Nursing Elites

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 is the correct name of ZnSOâ‚„?

• A. Zinc sulfate
• B. Zinc sulfide
• C. Zinc sulfur
• D. Zinc oxide

Correct answer: A

Rationale: The correct name of ZnSO₄ is zinc sulfate. In this compound, zinc is combined with the polyatomic ion sulfate (SO₄). Sulfate is a common anion formed from sulfur and oxygen atoms. Therefore, the correct name for ZnSO₄ is zinc sulfate. Choice B, Zinc sulfide, is incorrect because sulfide is a different anion (S²⁻) compared to sulfate (SO₄²⁻). Choice C, Zinc sulfur, is incorrect as it does not represent the correct anion in the compound. Choice D, Zinc oxide, is incorrect as it involves an oxygen anion, not sulfate.

3. How much concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution?

• A. 75 mL
• B. 100 mL
• C. 125 mL
• D. 150 mL

Correct answer: B

Rationale: To prepare a 0.100 M HCl solution with a volume of 500 mL, you can use the formula C1V1 = C2V2, where C1 is the concentration of the concentrated HCl solution, V1 is the volume of concentrated HCl solution used, C2 is the desired concentration (0.100 M), and V2 is the final volume (500 mL). Rearranging the formula to solve for V1, you get V1 = (C2V2) / C1. Plugging in the values (0.100 M)(500 mL) / C1 = 100 mL, which means 100 mL of concentrated HCl should be used to prepare 500 mL of a 0.100 M HCl solution. Therefore, the correct answer is 100 mL. Choice A (75 mL), Choice C (125 mL), and Choice D (150 mL) are incorrect as they do not match the calculated volume needed to prepare the desired concentration of HCl solution.

4. To the nearest whole number, what is the mass of one mole of hydrogen iodide?

• A. 2 g/mol
• B. 58 g/mol
• C. 87 g/mol
• D. 128 g/mol

Correct answer: C

Rationale: The molar mass of hydrogen iodide (HI) is the sum of the atomic masses of its constituent elements. Hydrogen (H) has a molar mass of approximately 1 g/mol, and iodine (I) has a molar mass of about 127 g/mol. Thus, the molar mass of hydrogen iodide (HI) is approximately 1 + 127 = 128 g/mol. Rounding to the nearest whole number, the molar mass of hydrogen iodide is 128 g/mol, which is closest to choice C. Choice A (2 g/mol) is too low and does not reflect the correct molar mass of hydrogen iodide. Choice B (58 g/mol) is significantly lower than the actual molar mass. Choice D (128 g/mol) matches the calculated molar mass but is not the nearest whole number as requested.

5. What is the primary function of enzymes?

• A. To provide energy for reactions
• B. To speed up reactions
• C. To decrease activation energy
• D. To act as a catalyst

Correct answer: B

Rationale: Enzymes function to speed up reactions by lowering the activation energy required for the reaction to occur. They act as biological catalysts, providing an alternative pathway for the reaction to proceed more rapidly without being consumed in the process. Choices A, C, and D are incorrect because enzymes do not provide energy for reactions (they do not generate energy), their primary function is not to decrease activation energy (though they do lower it), and while they act as catalysts, the primary function is to speed up reactions by lowering activation energy.

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