Physical Chemistry - Solutions for Competitive Exams

Physical Chemistry - Solutions

Q. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)

A. 0.26 °C
B. 0.52 °C
C. 1.04 °C
D. 0.78 °C

Solution

Boiling point elevation = i * Kb * m = 1 * 0.52 * 0.5 = 0.26 °C, so change = 0.52 °C.

Correct Answer: C — 1.04 °C

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Q. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2021)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = X_solute * P0 = (0.5 / (0.5 + 55.5)) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = (moles of solute / moles of solvent) * P0 = (0.5 / 55.5) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 1 kg of water is mixed with 1 kg of sugar, what is the expected change in boiling point? (2023)

A. No change
B. Increase
C. Decrease
D. Depends on the sugar type

Solution

The addition of sugar (a non-volatile solute) will raise the boiling point of water due to boiling point elevation.

Correct Answer: B — Increase

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Q. If 1 mole of a non-volatile solute is dissolved in 1 kg of water, what is the expected freezing point depression? (2019)

A. 1.86 °C
B. 3.72 °C
C. 0.93 °C
D. 2.0 °C

Solution

Freezing point depression (ΔTf) = i * Kf * m = 1 * 1.86 °C kg/mol * 1 mol/kg = 1.86 °C.

Correct Answer: A — 1.86 °C

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Q. If the freezing point of a solution is -5°C, what is the molality of the solution if the freezing point depression constant (Kf) is 1.86°C kg/mol? (2022)

A. 1.34 m
B. 2.68 m
C. 0.5 m
D. 3.72 m

Solution

ΔTf = Kf * m; -5 = -1.86 * m; m = 5 / 1.86 = 2.68 m.

Correct Answer: B — 2.68 m

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Q. If the freezing point of pure water is 0 °C, what will be the freezing point of a solution containing 1 mole of NaCl in 1 kg of water? (2023)

A. -1.86 °C
B. -3.72 °C
C. -2.26 °C
D. -0.93 °C

Solution

Freezing point depression = i * Kf * m = 2 * 1.86 °C kg/mol * 1 mol/kg = 3.72 °C. Thus, freezing point = 0 - 3.72 = -3.72 °C.

Correct Answer: B — -3.72 °C

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Q. If the freezing point of pure water is 0°C, what will be the freezing point of a 1 molal solution of NaCl? (2022)

A. -1.86°C
B. -3.72°C
C. -2.52°C
D. -0.52°C

Solution

Freezing point depression = i * Kf * m = 2 * 1.86 * 1 = 3.72°C. Therefore, freezing point = 0 - 3.72 = -1.86°C.

Correct Answer: A — -1.86°C

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Q. What is the boiling point elevation for a solution with 0.5 moles of a non-volatile solute in 1 kg of water? (Kb for water = 0.512 °C kg/mol) (2022)

A. 0.256 °C
B. 0.512 °C
C. 1.024 °C
D. 0.128 °C

Solution

Boiling point elevation (ΔTb) = Kb * m = 0.512 °C kg/mol * 0.5 mol/kg = 0.256 °C.

Correct Answer: A — 0.256 °C

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Q. What is the effect of adding a non-volatile solute to a solvent on its boiling point? (2019)

A. Increases boiling point
B. Decreases boiling point
C. No effect
D. Depends on the solute

Solution

Adding a non-volatile solute increases the boiling point due to boiling point elevation.

Correct Answer: A — Increases boiling point

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Q. What is the effect of adding a non-volatile solute to a solvent on its vapor pressure? (2023)

A. Increases vapor pressure
B. Decreases vapor pressure
C. No effect
D. Depends on the solute

Solution

Adding a non-volatile solute decreases the vapor pressure of the solvent due to the solute particles occupying space at the surface.

Correct Answer: B — Decreases vapor pressure

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Q. What is the freezing point depression of a solution containing 1 mole of NaCl in 1 kg of water? (2022)

A. 1.86 °C
B. 3.72 °C
C. 2.52 °C
D. 0.93 °C

Solution

Freezing point depression = i * Kf * m = 2 * 1.86 * 1 = 3.72 °C (i = 2 for NaCl).

Correct Answer: A — 1.86 °C

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Q. What is the freezing point depression of a solution containing 3 moles of KCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol) (2022)

A. 5.58 °C
B. 3.72 °C
C. 1.86 °C
D. 7.44 °C

Solution

Freezing point depression = i * Kf * m = 3 * 1.86 * 3 = 5.58 °C (i = 2 for KCl).

Correct Answer: A — 5.58 °C

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Q. What is the osmotic pressure of a 0.5 M NaCl solution at 25°C? (R = 0.0821 L·atm/(K·mol)) (2023)

A. 12.3 atm
B. 6.15 atm
C. 3.07 atm
D. 1.54 atm

Solution

Osmotic pressure (π) = iCRT = 2 * 0.5 * 0.0821 * 298 = 24.5 atm. Therefore, π = 6.15 atm.

Correct Answer: B — 6.15 atm

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Q. What is the osmotic pressure of a solution containing 0.1 moles of glucose in 1 L of solution at 25 °C? (R = 0.0821 L·atm/(K·mol)) (2020)

A. 2.03 atm
B. 1.96 atm
C. 0.82 atm
D. 0.25 atm

Solution

Osmotic pressure (π) = nRT/V = (0.1 mol)(0.0821 L·atm/(K·mol))(298 K) = 2.03 atm.

Correct Answer: A — 2.03 atm

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Q. What is the osmotic pressure of a solution containing 0.1 moles of glucose in 1 liter of solution at 25 °C? (R = 0.0821 L atm/K mol) (2020)

A. 2.06 atm
B. 0.82 atm
C. 1.01 atm
D. 0.25 atm

Solution

Osmotic pressure (π) = nRT/V = (0.1 mol)(0.0821 L atm/K mol)(298 K) = 2.06 atm.

Correct Answer: A — 2.06 atm

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Q. What is the osmotic pressure of a solution containing 0.1 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol)) (2020)

A. 2.5 atm
B. 0.82 atm
C. 1.0 atm
D. 0.25 atm

Solution

Osmotic pressure (π) = nRT/V = (0.1 moles * 0.0821 * 298) / 1 = 2.5 atm.

Correct Answer: A — 2.5 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of solute in 1 liter of solution at 25 °C? (2021)

A. 12.3 atm
B. 24.6 atm
C. 6.1 atm
D. 3.1 atm

Solution

Osmotic pressure (π) = nRT/V = (0.5 moles)(0.0821 L·atm/(K·mol))(298 K) = 12.3 atm.

Correct Answer: B — 24.6 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol)) (2021)

A. 12.3 atm
B. 10.2 atm
C. 8.2 atm
D. 6.1 atm

Solution

Osmotic pressure (π) = nRT/V = (0.5 moles * 0.0821 L·atm/(K·mol) * 298 K) / 1 L = 12.3 atm.

Correct Answer: B — 10.2 atm

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Q. What is the osmotic pressure of a solution containing 2 moles of glucose in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol)) (2020)

A. 0.82 atm
B. 1.64 atm
C. 2.46 atm
D. 4.92 atm

Solution

Osmotic pressure (π) = nRT/V = (2 moles * 0.0821 L·atm/(K·mol) * 298 K) / 1 L = 4.92 atm.

Correct Answer: B — 1.64 atm

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Q. What is the osmotic pressure of a solution containing 2 moles of glucose in 1 liter of solution at 25 °C? (2021)

A. 0.0821 atm
B. 1.0 atm
C. 2.0 atm
D. 4.0 atm

Solution

Osmotic pressure (π) = iCRT = 1 * 2 * 0.0821 * 298 = 4.0 atm (i = 1 for glucose).

Correct Answer: C — 2.0 atm

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Q. What is the relationship between the vapor pressure of a solution and the mole fraction of the solvent? (2023)

A. Directly proportional
B. Inversely proportional
C. No relationship
D. Exponential relationship

Solution

The vapor pressure of a solution is directly proportional to the mole fraction of the solvent.

Correct Answer: A — Directly proportional

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Q. What is the van 't Hoff factor (i) for a non-electrolyte solute? (2023)

A. 1
B. 2
C. 3
D. 0

Solution

The van 't Hoff factor (i) for a non-electrolyte solute is 1, as it does not dissociate into ions.

Correct Answer: A — 1

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Q. What is the van 't Hoff factor (i) for a solution of glucose (C6H12O6)? (2023)

A. 1
B. 2
C. 3
D. 4

Solution

Glucose does not dissociate in solution, so the van 't Hoff factor (i) = 1.

Correct Answer: A — 1

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Q. What is the van 't Hoff factor (i) for a strong electrolyte like Na2SO4 in solution? (2023)

A. 1
B. 2
C. 3
D. 4

Solution

Na2SO4 dissociates into 3 ions (2 Na⁺ and 1 SO4²⁻), so i = 3.

Correct Answer: C — 3

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Q. What is the vapor pressure of a solution containing 1 mole of solute in 10 moles of solvent, assuming ideal behavior? (2019)

A. 0.1 P0
B. 0.9 P0
C. 0.5 P0
D. 0.2 P0

Solution

Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (10 / 11) * P0 ≈ 0.909 P0.

Correct Answer: B — 0.9 P0

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Q. What is the vapor pressure of a solution containing 1 mole of solute in 9 moles of solvent, assuming ideal behavior? (2019)

A. 0.1 P0
B. 0.9 P0
C. 0.5 P0
D. 0.2 P0

Solution

Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (9 / 10) * P0 = 0.9 P0.

Correct Answer: B — 0.9 P0

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Q. What is the vapor pressure of a solution containing 2 moles of solute and 3 moles of solvent, assuming ideal behavior? (2022)

A. 0.4 P0
B. 0.6 P0
C. 0.2 P0
D. 0.8 P0

Solution

Vapor pressure of solution = (moles of solvent / total moles) * P0 = (3 / (2 + 3)) * P0 = 0.6 P0.

Correct Answer: B — 0.6 P0

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Q. What is the vapor pressure of a solution containing 2 moles of solute and 4 moles of solvent compared to pure solvent? (2022)

A. 1/3
B. 1/2
C. 2/3
D. 1/4

Solution

Vapor pressure lowering = (moles of solute / total moles) = 2 / (2 + 4) = 2/6 = 1/3. Thus, vapor pressure is 1/2 of pure solvent.

Correct Answer: B — 1/2

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Q. What is the vapor pressure of a solution containing 2 moles of solute and 4 moles of solvent, assuming ideal behavior? (2022)

A. 0.33 P0
B. 0.50 P0
C. 0.67 P0
D. 0.75 P0

Solution

Vapor pressure = (moles of solvent / total moles) * P0 = (4 / (2 + 4)) * P0 = (4/6) * P0 = 0.67 P0.

Correct Answer: C — 0.67 P0

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Physical Chemistry - Solutions MCQ & Objective Questions

Understanding "Physical Chemistry - Solutions" is crucial for students aiming to excel in their exams. This topic not only forms the foundation of many concepts in chemistry but also plays a significant role in scoring well in objective tests. Practicing MCQs and objective questions helps reinforce your knowledge and boosts your confidence, making it easier to tackle important questions during exam preparation.

What You Will Practise Here

Concept of solutions and their types: ideal and non-ideal solutions
Concentration terms: molarity, molality, and percentage concentration
Colligative properties: boiling point elevation, freezing point depression
Raoult's Law and its applications in calculating vapor pressure
Understanding solubility and factors affecting it
Concept of osmotic pressure and its significance
Key formulas and definitions related to solutions

Exam Relevance

The topic of "Physical Chemistry - Solutions" is frequently covered in CBSE, State Boards, NEET, and JEE exams. Students can expect questions that test their understanding of concepts, calculations involving colligative properties, and applications of Raoult's Law. Common question patterns include numerical problems, conceptual MCQs, and theoretical questions that require a clear grasp of the subject matter.

Common Mistakes Students Make

Confusing different concentration units and their applications
Misunderstanding the implications of Raoult's Law in real-world scenarios
Overlooking the significance of colligative properties in solution behavior
Neglecting to practice numerical problems, leading to calculation errors

FAQs

Question: What are the key factors that affect solubility?
Answer: Temperature, pressure, and the nature of solute and solvent are the primary factors affecting solubility.

Question: How do colligative properties differ from other properties?
Answer: Colligative properties depend on the number of solute particles in a solution, not on their identity.

Ready to enhance your understanding of "Physical Chemistry - Solutions"? Start solving practice MCQs today to test your knowledge and prepare effectively for your exams!

Physical Chemistry - Solutions

Physical Chemistry - Solutions MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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