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. How many moles of potassium bromide are in 25 mL of a 4 M KBr solution?

A. 0.035 mol
B. 0.1 mol
C. 0.18 mol
D. 1.6 mol

Correct answer: B

Rationale: To find the moles of potassium bromide in 25 mL of a 4 M KBr solution, we first need to convert the volume from milliliters to liters. 25 mL is equal to 0.025 L. Then, we use the formula moles = molarity x volume in liters. Substituting the values, moles = 4 M x 0.025 L = 0.1 mol. Therefore, there are 0.1 moles of KBr in 25 mL of a 4 M solution. Choice A, 0.035 mol, is incorrect as it does not properly calculate the moles. Choice C, 0.18 mol, and choice D, 1.6 mol, are also incorrect as they are not the result of the correct calculation based on the given molarity and volume.

3. A salt solution has a molarity of 5 M. How many moles of this salt are present in 0 L of this solution?

A. 0
B. 1.5
C. 2
D. 3

Correct answer: A

Rationale: Molarity is defined as the number of moles of solute per liter of solution. A molarity of 5 M indicates there are 5 moles of salt in 1 liter of the solution. Since the volume of the solution is 0 liters, multiplying the molarity by 0 liters results in 0 moles of salt (5 moles/L x 0 L = 0 moles). Therefore, the correct answer is 0. Option B, 1.5, is incorrect because it doesn't consider the volume being 0 liters. Options C and D, 2 and 3 respectively, are also incorrect as they do not account for the zero volume of the solution. Hence, there are no moles of salt present in 0 liters of the solution.

4. Why does the diffusion rate increase as a substance is heated?

A. The kinetic energy of particles increases.
B. The space between particles increases.
C. The density of particles decreases.
D. The size of particles increases.

Correct answer: A

Rationale: The correct answer is A. When a substance is heated, the kinetic energy of particles increases, causing them to move faster. This increased movement allows the particles to spread out more rapidly, leading to a higher diffusion rate. Choice B is incorrect because heating does not directly affect the space between particles. Choice C is incorrect because heating does not necessarily lead to a decrease in the density of particles. Choice D is incorrect because the size of particles does not necessarily increase when a substance is heated. Therefore, the correct explanation for the increase in diffusion rate is the rise in kinetic energy of particles.

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.