas solute and solid solvent can form gas-in-solid solution.
3. • Option C → Liquid solute and solid solvent can form liquid-in-solid solution.

🎯 Strategy

Used

4. Elimination

Application:

�� Eliminate all options where the solvent is solid because they can form solid solutions.

Final Logic:

�� Solid final solution needs solid solvent.

🧠 Memory Trick

1. → Final state follows solvent.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Classification of Solutions by Solvent State

5 Match the gaseous mixture concepts with their corresponding principles:

List 1 (Gaseous Mixture Concept)	List 2 (Corresponding Principle)
1. Mixture of O₂ and N₂	a. Form homogeneous solutions in all proportions
2. Dalton's Law of partial pressures	b. Total pressure equals sum of partial pressures
3. Mole fraction in a binary gas mix	c. Useful in describing calculations involving gas mixtures
4. Gas solvent properties	d. Determines the gaseous state of the final solution

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

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

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

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

💡 Short Explanation

�� O₂ and N₂ form homogeneous gas mixtures. �� Dalton's law relates total pressure to partial pressures. 1. • Mole fraction helps in gas mixture calculations.

✅ Correct Answer Explanation

2. → A mixture of oxygen and nitrogen forms a homogeneous gaseous solution. Dalton's law states that total pressure equals the sum of partial pressures. Mole fraction is useful in gas mixture calculations, especially where partial pressure is involved. The gas solvent determines the gaseous state of the final solution.

❓ Why Other Options Are Incorrect

�� Option B → Dalton's law and mole fraction are incorrectly matched.
�� Option C → Mixture of O₂ and N₂ is not mainly defined as a calculation unit.
3. • Option D → Most matches are incorrectly paired.

🎯 Strategy

Used

4. Option Grouping

Application:

�� First match Dalton's law with partial pressure, then match mole fraction with gas calculations.

Final Logic:

�� O₂ + N₂ = homogeneous gas solution; Dalton = partial pressures.

🧠 Memory Trick

1. → Dalton adds pressures.

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

6 When analyzing camphor subliming and mixing into nitrogen gas:
1. It forms a solid-in-gas solution.
2. Nitrogen acts as the solvent.
3. The final state is a liquid solution due to camphor's melting point.
4. It is a binary mixture.

A 1, 2, 4

B 1, 3, 4

C 2, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Camphor is solid solute. �� Nitrogen is gaseous solvent. 1. • It is a binary solid-in-gas solution.

✅ Correct Answer Explanation

2. → Camphor subliming and mixing uniformly into nitrogen gas forms a solid-in-gas solution. Nitrogen acts as the gaseous solvent and camphor acts as the solute. Since only two components are involved, it is a binary mixture.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 3, which is incorrect.
�� Option C → Omits Statement 1, which is correct.
3. • Option D → Includes Statement 3, which is false.

🎯 Strategy

Used

4. Elimination

Application:

�� Identify the incorrect statement about liquid solution and remove options containing it.

Final Logic:

�� Camphor + nitrogen = solid-in-gas binary solution.

🧠 Memory Trick

1. → Camphor in nitrogen = solid in gas.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solid in Gas Solution

7 When assessing the solubility of a gas in a liquid, Henry's Law constant (K_H) is utilized. If the partial pressure of the gas is measured in bar, and solubility is measured in mole fraction, what is the expected unit for K_H?

A mol L⁻¹

B bar

C bar⁻¹

D Dimensionless

💡 Short Explanation

�� Henry's law: p = KH x. �� Mole fraction is dimensionless. 1. • Therefore, KH has unit of pressure.

✅ Correct Answer Explanation

2. → According to Henry's law, partial pressure of gas is proportional to its mole fraction in solution: p = KH x. Since mole fraction has no unit, KH must have the same unit as pressure. If pressure is measured in bar, KH is also expressed in bar.

❓ Why Other Options Are Incorrect

�� Option A → mol L⁻¹ is a concentration unit, not KH here.
�� Option C → bar⁻¹ would apply to inverse pressure, not KH in p = KHx.
3. • Option D → KH is not dimensionless when pressure is given in bar.

🎯 Strategy

Used

4. Dimensional/Unit Analysis

Application:

�� Compare units on both sides of Henry's law equation.

Final Logic:

�� p = KHx; x has no unit, so KH = pressure unit.

🧠 Memory Trick

1. → Henry's constant carries pressure.

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

8 When a solid dissolves in a liquid dynamically, and the dissolution process is endothermic (ΔsolH > 0), the solubility strictly increases with rise in temperature according to which thermodynamic principle?

A Raoult's Law

B Le Chatelier's Principle

C Henry's Law

D Dalton's Law

💡 Short Explanation

�� Endothermic dissolution absorbs heat. �� Temperature rise favours heat-absorbing process. 1. • Solubility increases according to Le Chatelier's principle.

✅ Correct Answer Explanation

2. → If dissolution is endothermic, heat acts like a reactant. According to Le Chatelier's principle, increasing temperature favours the forward process that absorbs heat. Hence, solubility increases with rise in temperature.

❓ Why Other Options Are Incorrect

�� Option A → Raoult's law relates vapour pressure and mole fraction.
�� Option C → Henry's law relates gas solubility to pressure.
3. • Option D → Dalton's law relates total pressure to partial pressures.

🎯 Strategy

Used

4. Contextual/Tonal Matching

Application:

�� Link endothermic dissolution and temperature change to equilibrium shift.

Final Logic:

�� Heat added favours endothermic dissolution.

🧠 Memory Trick

1. → Endothermic loves heat.

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

9 Regarding the solution of hydrogen in palladium:
1. It is classified as a gas-in-solid solution.
2. Palladium retains its solid state and acts as the solvent.
3. The overall physical state of the solution is gaseous.
4. It is a homogeneous mixture.

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Hydrogen is gas solute. �� Palladium is solid solvent. 1. • The final solution is homogeneous and solid.

✅ Correct Answer Explanation

2. → Hydrogen in palladium is a gas-in-solid solution. Palladium remains solid and acts as the solvent, so the final solution is solid, not gaseous. It is homogeneous because hydrogen is uniformly distributed in palladium.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

4. Elimination

Application:

�� Remove all options containing the false statement that the solution is gaseous.

Final Logic:

�� Solid solvent palladium gives a solid solution.

🧠 Memory Trick

1. → Hydrogen hides in palladium.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Gas in Solid Solution

10 Analytically, why are alloys like bronze and brass considered solutions rather than simple mechanical mixtures?

A They have distinct, visible boundaries between different metal phases.

B Their composition and metallic properties are uniform throughout on a macroscopic level.

C They can be separated into their constituent elements by simple physical filtration.

D They are always composed of a metal dissolved in a non-metal.

💡 Short Explanation

�� Alloys are homogeneous solid solutions. �� Their properties are uniform throughout. 1. • They are not simple visible mixtures.

✅ Correct Answer Explanation

2. → Alloys like bronze and brass are considered solid solutions because their components are uniformly distributed at the macroscopic level. They do not show visible separation like mechanical mixtures and have uniform metallic properties throughout.

❓ Why Other Options Are Incorrect

�� Option A → Visible boundaries indicate heterogeneity, not solution behavior.
�� Option C → Alloys cannot be separated by simple filtration.
3. • Option D → Alloys are not always metal and non-metal combinations.

🎯 Strategy

Used

4. Odd One Out

Application:

�� Select the option that expresses homogeneity, the key property of solutions.

Final Logic:

�� Uniform composition and properties make alloys solid solutions.

🧠 Memory Trick

1. → Alloy = uniform metal mix.

📚 2. Class XII – Chemistry Part I – Chapter: Solutions – Topic: Solid Solutions and Alloys

11 Consider a dilute solution of sodium chloride in water:
1. The number of solvent molecules vastly outnumbers the solute ions.
2. Adding a small amount of solvent significantly alters its qualitative "dilute" status.
3. Quantitative descriptions are preferred to avoid confusion over its exact concentration.
4. It implies a relatively very small quantity of solute.

A 1, 3, 4

B 1, 2, 3

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

Dilute solutions contain relatively less solute. Solvent molecules greatly outnumber solute particles. Quantitative concentration is preferred for accuracy.

✅ Correct Answer Explanation

A dilute solution contains a relatively small amount of solute compared to the solvent, so Statement 1 is correct. NCERT emphasizes that qualitative terms like dilute and concentrated can create confusion; therefore quantitative descriptions are preferred, making Statement 3 correct. Statement 4 correctly defines a dilute solution. Statement 2 is incorrect because adding a small amount of solvent does not necessarily change its qualitative classification significantly.

❓ Why Other Options Are Incorrect

Option B → Includes Statement 2, which is incorrect.
Option C → Includes Statement 2 and omits Statement 1.
Option D → Includes Statement 2 while excluding Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statement and eliminate all options containing it.

Final Logic:

Only Statements 1, 3 and 4 correctly describe a dilute solution.

🧠 Memory Trick

Dilute = Less Solute, More Solvent.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Qualitative Description of Concentration

12 Match the term to its analytical description in the context of concentration.

List 1 (Concentration Term)	List 2 (Analytical Description)
1. Saturated solution	a. No more solute can be dissolved at the given temperature and pressure
2. Unsaturated solution	b. More solute can still be dissolved at the same temperature
3. Concentrated solution	c. Contains a relatively very large quantity of solute
4. Dynamic equilibrium	d. Dissolution rate equals crystallization rate

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

Saturated solutions cannot dissolve more solute. Unsaturated solutions can dissolve additional solute. Dynamic equilibrium means equal dissolution and crystallization rates.

✅ Correct Answer Explanation

A saturated solution contains the maximum amount of dissolved solute at a given temperature and pressure. An unsaturated solution can dissolve more solute. A concentrated solution contains a relatively large amount of solute, while dynamic equilibrium exists when dissolution and crystallization occur at equal rates.

❓ Why Other Options Are Incorrect

Option B → Saturated and unsaturated solutions are interchanged.
Option C → Concentrated solution and dynamic equilibrium are incorrectly matched.
Option D → All four matches are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

�� Match the fixed definition of dynamic equilibrium first and then identify saturated and unsaturated solutions.

Final Logic:

Saturated → no more dissolution; Unsaturated → more dissolution possible.

🧠 Memory Trick

Saturated Stops, Unsaturated Starts.

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

13 If you mix 22 g of benzene with 122 g of carbon tetrachloride, what is the analytical formula used to calculate the mass percentage of benzene?

A (22 / 122) × 100

B (122 / 144) × 100

C (22 / 144) × 100

D (22 / 100) × 122

💡 Short Explanation

Mass percentage uses total solution mass. Total mass = 22 + 122 = 144 g. Solute mass divided by total mass × 100.

✅ Correct Answer Explanation

Mass percentage is calculated as Mass of solute / Total mass of solution × 100 Therefore, 22/(22+122) ×100 = 22/144 ×100.

❓ Why Other Options Are Incorrect

Option A → Uses solvent mass instead of total solution mass.
Option B → Calculates solvent percentage.
Option D → Uses an incorrect mathematical expression.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Apply the standard NCERT formula for mass percentage.

Final Logic:

Mass percentage always uses total solution mass.

🧠 Memory Trick

w/w = Solute ÷ Solution ×100

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Mass Percentage

14 For a 35% (v/v) solution of ethylene glycol used as antifreeze, how is the total volume of the solution analytically constrained?

A It must be exactly 100 mL, formed by adding 65 mL of water to 35 mL of ethylene glycol.

B It is 100 volumes total, containing 35 volumes of ethylene glycol.

C Volume percentage implies adding 100 mL of solvent to 35 mL of solute.

D It implies 35 grams of ethylene glycol in 100 mL of solvent.

💡 Short Explanation

Volume percentage uses total solution volume. 35% v/v means 35 volumes per 100 volumes solution. Commonly used for liquid-liquid solutions.

✅ Correct Answer Explanation

Volume percentage is defined as Volume of solute / Volume of solution ×100. Thus, 35% (v/v) means 35 volumes of ethylene glycol are present in every 100 volumes of the final solution.

❓ Why Other Options Are Incorrect

Option A → Water addition is not necessarily 65 mL because solution volumes are not always additive.
Option C → Uses solvent volume instead of solution volume.
Option D → Refers to mass rather than volume.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Identify that v/v uses solution volume, not solvent volume.

Final Logic:

35% v/v = 35 volumes in 100 volumes solution.

🧠 Memory Trick

v/v = Volume in Volume.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Volume Percentage

15 When preparing a mass by volume percentage solution in a pharmacy, the physical process of incorporating the solid active pharmaceutical ingredient into the liquid vehicle until a uniform single phase is reached is termed:

A Fractional distillation

B Sublimation

C Dissolution

D Coagulation

💡 Short Explanation

Solid particles disperse into liquid. A homogeneous single phase is formed. This process is dissolution.

✅ Correct Answer Explanation

Dissolution is the physical process in which solute particles separate and become uniformly distributed throughout the solvent, producing a homogeneous solution. Pharmaceutical preparations commonly rely on this process.

❓ Why Other Options Are Incorrect

Option A → Fractional distillation separates liquids based on boiling points.
Option B → Sublimation converts solid directly into gas.
Option D → Coagulation causes particles to aggregate instead of forming a homogeneous solution.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Focus on the phrase "uniform single phase."

Final Logic:

Uniform solid-liquid mixing is dissolution.

🧠 Memory Trick

Dissolution = Disappear into Solution.

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Formation of Solutions

16 Arrange the following saline solutions in decreasing order of their NaCl mass by volume percentage (w/V):
1. Hypertonic saline used to draw out fluid (>0.9% w/V)
2. Normal saline matching blood plasma (0.9% w/V)
3. Hypotonic saline causing cells to swell (<0.9% w/V)

A 1, 2, 3

B 3, 2, 1

C 2, 1, 3

D 1, 3, 2

💡 Short Explanation

Hypertonic saline has the highest NaCl concentration. Normal saline contains 0.9% (w/V) NaCl. Hypotonic saline contains less than 0.9% (w/V) NaCl.

✅ Correct Answer Explanation

Hypertonic saline has a sodium chloride concentration greater than 0.9% (w/V), making it the highest concentration. Normal saline contains exactly 0.9% (w/V) NaCl and is isotonic with blood plasma. Hypotonic saline contains less than 0.9% (w/V) NaCl, making it the lowest concentration. Therefore, the decreasing order is Hypertonic > Normal > Hypotonic.

❓ Why Other Options Are Incorrect

Option B → Arranges the concentrations in increasing order instead of decreasing order.
Option C → Places normal saline before hypertonic saline.
Option D → Places hypotonic saline above normal saline.

🎯 Strategy

Used

Option Grouping

Application:

�� Compare each saline solution with the standard 0.9% (w/V) concentration.

Final Logic:

Greater than 0.9% > 0.9% > Less than 0.9%.

🧠 Memory Trick

Hyper > Normal > Hypo

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Mass by Volume Percentage and Clinical Solutions

17 Analytically, the formula for calculating parts per million (ppm) of component A is
(Number of parts of A / Total number of parts of all components) × 10⁶.
This ratio can fundamentally represent which of the following?

A Mass to mass only

B Volume to volume only

C Mass to volume only

D Mass to mass, volume to volume, or mass to volume

💡 Short Explanation

ppm expresses extremely small concentrations. It can be based on mass or volume ratios. It is widely used for trace substances.

✅ Correct Answer Explanation

Parts per million (ppm) expresses one part of a component per one million parts of the mixture. Depending on the system being studied, it may represent mass/mass, volume/volume, or mass/volume ratios. Therefore, ppm is a flexible concentration unit applicable to different types of measurements.

❓ Why Other Options Are Incorrect

Option A → ppm is not restricted to mass/mass measurements.
Option B → ppm is not limited to volume/volume measurements.
Option C → ppm is not limited to mass/volume measurements.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Identify the possible numerator and denominator combinations used in ppm calculations.

Final Logic:

ppm can represent multiple ratio types depending on the application.

🧠 Memory Trick

ppm = Parts in a Million (Any Ratio)

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Parts Per Million (ppm)

18 Identify the alternative unit commonly used interchangeably with parts per million (ppm) to express the trace concentration of pollutants dissolved in a liquid medium like water.

A mol L⁻¹

B mg mL⁻¹

C g L⁻¹

D mol kg⁻¹

💡 Short Explanation

ppm is used for trace concentrations. Very small pollutant concentrations are often expressed in mg mL⁻¹ or ppm in NCERT context. Both represent extremely low concentrations.

✅ Correct Answer Explanation

NCERT mentions that trace concentrations of pollutants in water are commonly expressed in parts per million (ppm) or mg mL⁻¹. These units are convenient for representing very small quantities of dissolved substances in large amounts of solvent.

❓ Why Other Options Are Incorrect

Option A → mol L⁻¹ represents molarity, not trace concentration.
Option C → g L⁻¹ is used for relatively larger concentrations.
Option D → mol kg⁻¹ represents molality.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the NCERT example of pollutant concentration units.

Final Logic:

Trace pollutants are commonly expressed in ppm or mg mL⁻¹.

🧠 Memory Trick

Pollutants → ppm or mg mL⁻¹

📚 Class XII – Chemistry Part I – Chapter: Solutions – Topic: Trace Concentration Units

📄 Passage

Each method of expressing concentration of the solutions has its own merits and demerits. Mass percentage, ppm, mole fraction and molality are independent of temperature. These methods are highly reliable for experiments conducted across varying thermal conditions because mass does not depend on temperature. Whereas molarity is a function of temperature. This dependence arises because molarity is defined using the volume of the solution, and volume depends on temperature while mass does not.

According to the passage, why is molality preferred over molarity in experiments involving significant temperature fluctuations?

A Molality uses liters of solvent, preventing thermal expansion.

B Molality relies solely on mass, which does not change with temperature.

C Molarity becomes independent of temperature at high pressures.

D Molality calculations automatically adjust for volume changes.

💡 Short Explanation

Molality depends on mass of solvent. Mass is temperature independent. Therefore molality remains constant with temperature.

✅ Correct Answer Explanation

Molality is defined as moles of solute per kilogram of solvent. Since mass does not change with temperature, molality remains constant even when temperature varies. Molarity, however, depends on solution volume, which changes with temperature.

❓ Why Other Options Are Incorrect

Option A → Molality uses kilograms of solvent, not litres.
Option C → Molarity remains temperature dependent.
Option D → Molality does not automatically adjust for volume changes; it simply does not depend on volume.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the sentence in the passage explaining why molality is temperature independent.

Final Logic:

Mass remains constant while volume changes.

🧠 Memory Trick

Molality = Mass = Stable

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

📄 Passage

Based on the passage, what occurs to the molarity of a 1.0 M aqueous solution if it is heated, causing the liquid to expand?

A The molarity increases because the kinetic energy of the solute increases.

B The molarity decreases because the volume of the solution increases while moles remain constant.

C The molarity remains perfectly constant.

D The molarity decreases because solute mass is lost to evaporation.

💡 Short Explanation

Molarity = moles per litre. Heating increases solution volume. Moles remain unchanged, so molarity decreases.

✅ Correct Answer Explanation

Molarity is defined as the number of moles of solute per litre of solution. When the solution is heated, its volume increases due to thermal expansion, while the number of moles of solute remains constant. Consequently, the molarity decreases.

❓ Why Other Options Are Incorrect

Option A → Increased kinetic energy does not increase molarity.
Option C → Molarity is temperature dependent and therefore does not remain constant.
Option D → The decrease is due to volume expansion, not loss of solute mass.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Apply the definition of molarity (moles/volume) and analyze the effect of increasing volume.

Final Logic:

Constant moles + increased volume = decreased molarity.

🧠 Memory Trick

Heat ↑ → Volume ↑ → Molarity ↓

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

-CUET UG Chemistry Booster Test – 2 Fundamentals of Solutions

CUET UG Chemistry Booster Test - 2 Fundamentals of Solutions

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