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Example Questions
Example Question #1 : Colligative Properties
Two moles of sodium chloride (NaCl) are added to 1kg of a mystery solvent. The addition of the NaCl caused an increase of 6K to the solvent's boiling point.
Based on this information, what is the boiling constant for the solvent?
In order to solve this problem, we can use the boiling point elevation equation: .
We know the temperature change, we can compute molality from the given information, and we know the van't Hoff factor (expected to be 2 in this scenario due to NaCl becoming 2 ions in solution). We can calculate the boiling point constant for the solvent.
Example Question #11 : Boiling 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).
Which of the following is true regarding the boiling point of a sodium chloride solution and calcium chloride solution?
The boiling point of sodium chloride solution will be higher
The boiling point of calcium chloride solution will be higher
The relative boiling points cannot be determined because the concentration is not given
The boiling point of both solutions will be equal
The relative boiling points cannot be determined because the concentration is not given
Recall that the boiling point of a solution depends on the number of ions in the solution and the concentration of the solution. Increase in both of these factors will elevate the boiling point to higher temperatures. The equation describing this is given below.
where is change in boiling point, is the boiling point elevation constant, is the number of ions, and is the molality. Sodium chloride, or , will produce two ions in solution whereas calcium chloride, or , will produce three ions. If the concentrations were the same, the calcium chloride solution would have a higher boiling point; however, since we are not given the concentration we cannot determine the relative boiling points of the solution.
Example Question #42 : 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).
How does the boiling point of a solution change after adding potassium bromide? The solution has a density of and .
Increases by
Decreases by
Increases by
Decreases by
Increases by
Recall that adding solutes to a solution increases the boiling point. This phenomenon is called the boiling point elevation. Knowing this information, we can eliminate two choices immediately. The equation to calculate the boiling point elevation is as follows.
where is change in boiling point, is the boiling point elevation constant, is the number of ions, and is the molality. The question gives us the concentration in molarity; therefore, we need to convert the concentration to molality.
Molarity =
Molality =
We need to find the mass of solvent in kilograms. The molarity is ; therefore, let’s assume we have of solute and of solution. The density of the solution is or . We can calculate the mass of the total solution using density.
mass of total solution =
We need the mass of the solvent only. To find this, we need to first calculate the mass of solute. The MW of potassium bromide is (this can be calculated by obtaining values from the periodic table). We have of solute; therefore, the mass of solute is
This means that the mass of solvent is equal to:
mass of solvent = total mass of solution - mass of solute = =
Molality of solution is
, or the number of ions, for is 2 (because will produce two ions in solution). Now we have all the information to calculate the boiling point elevation.
Therefore, boiling point increases by .
Example Question #1 : Freezing Point
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?
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?
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 K2SO4 in water? (kf = 1.9oC/mol)
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. kf is the freezing point constant.
The question gives us m as 3mol, kf as 1.9, and i is found to be 3 (2K+ and SO42- ). Whe then plug the values into the equation.
Example Question #1 : Freezing Point
Five solutions are made, each comprised of 1L of water and 1mol of C6H12O6, NaCl, H2SO4, Al(NO3)3, or CaCl2.
Which solution will show the smallest amount of freezing point depression?
C6H12O6
H2SO4
CaCl2
Al(NO3)3
NaCl
C6H12O6
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 #2 : Freezing Point
What is the freezing point of a aqueous solution of ?
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 #3 : Freezing Point
What is the freezing point of a aqueous solution of ?
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 #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.
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