Solutions and States of Matter - AP Chemistry

In this question, we're presented with a phase diagram and are asked to determine where on the graph the rate of vaporization equals the rate of condensation.

First, it's important to realize that when the rate of vaporization and condensation are equal, we have an equilibrium of liquid and gas phases. In other words, for a given temperature and pressure, the rate at which the liquid evaporates into a gas is exactly equal to the rate at which the gas condenses into a liquid.

On a phase diagram, the area of the upper left portion of the diagram represents the solid state. The middle portion of the diagram represents the liquid state. The bottom and right most part of the diagram represents the gas phase.

Furthermore, each line on the diagram represents the specific combination of temperature and pressure in which a given compound will exist in equilibrium between two phases. The point where all three lines intersect, however, represents the triple point. This tells us the temperature and pressure in which the compound will exist in an equilibrium between all three states.

Because we are looking for the equilibrium line that represents equilibrium of vaporization and condensation, we want the line that separates the liquid portion of the diagram from the gas portion. Based on the identification of regions on the diagram discussed above, that would be line C as shown in the diagram. Line A represents equilibrium between solid and gas (sublimation rate = deposition rate). Line B represents equilibrium between solid and liquid (melting rate = freezing rate).

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.

This equation explicitly shows how the rate of effusion is inversely proportional to the molar mass of a gas in a gaseous solution.

Because , time relates to molar mass by:

Simplifying this equation, we see that:

As a result, time relates directly to molar mass

Recall Graham's Law of Effusion for two gases, A and B:

From the equation, we know the following:

Thus, we can solve for the molar mass of the unknown gas. Let be the molar mass of the unknown gas.

Make sure that your answer has significant figures.

We're being asked to compare the effusion rates of oxygen and carbon dioxide.

Remember that effusion is the spontaneous movement of a gas through a small hole from one area to another. It's worth noting that at a given temperature, the average speed of all gas molecules in a system is used to calculate the average kinetic energy of the gas particles. This dependence of kinetic energy on temperature means that at a given temperature, any gas particle will have the same kinetic energy.

In this case, we can say that the kinetic energy of oxygen molecules in one system is equal to the kinetic energy of carbon dioxide molecules in another system. Furthermore, since mass is inversely proportional to velocity, identical kinetic energies would mean that as the mass of the gas particles in a system decreases, their velocity (and thus, effusion rates) would increase.

We can use this information to solve for the relative effusion rates between oxygen and carbon dioxide. By setting their kinetic energies equal to each other, we can derive an expression that relates their relative speeds to their relative masses.

Generally speaking, this expression shows how the velocity of any two gasses depends on their mass. In this case, the gasses are oxygen and carbon dioxide.

We can use the periodic table of the elements to find out the mass of each gas, and use that information to calculate the relative effusion rates.

This shows that oxygen will effuse at a rate that is about faster than carbon dioxide.

To solve this problem use Graham's Law of Effusion

By plugging in the values we can rewrite the equation as

For this question, we're given the relative effusion rates for two gasses. We're then told how the mass of each of these gasses is changed, and then we're asked to determine the new relative effusion rates of the two gasses.

First, we can recall the expression that describes the dependence of the effusion rates of two gasses on their mass. Since we're told that the rate of gas A is twice that of gas B, we can write the following expression.

Furthermore, since we're told that the mass of gas B is doubled and the mass of gas A is halved, we can determine how the rate will change.

Thus, we can see that the rate will change by a factor of two. Hence, the new rate will be . Thus, gas A will now effuse at a rate times that of gas B.

Use P1V1 = P2V2

P1 = 10atm; V1 = 50L

P2 = X; V2 = 25L

(10atm)(50L) = (x)(25L)

500 = 25x

x = 20

Olive oil, like most oils, is non-polar, while aqueous acetic acid is very polar. These two phases do not mix because of their different solvent polarities.

use PV = nRT

V = nRT / P ; must covert T into K

V = (2)(0.0821)(273)/ 1

= 44.8 L

At high pressure and low temperature two things are happening that will cause gases to deviate from ideal behavior. At low temperature the individual gas molecules are moving slower. As the pressure is increased the individual molecules are being pushed closer to one another. Thus, when the gas molecules are closer together and moving at reduced speeds they are more likely to interact with one another. Ideal gas behavior is dependent upon an absence of intermolecular interaction between the gas molecules. Therefore, the increased likelihood of intermolecular interactions between the gas molecules at increased pressure and decreased temperature is likely to cause gases to deviate from ideal behavior.

Hydrogen gas is a diatomic gas, so the molecules that are filling the balloon exist as H2. Neon since it is a noble gas exists as a monoatomic gas, so the molecules that are filling the neon balloon exist as Ne. The volumes given allow us to calculate the amount of each gas in moles. At STP one mole of gas occupies 22.4 L, so there is one mole of hydrogen gas and there are two moles of neon gas. One mole of hydrogen gas indicates that there are actually two moles of hydrogen atoms in the balloon. Two moles of neon gas indicates that there are two moles of neon gas present in the balloon because neon exists as a monoatomic gas. Thus there are two moles of atoms in each balloon.

An ideal gas is most likely low concentration of identical molecules and low pressure. The molecules move randomly and collisions are completely elastic.

When dealing with gas problems that have multiple factors but very little data is given, it is a good idea to use the ideal gas law to compare factors.

The ideal gas law is written as PV = nRT

This allows us to see how factors could affect the pressure in the container. By lowering volume, we would see that pressure increased. If more gas was allowed into the container (increasing n), the pressure would increase as well. If we lower the temperature (lower T), then pressure should decrease as well. As a result, pressure rising due to a drop in temperature does not make sense in this question.

At STP, one mole of an ideal gas occupies 22.4L. You should know this value for the exam.

Another approach would use the ideal gas law: . Since we know that the container is at STP, we know that the container has a temperature of 273.15K and has a pressure of 1atm.

Adding the other known variables, the equation becomes

Solving for n, we find that there is 0.13 moles of hydrogen gas in the container.

Ideal gas pressure only takes the repulsive forces between gas molecules into consideration. The truth is, gas molecules can also exhibit attractive forces with one another (London dispersion forces). This attractive force pulls the molecules inward and slows their velocity before striking the wall of the container. As a result, real gases exert slightly less pressure compared to the ideal pressure.

One of the key charactersitics of an ideal gas is that gas molecules have no volume. This is obviously not the case, and the volume of the molecules must be added to the ideal volume. As a result, the real volume is slightly larger than the ideal volume.

Boiling occurs when the vapor pressure exceeds the air pressure. There is no air pressure in a vacuum, so water at any temperature will boil in a vacuum.