1-7 2026 Complete Practice Examination - 200
Verified Questions Aligned to UNE General
Chemistry II Course Objectives
Published by: Chemistry Exam Preparation Resources
Edition: 2026
Total Questions: 200
Domains Covered: Solutions and Colligative Properties, Chemical Kinetics,
Chemical Equilibrium, Acid-Base Equilibrium, Buffers and Titrations, Solubility
Equilibrium, Thermodynamics and Electrochemistry
1. A solution contains 5.00 g of glucose (C₆H₁₂O₆, molar mass 180.16 g/mol) in
95.0 g of water. What is the molality of the solution?
A. 0.028 m
B. 0.292 m
C. 0.526 m
D. 0.052 m
Correct Answer: B
Rationale: Molality = moles of solute / kg of solvent. Moles glucose = 5.00 g /
180.16 g/mol = 0.0278 mol. Mass water = 95.0 g = 0.0950 kg. Molality = 0.0278
mol / 0.0950 kg = 0.292 m.
2. Which of the following concentration units is temperature dependent?
A. Molality
B. Molarity
C. Mole fraction
D. Mass percent
Correct Answer: B
,Rationale: Molarity (moles solute / liters solution) is temperature dependent
because volume changes with temperature. Molality, mole fraction, and mass
percent are temperature independent.
3. What is the mole fraction of ethanol in a solution containing 46.0 g of
ethanol (C₂H₅OH, molar mass 46.0 g/mol) and 72.0 g of water (molar mass
18.0 g/mol)?
A. 0.500
B. 0.200
C. 0.800
D. 0.333
Correct Answer: B
Rationale: Moles ethanol = 46.0 g / 46.0 g/mol = 1.00 mol. Moles water = 72.0 g /
18.0 g/mol = 4.00 mol. Total moles = 5.00 mol. Mole fraction ethanol = 1..00
= 0.200.
4. According to Raoult's Law, the vapor pressure of a solution is equal to:
A. The vapor pressure of the pure solvent
B. The mole fraction of the solvent multiplied by the vapor pressure of the
pure solvent
C. The mole fraction of the solute multiplied by the vapor pressure of the pure
solvent
D. The vapor pressure of the solute
Correct Answer: B
Rationale: Raoult's Law states that P_solution = X_solvent × P°_solvent, where
X_solvent is the mole fraction of the solvent and P°_solvent is the vapor pressure
of the pure solvent.
5. A nonvolatile solute is dissolved in a solvent. Which of the following
statements is true?
,A. The vapor pressure of the solution is greater than that of the pure solvent
B. The vapor pressure of the solution is lower than that of the pure solvent
C. The boiling point of the solution is lower than that of the pure solvent
D. The freezing point of the solution is higher than that of the pure solvent
Correct Answer: B
Rationale: A nonvolatile solute lowers the vapor pressure of the solvent
(colligative property). This results in boiling point elevation and freezing point
depression.
6. What is the boiling point of a 0.50 m aqueous glucose solution? (Kb for
water = 0.512 °C/m, boiling point of pure water = 100.00 °C)
A. 100.00 °C
B. 100.51 °C
C. 100.26 °C
D. 101.02 °C
Correct Answer: C
Rationale: ΔTb = Kb × m = 0.512 °C/m × 0.50 m = 0.256 °C. Boiling point =
100.00 + 0.256 = 100.26 °C.
7. What is the freezing point of a 0.50 m aqueous NaCl solution? (Kf for water
= 1.86 °C/m, i for NaCl = 2)
A. -0.93 °C
B. -0.47 °C
C. -1.86 °C
D. -3.72 °C
Correct Answer: C
Rationale: ΔTf = i × Kf × m = 2 × 1.86 °C/m × 0.50 m = 1.86 °C. Freezing point =
0.00 - 1.86 = -1.86 °C.
8. Osmotic pressure (π) is calculated using which equation?
, A. π = MRT
B. π = iMRT
C. π = MRT
D. π = iMRT
Correct Answer: D
Rationale: Osmotic pressure is given by π = iMRT, where i is the van't Hoff
factor, M is molarity, R is the gas constant, and T is temperature in Kelvin.
9. A solution contains 2.0 mol of sucrose in 1000 g of water. What is the
molality of the solution?
A. 0.20 m
B. 2.00 m
C. 0.50 m
D. 5.00 m
Correct Answer: B
Rationale: Molality = moles solute / kg solvent = 2.0 mol / 1.000 kg = 2.00 m.
10. Which of the following aqueous solutions will have the lowest freezing
point?
A. 0.10 m glucose (non-electrolyte)
B. 0.10 m NaCl
C. 0.10 m CaCl₂ (i = 3)
D. All will have the same freezing point
Correct Answer: C
Rationale: Freezing point depression depends on the total particle concentration (i
× m). CaCl₂ (i = 3) gives 0.30 m particles; NaCl (i = 2) gives 0.20 m; glucose (i =
1) gives 0.10 m. The highest particle concentration gives the lowest freezing point.