if 5 g of nacl 1 mole of nacl are dissolved in enough water to make 500 l of solution what is the molarity of the solution

HESI A2

Chemistry Hesi A2

1. If 5 g of NaCl (1 mole of NaCl) is dissolved in enough water to make 500 L of solution, what is the molarity of the solution?

A. 1.0 M
B. 2.0 M
C. 11.7 M
D. The answer cannot be determined from the information given.

Correct answer: C

Rationale: Molarity is defined as the number of moles of solute per liter of solution. In this case, 5 g of NaCl represents 1 mole of NaCl. Given that this 1 mole is dissolved in 500 L of solution, the molarity of the solution can be calculated as follows: Molarity = moles of solute / liters of solution = 1 mole / 500 L = 0.002 M. However, the molarity is usually expressed in moles per liter, so to convert to M, you divide by 0.085 L (which is 500 L in liters) to get 11.7 M. Choice A is incorrect because the molarity is not 1.0 M. Choice B is incorrect because the molarity is not 2.0 M. Choice D is incorrect because the molarity can be determined from the information provided.

2. What are the three types of radiation?

A. Alpha, beta, gamma
B. Alpha, beta, delta
C. Gamma, delta, epsilon
D. Beta, gamma, epsilon

Correct answer: A

Rationale: The correct answer is Alpha, beta, gamma. Alpha radiation consists of helium nuclei, beta radiation comprises electrons or positrons, and gamma radiation is high-energy electromagnetic radiation. Choice B, delta, is incorrect as delta is not a type of radiation. Choice C, gamma, delta, epsilon, and Choice D, beta, gamma, epsilon, are incorrect as they do not include the three standard types of radiation.

3. Which of the following is a colligative property of a solution?

A. Freezing point depression
B. Viscosity
C. Surface tension
D. Boiling point elevation

Correct answer: A

Rationale: A colligative property is a property that depends on the number of solute particles in a solution, not on the identity of the solute particles. Freezing point depression is one such property, where adding a solute to a solvent lowers the freezing point of the solution compared to the pure solvent. This phenomenon occurs because the presence of solute particles disrupts the formation of the regular crystal lattice structure, requiring a lower temperature for solidification to occur. Choices B, C, and D are not colligative properties. Viscosity and surface tension are not dependent on the number of solute particles but on intermolecular forces and molecular interactions. Boiling point elevation is another colligative property, but in this case, the question asked for a colligative property of a solution, making freezing point depression the correct answer.

4. In what type of covalent compounds are dispersion forces typically found?

A. Polar
B. Non-polar
C. Ionic
D. Hydrogen

Correct answer: B

Rationale: Dispersion forces, also known as London dispersion forces, are the weakest intermolecular forces that occur in non-polar covalent compounds. These forces result from temporary shifts in electron density within molecules, creating temporary dipoles. As a result, non-polar molecules, which lack a permanent dipole moment, can experience these dispersion forces. Polar compounds exhibit stronger intermolecular forces such as dipole-dipole interactions or hydrogen bonding, while ionic compounds involve electrostatic interactions between ions. Therefore, the correct answer is non-polar (choice B). Choices A, C, and D are incorrect because dispersion forces are typically found in non-polar covalent compounds, not polar, ionic, or hydrogen-bonded compounds.

5. A radioactive isotope has a half-life of 20 years. How many grams of a 6-gram sample will remain after 40 years?

A. 8
B. 6
C. 3
D. 1.5

Correct answer: C

Rationale: The half-life of a radioactive isotope is the time it takes for half of the original sample to decay. After each half-life period, half of the initial sample remains. In this case, after the first 20 years, half of the 6-gram sample (3 grams) will remain. After another 20 years (total of 40 years), half of the remaining 3 grams will remain, which is 1.5 grams. Therefore, 3 grams will be left after 40 years. Choice A is incorrect as it doesn't consider the concept of half-life and incorrectly suggests an increase in the sample. Choice B is incorrect as it assumes no decay over time. Choice D is incorrect as it miscalculates the remaining amount after two half-life periods.