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 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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

Showing 1 to 30 of 32 (2 Pages)

Physical Chemistry - Solutions

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?