Colligative Properties - AP Chemistry

Colligative properties are properties of solutions which depend on the number of dissolved particles in solution. The four main colligative properties are:

1. Freezing point depression: The presence of a solute lowers the freezing point of a solution as compared to that of the pure solvent.

2. Boiling point elevation: The presence of a solute increases the boiling point of a solution as compared to that of the pure solvent.

3. Vapor pressure depression: The vapor pressure of a pure solvent is greater than that of a solution containing a non-volatile liquid. The lowering of vapor pressure leads to boiling point elevation.

4. Osmotic pressure: The osmotic pressure of a solution is the pressure difference between the solution and pure solvent when the two are in equilibrium across a semipermeable membrane. Because it depends on the concentration of solute particles in solution, it is a colligative property.

Electronegativity is not a property of solutions reliant on the number of dissolved particles, but a property of atoms themselves.

This question is asking us to identify a solution that increases the boiling point of water by the greatest amount.

To answer this, we need to understand the concept of colligative properties. When a solute dissolves in a solvent such as water, various physical properties are affected. The four colligative properties that change as a result of the addition of solute are freezing point, boiling point, vapor pressure, and osmotic pressure.

With regards to boiling point, as more solute is added to a solution, the boiling point increases. This is due to the fact that addition of solute makes it more difficult for the solute molecules to gain enough kinetic energy at the solution's surface to escape as a gas.

Furthermore, the identity of the solute does not matter. Thus, we need to look only at the number of dissolved solute particles rather than their identity. A compound such as sucrose will not dissociate in solution, which means that the osmotic pressure of the solution is the same as the concentration of sucrose.

Compounds that can dissociate into two or more particles will increase the osmolarity of the solution further. In this case, will double the stated osmolarity. , on the other hand, will dissociate completely because it is a strong acid, however the protons will not contribute to the osmolarity.

is able to dissociate into three equivalents of particles in solution. Thus, its initial concentration will be tripled, which gives it the highest osmolarity of any of the choices shown and will thus increase the boiling point by the greatest amount.

The correct answer choice is the equation for boiling point elevation when solute is added to a solvent.

The equation for freezing point depression is , where is the change in temperature, is a constant related to the solvent, is molality, and is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.

We know our constant is . Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is .

The van't Hoff factor is . Sodium chloride creates only two ions when dissolved, one and one .

Using these values, we can solve for the freezing point depression.

The freezing point will be decreased by . The normal freezing point is , making the new freezing point .

First, we need to calculate the molality because that is what we use in our equation for freezing point depression. We can get that from the molarity without knowing exactly how many liters or grams we have. We just have to know what we have one mole per liter. The weight of water is one kilogram per liter, so this allows us to make this conversion.

The molality is 2m. The van't Hoff factor is 3, as we get one calcium ion and two chloride ions per molecule during dissociation.

We can now plug the values into the equation for freezing point depression.

This gives us our depression of . The normal freezing point of pure water is , which means our new freezing point is .

The important thing to remember for this question is that it doesn't matter what the solutes are in freezing point depression, just how many ions are created during dissociation. First, we need to convert our solute amounts to moles.

The molality of the solution is the moles of solute per kilogram of solvent. Since water has a density of one kilogram per liter, we can simply divide the total moles by the liters of solution. The molality of the solution is .

We can treat this solution as 3L of water containing 4mol of a solute that dissolves into a total of 5 ions. It does not matter which compound the ions come from, only that they end up in solution in the correct proportion. 2mol of calcium chloride will contribute 3 ions per molecule and 2mol of sodium chloride will contribute 2 ions per molecule, for a total of 5 ions per mole of solution.

Using these conclusions, we can solve the freezing point depression.

Our depression is then . Since the freezing point of pure water is , our new temperature for freezing point is .

A nonvolatile solute will not contribute to the vapor pressure of a solution, and will only act to decrease the vapor pressure of the pure solvent. The molar fraction of the solvent in the solution can be determined using Raoult's law.

The solution's vapor pressure is equal to the vapor pressure of the pure solvent multiplied by the molar fraction of solvent in solution.

This means that the molar fraction of solvent in the solution is 0.76. As a result, we conclude that the molar fraction of solute in the solution is 0.24, since the sum of the mole fractions must equal 1.

Since the solute is volatile, it will contribute to the total vapor pressure of the solution. As a result, we must incorporate it when solving for the total vapor pressure.

We can find the vapor pressures using Raoult's law.

Since we are solving for the molar fraction of the solvent, we will designate its molar fraction as . The sum of the molar fractions of each component must be equal to 1; thus, the molar fraction of the solute must be . Using these variables and the information given, we can solve for the molar fraction of the solvent.

Since the glucose is nonvolatile, we can use the boiling point elevation equation to solve for the new boiling point.

Since glucose does not ionize in water, the van't Hoff factor is simply 1 for this problem. The molality can be found by the moles of solute per kilogram of solvent.

This means that the boiling point for the water will be elevated by 1.03oC with the addition of the glucose. Since pure water has a boiling point of 100oC, the boiling point for this solution is 101.03oC.

In order to solve for the molar mass of the unknown compound, we need to use the freezing point depression equation.

Since molality is equal to the moles of solute divided by the kilograms of solvent, we can substitute the moles of solute with the mass of the solute divided by the molar mass of the solute.

This allows us to solve for the molar mass of the compound using a substituted equation.

We can use the values given in the question to solve for the molar mass.

The equation for boiling point elevation is .

Since all of the solutions are aqueous, we do not need to consider the boiling point elevation constant () when comparing the solutions. The two factors we need to consider are molality () and the van't Hoff factor () of the solute.

Glucose will not ionize in solution, sodium chloride will make two ions in solution, and magnesium chloride will make three ions in solution.

When multiplying the molality by the van't Hoff factor, we can determine that the magnesium chloride solution will elevate the boiling point by the highest number.

When a solid is dissolved into a liquid, in our case water (snow), the freezing point will decrease. This phenomenon is called freezing point depression. In order to freeze, the molecules of the liquid must organize into a relatively static lattice. This highly organized structure is disrupted by the presence of extra molecules and ions that are dissolved in the liquid, making it harder to transition to a solid.

With enough salt in the snow, the freezing point will depress below the current temperature, leading the snow to melt and be in the liquid phase. The salt ions will infiltrate the ice/snow lattice, breaking apart the solid and cause the water to remain liquid.

Dissolving solutes into a solvent will influence the movement of the molecules in solution. In a pure solvent, the molecules are essentially in uniform motion. When an impurity is added, such as salt in seawater, the foreign particles begin to interact with the solvent. Different masses, velocities, and attractive forces alter the patterns of the liquid molecules and cause an overall decrease in their energy. Transitioning to a gaseous state from a liquid state requires a reduction of intermolecular forces, while adding solute increases these forces in the solution. Adding solute thus inhibits the transition from liquid to gas, resulting in boiling point elevation.

Filtering the water to remove some solute could help to distill the water by removing some of the boiling point elevation. Moving to the lower elevation or increasing the pressure will increasing the boiling point.

A nonvolatile solute will not contribute to the vapor pressure of the solution, but will decrease the vapor pressure of the pure solvent. We can solve for the vapor pressure of the solution by using Raoult's law:

In other words, the new vapor pressure is equal to the molar fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Since both of the solvents have a vapor pressure, we can find the percentage of ethanol in the solution by using Raoult's law, including both solvents in the equation:

Since we want to find the percentage of ethanol in the solution, we will designate the molar fraction of ethanol as . Since the sum of the molar fractions equals one, the molar fraction of hexane will be designated as .

So, 63.4% of the solution is ethanol.

Since a salt has been added to the pure water, we can find the new melting point of the solution by using the freezing point depression equation:

The change in temperature is equal to the freezing point constant for the solvent multiplied by the molality of the solution multiplied by the van't Hoff factor. The van't Hoff factor is the number of ions that a salt will dissociate into when in solution.

For this particular salt, the van't Hoff factor is three. The molality will be equal to the moles of solute over the mass of the solvent.

Using these terms together in the original equation, we can find the freezing point depression.

This is the change in temperature from the regular freezing point. Since the freezing point for pure water is , the new melting point is .

For reference, the freeing point depression equation is:

Since we want to triple the melting point change, we need to manipulate the freezing point depression equation so that the value of is three times as large.

Replacing glucose with magnesium chloride will make the van't Hoff factor three, so the melting point change will triple.

Reducing the amount of water by a factor of three will also triple the melting point change, as will tripling the glucose amount.

Six moles of sodium chloride will triple the solute concentration and make the van't Hoff factor two. This multiplies the melting point change by six.

The correct answer choice is the equation for boiling point elevation when solute is added to a solvent.

The equation for freezing point depression is , where is the change in temperature, is a constant related to the solvent, is molality, and is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.

We know our constant is . Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is .

The van't Hoff factor is . Sodium chloride creates only two ions when dissolved, one and one .

Using these values, we can solve for the freezing point depression.

The freezing point will be decreased by . The normal freezing point is , making the new freezing point .