MCAT Physical : Solution Chemistry
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Example Question #51 : Solution Chemistry
What is the freezing point of a 
Freezing-point depression is a colligative property and is calculated with the formula:
The van't Hoff factor, 
Using the equation, we can calculate the freezing point depression using the values from the question.
The standard freezing point of water is 
Example Question #3 : Freezing Point
What is the freezing point of a 
Freezing-point depression is a colligative property and is calculated with the formula:
The van't Hoff factor, 
Using the equation, we can calculate the freezing point depression using the values from the question.
The standard freezing point of water is 
Example Question #4 : 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?
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 #1 : 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?
It will become more negative
Cannot be determined
It will become less negative
It will not change
It will approach a horizontal asymptote
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.
kf 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 kf 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 #3 : 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?
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 #2 : Freezing Point
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 
increases . . . increases
increases . . . decreases
decreases . . . increases
decreases . . . decreases
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.
Example Question #51 : Solution Chemistry
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).
Consider the following two solutions:
Solution A:
Solution B:
Which of the following is true regarding these two solutions?
Solution A has 2 times higher freezing point
Solution B has 2 times higher freezing point
Solution B has 3 times higher freezing point
Solution A has 3 times higher freezing point
Solution A has 3 times higher freezing point
The freezing point depression can be calculated by the following equation.
where 
Example Question #1 : Vapor Pressure
Boiling point is the temperature a liquid needs to achieve in order to begin its transformation into a gaseous state. Campers and hikers who prepare food during their trips have to account for differences in atmospheric pressure as they ascend in elevation. During the ascent, the decrease in atmospheric pressure changes the temperature at which water boils.
Further complicating the matter is the observation that addition of a solute to a pure liquid also changes the boiling point. Raoult’s Law can be used to understand the changes in boiling point if a non-volatile solute is present, as expressed here.
In this law, 
If a non-volatile solute is added to a solvent, the freezing point of the solution tends to __________ relative to the pure solvent's freezing point.
decrease
be elevated only at low pressures
increase
either decrease or increase
remain unchanged
decrease
Freezing points are decreased, or depressed, with the addition of non-volatile solutes in a similar manner to boiling point elevation. The addition of a solute makes phase changes more difficult, and thus solutions with non-volatile solutes require more heat to boil, or a colder environment to freeze. The solute ions and particles in solution disrupt the forces between solvent molecules, preventing the formation of a solid frozen lattice.
Example Question #2 : Vapor Pressure
The addition of a nonvolatile solute to a solvent will raise its boiling point. This is because __________.
the solute increases the vapor pressure
the solute causes more heat to be needed in order for the solution to have the same vapor pressure as the atmospheric pressure
the solute increases the heat capacity of the solvent
the solute reacts with the solvent and creates a product more resistant to temperature change
the solute causes more heat to be needed in order for the solution to have the same vapor pressure as the atmospheric pressure
Remember that a nonvolatile solute will lower the vapor pressure of a solvent in proportion to the mole fraction of the solvent (Raoult's Law). Since vapor pressure must equal the atmospheric pressure in order to boil, a greater amount of heat is required to increase the lowered vapor pressure of the solution.
Example Question #51 : Solution Chemistry
A nonvolatile solute is added to a solution so that it makes up 5% of the molecules in the solution. Which of the following is true?
The solution's vapor pressure is less than 95% of the pure solvent's vapor pressure
The vapor pressure of the solution is greater than the pure solvent's vapor pressure
The solution's vapor pressure is somewhere between 95% and 100% of the pure solvent's vapor pressure
The vapor pressure is unaffected by the addition of the solute
The solution has 95% of the vapor pressure of the pure solvent
The solution has 95% of the vapor pressure of the pure solvent
When a nonvolatile solute is added to a solvent, it will not contribute to the molecules which exert pressure on the container. It will, however, take up some of the surface area interacting with the air in the container. This reduces the number of solvent molecules that are able to break from their bonds and become gas molecules in the container. The reduction of vapor pressure is dependent on the percentage of solute molecules in the solution. Since 5% of the molecules in this solution come from the solute, the vapor pressure will be 95% of the pure solvent's vapor pressure.
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