Liquids and Solutions - AP Chemistry

Start by writing out the the chemical formulas for the reactants:

In order to figure out the possible products, combine the cation of one molecule with the anion of the other molecule.

For this reaction, the possible products are and .

Next, use solubility rules to figure out if any precipitate is formed. Since compounds with are soluble, is soluble. Since is only soluble when paired with , is insoluble.

Thus, we can write the final chemical equation:

Recall the solubility rules:

is generally soluble, except when paired with .

Solubility rules can help us find the answer to this question.

First, solubility rules state that all compounds of Group 1A elements on the periodic table are soluble in aqueous solution. This means that all of the alkali metals, including potassium, form compounds which are soluble in aqueous solution; thus, is soluble in aqueous solution.

Solubility rules also tell us that all ammonium salts (salts of ) are soluble. This means that is soluble.

Next, solubility rules tell us that all bromide salts are soluble, except for those of , , and . Thus, is not soluble in aqueous solution. However, as lithium is not included in that list, is soluble in aqueous solution.

Remember the solubility rules for ionic solids in water:

1. Salts of group 1 (with few exceptions) and NH4+ are soluble

2. Nitrates, acetates, and perchlorates are soluble

3. Salts of silver, lead, and mercury (I) are insoluble

4. Chlorides, iodides, and bromides are soluble

5. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble. Exceptions: sulfides of group 2 cations and hydroxides of calcium, strontium, and barium are slightly soluble

6. Sulfates are soluble except for those of calcium, strontium, and barium

Following these rules, we see that MgOH is insoluble in water

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by , and fluoride ions increase by .

End: The equilibrium expression for this salt is , which will allow us to calculate the solubility based on the change in ion concentration.

The solubility for barium fluoride is .

Definition of :

for .

For :

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

Solve for the unknown variable:

Multiply times the given volume:

Multiply times the molar mass of :

In this question, we're asked to identify an answer choice that would be expected to give us a solution with the greatest osmotic pressure. Remember that osmotic pressure is proportional to the total number of dissolved solute particles in solution, regardless of the identity of those solute particles.

When looking at the answer choices, we need to keep in mind two things. First, we need to recognize the numerical value given for the concentration of the compound given. Secondly, we need to identify if the compound shown is capable of dissociating in solution to give rise to even more solute particles. This is important, as it would affect the osmotic pressure.

would be expected to have the largest osmotic pressure because, in total, this would be a solution after dissociation occurs.

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.

In this question, we're asked to identify an answer choice that would be expected to give us a solution with the greatest osmotic pressure. Remember that osmotic pressure is proportional to the total number of dissolved solute particles in solution, regardless of the identity of those solute particles.

When looking at the answer choices, we need to keep in mind two things. First, we need to recognize the numerical value given for the concentration of the compound given. Secondly, we need to identify if the compound shown is capable of dissociating in solution to give rise to even more solute particles. This is important, as it would affect the osmotic pressure.

would be expected to have the largest osmotic pressure because, in total, this would be a solution after dissociation occurs.

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.