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.  

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.

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 SO₄^2- ). Whe then plug the values into the equation.

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.

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 .

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 

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.

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.

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.

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.