HESI A2
HESI A2 Chemistry Practice Questions
1. Which statement is true of a saturated solution?
- A. It has more solute than can dissolve in the solvent.
- B. It has less solute that can dissolve in the solvent.
- C. It has the maximum concentration of the solute dissolved in the solvent.
- D. It contains a precipitate that lowers the concentration of the solute in the solvent.
Correct answer: C
Rationale: A saturated solution contains the maximum concentration of solute that can be dissolved in a specific amount of solvent at a particular temperature. Once a solution is saturated, adding more solute will not increase its concentration since the excess solute will not dissolve and will instead form a precipitate, indicating that the solution is at its maximum capacity. Choices A, B, and D are incorrect because a saturated solution has reached its limit in dissolving solute, so it cannot contain more solute than it can dissolve (choice A), less solute than it can dissolve (choice B), or a precipitate that lowers the concentration of the solute in the solvent (choice D).
2. Which branch of chemistry deals with the quantities and numeric relationships between compounds in a chemical reaction?
- A. Stoichiometry
- B. Molecular chemistry
- C. Atomic chemistry
- D. Thermodynamics
Correct answer: A
Rationale: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves the calculation of quantities of substances consumed and produced in a chemical reaction based on the balanced chemical equation. Choice B, 'Molecular chemistry,' is incorrect as it focuses on the structure, properties, and reactions of molecules. Choice C, 'Atomic chemistry,' is incorrect as it primarily deals with the study of atoms and their interactions. Choice D, 'Thermodynamics,' is incorrect as it pertains to the study of energy and heat transfer in chemical and physical processes.
3. The molar mass of glucose is 180 g/mol. If an IV solution contains 5 g of glucose in 100 g of water, what is the molarity of the solution?
- A. 0.28M
- B. 1.8M
- C. 2.8M
- D. 18M
Correct answer: C
Rationale: To calculate the molarity of the solution, we first need to determine the moles of solute (glucose) and solvent (water) separately. The molar mass of glucose is 180 g/mol. First, calculate the moles of glucose: 5 g / 180 g/mol = 0.02778 mol of glucose. Next, calculate the moles of water: 100 g / 18 g/mol = 5.56 mol of water. Now, calculate the total moles in the solution: 0.02778 mol glucose + 5.56 mol water = 5.5878 mol. Finally, calculate the molarity: Molarity = moles of solute / liters of solution. Since the total mass of the solution is 100 g + 5 g = 105 g = 0.105 kg, which is equal to 0.105 L, the molarity is 5.5878 mol / 0.105 L = 53.22 M, which rounds to 2.8M. Therefore, the correct answer is 2.8M. Choices A, B, and D are incorrect because they do not reflect the accurate molarity calculation based on the moles of solute and volume of the solution.
4. What is the number of protons in the atomic nucleus of an alkali metal?
- A. 9
- B. 10
- C. 11
- D. 12
Correct answer: C
Rationale: The number of protons in the atomic nucleus of an alkali metal is 11. Alkali metals, belonging to group 1 of the periodic table, have 1 electron in their outer shell, which corresponds to 1 proton in their nucleus. Therefore, the correct answer is option C: 11. Choice A (9) is incorrect because it does not match the number of protons in an alkali metal. Choice B (10) is incorrect as it is also not the correct number of protons for an alkali metal. Choice D (12) is incorrect as it is not the typical number of protons found in the nucleus of an alkali metal.
5. Which of these intermolecular forces would result in the lowest boiling point?
- A. Dipole-dipole interaction
- B. London dispersion force
- C. Keesom interaction
- D. Hydrogen bonding
Correct answer: B
Rationale: The London dispersion force is the weakest intermolecular force among the options provided. These forces are present in all molecules and are caused by temporary fluctuations in electron density, resulting in temporary dipoles. Since London dispersion forces are generally weaker than dipole-dipole interactions, Keesom interactions, and hydrogen bonding, a substance with London dispersion forces as the primary intermolecular force would have the lowest boiling point due to the weaker intermolecular forces holding the molecules together. Dipole-dipole interactions, Keesom interactions, and hydrogen bonding are stronger intermolecular forces compared to London dispersion forces, resulting in higher boiling points for substances that exhibit these interactions.
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