### All MCAT Physical Resources

## Example Questions

### Example Question #11 : Colligative Properties

The values for normal boiling and freezing points, along with and values are given below for select solvents.

Which of the following will result in the least freezing point depression when added to of water?

**Possible Answers:**

**Correct answer:**

We are looking for the least amount of freezing point depression. Freezing point depression is calculated using the equation:

Each of these solutions has a different molality, which needs to be calculated. Molality is equal to moles of solute per kilogram solvent. To find this value, convert grams to moles (using molar mass) and divide by the mass of the solvent.

Next, find the van't Hoff factor for each compound.

Finally, use the initial equation to find the *smallest* freezing point depression (we are looking for the solution with the highest freezing point).

After calculating the change in freezing point for each solution, we find that the barium hydroxide solution has the smallest depression.

### Example Question #1 : Freezing Point

The values for normal boiling and freezing points, along with and values are given below for select solvents.

What is the freezing point of a solution of glucose in benzene?

**Possible Answers:**

**Correct answer:**

First, calculate the freezing point depression with the equation:

The van't Hoff factor for glucose is 1, since it does not dissociate in solution. The freezing point depression constant is given in the table.

Next, subtract this value from the freezing point of pure benzene to find the freezing point of the final solution.

### Example Question #1 : Freezing Point

What is the freezing point of a 3m solution of K_{2}SO_{4} in water? (k_{f} = 1.9^{o}C/mol)

**Possible Answers:**

**Correct answer:**

Freezing point depression is given by the equation .

i is the Vant Hoff Factor that tells us how many particles (often ions) a solid produces when dissolved in solution. m is the molality of the solution. k_{f} is the freezing point constant.

The question gives us m as 3mol, k_{f} as 1.9, and i is found to be 3 (2K^{+} and SO_{4}^{2-} ). Whe then plug the values into the equation.

### Example Question #11 : Colligative Properties

Five solutions are made, each comprised of 1L of water and 1mol of C_{6}H_{12}O_{6}, NaCl, H_{2}SO_{4}, Al(NO_{3})_{3}, or CaCl_{2}.

Which solution will show the smallest amount of freezing point depression?

**Possible Answers:**

NaCl

CaCl_{2}

H_{2}SO_{4}

C_{6}H_{12}O_{6}

Al(NO_{3})_{3}

**Correct answer:**

C_{6}H_{12}O_{6}

Freezing point depression is a colligative property, meaning that it depends on the amount of particles present in solution. The more ions that a solute dissociates into, the larger the magnitude of freezing point depression. Here, we're looking for the smallest amount of depression, so we want to find the solute that dissociates into the fewest particles. The answer is glucose, which is not ionic and does not dissociate at all in solution.

### Example Question #1 : Freezing Point

What is the freezing point of a aqueous solution of ?

**Possible Answers:**

**Correct answer:**

Freezing-point depression is a colligative property and is calculated with the formula:

The van't Hoff factor,* *, represents the number of particles a compound will dissociate into when dissolving in a solution. is the freezing-point depression constant, which is given in the question. is molality.

Using the equation, we can calculate the freezing point depression using the values from the question.

The standard freezing point of water is . With the freezing point depression, the solution will freeze at .

### Example Question #1 : Freezing Point

What is the freezing point of a aqueous solution of ?

**Possible Answers:**

**Correct answer:**

Freezing-point depression is a colligative property and is calculated with the formula:

The van't Hoff factor,* *, represents the number of particles a compound will dissociate into when dissolving in a solution. is the freezing-point depression constant, which is given in the question. is molality.

Using the equation, we can calculate the freezing point depression using the values from the question.

The standard freezing point of water is . With the freezing point depression, the solution will freeze at

### Example Question #1 : Freezing Point

The city of New York is developing a new salt substitute to distribute on the streets during icy weather. Which solution would be the most useful for this purpose?

**Possible Answers:**

**Correct answer:**

The key to understanding this problem is realizing that it is asking about freezing point depression, or colligative properties. The main factor that concerns freezing point depression is the number of particles in solution, rather than other features, such as size or weight. The more solute dissolved, the greater the freezing point depression will be, meaning that a colder temperature will be needed for a solution to freeze.

In our answer choices we are looking for the solute that is dissociate into the greatest number of ions, or has the highest van't Hoff factor. Magnesium chloride and calcium bromide both generate three moles of ions per mole of solute. The answer choice with magnesium chloride is more concentrated, however, making this the correct option.

### Example Question #2 : Freezing Point

Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.

If a polar molecule were added to the solution, how would the slope of segment AD change?

**Possible Answers:**

It will become more negative

Cannot be determined

It will become less negative

It will approach a horizontal asymptote

It will not change

**Correct answer:**

It will become less negative

This question asks us to consider what would happen to the freezing point (also known as the melting point) of the solution if a polar molecule were added, for exampl adding NaCl to water. More directly, this question is a freezing point elevation/depression problem. After we determine what would happen to the freezing point of the solution, we can predict the change in the slope of the line.

The formula for freezing point depression is given below.

k_{f} is the cryoscopic constant that says how easily a solution freezes with a given change in temperature, m is the molality of the solution, and i is the Vant Hoff factor. (The Vant Hoff factor is the number of ion particles per individual molecule that has been added; i.e. for NaCl this is 2 because it will separate into exactly two ionic particles when dissociated).

Given that k_{f} and b will be constants for a given solution, we know that adding a molecule will decrease the freezing point because i has some positive value. Now that we know the freezing point is depressed (lowered), how will the slope of the line change?

At the same pressure, a lower temperature will be required for the solution with the added polar molecule to freeze, thus the slope will be less negative (segment AD will look more flat on the graph).

Why does the added molecule have to be polar to have an effect? First, notice that we are told that the phase diagram is of a polar solution. Next, let’s look at the i factor. i only has a value if the added molecules can be dissolved in the original solution. If, for example, we added a nonpolar molecule, it could not be dissolved by the solution and will not contribute to depressing the freezing point. “Like-dissolves-like” also applies here; the freezing point will only change if the solution and molecule added are both polar or both nonpolar.

### Example Question #1 : Freezing Point

A 2-liter aqueous solution of glucose has a freezing point of .

Given the solution's freezing point, how many moles of glucose were added to the water?

**Possible Answers:**

**Correct answer:**

Since we know that the freezing point of the water has changed by 3.5 degrees, we can solve for how many moles of glucose were added using the freezing point depression equation:

Since glucose does not ionize in solution, the vant Hoff factor is equal to 1. Keep in mind that we must use molality for the concentration.

### Example Question #21 : Colligative Properties

Colligative properties are properties of compounds that are altered by the amount of substance present. There are four main colligative properties: boiling point, freezing point, vapor pressure, and osmotic pressure. The change in each of these properties can be calculated using the amount of molecules/ions present in solution and the concentration or partial pressure of the compound. The boiling point is defined as the temperature at which the vapor pressure equals the atmospheric pressure. The freezing point is the temperature at which a liquid is converted to a solid. Vapor pressure is the pressure produced by the vapor above a solution. Osmotic pressure is the pressure required to prevent flow of water into a solution (across a membrane).

The freezing point of a solution __________ and the boiling point of the solution __________ after the addition of .

**Possible Answers:**

decreases . . . increases

decreases . . . decreases

increases . . . increases

increases . . . decreases

**Correct answer:**

decreases . . . increases

Freezing point and boiling point are both colligative properties that are altered by the addition of solutes. Addition of solutes decreases freezing point whereas addition of solutes increases boiling point. These phenomena are called freezing point depression and boiling point elevation, respectively.

The other two colligative properties are vapor pressure and osmotic pressure. Vapor pressure decreases and osmotic pressure increases after addition of solutes.

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