Nursing Elites

1. If gas A has four times the molar mass of gas B, you would expect it to diffuse through a plug ___________.

• A. at half the rate of gas B
• B. at twice the rate of gas B
• C. at a quarter the rate of gas B
• D. at four times the rate of gas B

Correct answer: A

Rationale: When comparing the diffusion rates of two gases, according to Graham's law of diffusion, the rate of diffusion is inversely proportional to the square root of the molar mass. If gas A has four times the molar mass of gas B, the square root of the molar masses ratio (4:1) is 2. This means that gas A would diffuse through a plug at half the rate of gas B. Therefore, the correct answer is A, at half the rate of gas B. Choices B, C, and D are incorrect because they do not reflect the correct relationship between the molar masses and the rates of diffusion according to Graham's law.

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. 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. Which of these intermolecular forces might represent attraction between atoms of a noble gas?

• A. Dipole-dipole interaction
• B. London dispersion force
• C. Keesom interaction
• D. Hydrogen bonding

Correct answer: B

Rationale: Noble gases are non-polar molecules without a permanent dipole moment. The only intermolecular force applicable to noble gases is the London dispersion force, also known as Van der Waals forces. This force is a temporary attractive force resulting from the formation of temporary dipoles in non-polar molecules. Dipole-dipole interactions, Keesom interactions, and hydrogen bonding involve significant dipoles or hydrogen atoms bonded to electronegative atoms, which do not apply to noble gases.

5. What type of intermolecular force is a dipole attraction?

• A. Strong
• B. Weak
• C. Medium
• D. Very strong

Correct answer: B

Rationale: A dipole attraction is considered a weak intermolecular force. It occurs between molecules with permanent dipoles, where the positive end of one molecule is attracted to the negative end of another molecule. While dipole-dipole interactions are stronger than dispersion forces, they are weaker than hydrogen bonding or ion-dipole interactions. Therefore, the correct answer is 'Weak.' Choices A, C, and D are incorrect because dipole attractions are not classified as strong, medium, or very strong intermolecular forces, but rather fall into the category of weak intermolecular forces.

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