Solutions, States of Matter, and Thermochemistry - AP Chemistry
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How much faster/slower the rate of effusion for oxygen gas compared to hydrogen gas?
How much faster/slower the rate of effusion for oxygen gas compared to hydrogen gas?
Rate of effusion:
and 
must be used because they exist as bimolecular molecules. The correct answer is that Oxygen gas will effuse 4 times slower than hydrogen gas.
Rate of effusion:
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Molecule A has twice the mass of molecule B. A sample of each molecule is released into separate, identical containers. Which compound will have a higher rate of diffusion?
Molecule A has twice the mass of molecule B. A sample of each molecule is released into separate, identical containers. Which compound will have a higher rate of diffusion?
According to Graham's law, the rate of diffusion of a gas molecule is inversely proportional to the root square of that molecule's mass. Because molecule B has a smaller mass than molecule A, it will have a higher rate of diffusion.
Note: this also applies to finding the rate of effusion.
According to Graham's law, the rate of diffusion of a gas molecule is inversely proportional to the root square of that molecule's mass. Because molecule B has a smaller mass than molecule A, it will have a higher rate of diffusion.
Note: this also applies to finding the rate of effusion.
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A 20cm tube holds two cotton balls, one in each end. The left cotton ball is saturated with undiluted HCl. The right cotton ball is soaked in an undiluted mystery compound. Vapors from the two cotton balls are allowed to mix within the tube.
Let us assume that the two compounds form a precipitate in the tube 6cm to the left of the right cotton ball. What is the molar mass of the mystery compound?
A 20cm tube holds two cotton balls, one in each end. The left cotton ball is saturated with undiluted HCl. The right cotton ball is soaked in an undiluted mystery compound. Vapors from the two cotton balls are allowed to mix within the tube.
Let us assume that the two compounds form a precipitate in the tube 6cm to the left of the right cotton ball. What is the molar mass of the mystery compound?
This question is notably difficult, as it may not be immediately apparent what concept is being tested. As the vapors of the compounds mix and react, we are able to establish the distance the each vapor has traveled from the cotton ball into the tube in the given amount of time. The tube is 20cm, and the reaction takes place 6cm from the mystery compound cotton ball. From this, we can establish that in an equal amount of time the HCl vapor traveled 14cm and the mystery compound traveled 6cm.
In order to solve this problem, we use Graham's law to compare molar masses to the rates of diffusion of the two gases.
Since HCl moved 14cm to the right before interacting with the mystery compound, we know that the mystery compound moved only 6cm to the left. As a result, the diffusion ratio is 2.33.
Now, we need to find the square root of the inversed molar masses, which equals this diffusion ratio.
So, the molar mass of the mystery compound is 198 grams per mole. This makes sense, because larger gases will move more slowly compared to lighter gases.
This question is notably difficult, as it may not be immediately apparent what concept is being tested. As the vapors of the compounds mix and react, we are able to establish the distance the each vapor has traveled from the cotton ball into the tube in the given amount of time. The tube is 20cm, and the reaction takes place 6cm from the mystery compound cotton ball. From this, we can establish that in an equal amount of time the HCl vapor traveled 14cm and the mystery compound traveled 6cm.
In order to solve this problem, we use Graham's law to compare molar masses to the rates of diffusion of the two gases.
Since HCl moved 14cm to the right before interacting with the mystery compound, we know that the mystery compound moved only 6cm to the left. As a result, the diffusion ratio is 2.33.
Now, we need to find the square root of the inversed molar masses, which equals this diffusion ratio.
So, the molar mass of the mystery compound is 198 grams per mole. This makes sense, because larger gases will move more slowly compared to lighter gases.
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What is represented by point D
What is represented by point D
the critical point is where the solid and gas cannot be distinguished
the critical point is where the solid and gas cannot be distinguished
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A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.
Which of the following statements is true after the pinhole is plugged?
A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.
Which of the following statements is true after the pinhole is plugged?
The rate of effusion for two gases can be compared to one another using the following equation:
Here, the effusion rates are inversely proportional to the square root of the molecular masses of the gases in question. Because the relationship is to the square roots of the molecular masses, we will not observe a 2:1 ratio of effusion for neon compared to argon.
We will, however, see that more neon effuses out of container A compared to the amount of argon because neon is the lighter gas and will thus have a faster effusion rate. As a result, there will be more argon than neon in container A after the pinhole is plugged. This results in argon having a larger partial pressure than neon in container A.
The rate of effusion for two gases can be compared to one another using the following equation:
Here, the effusion rates are inversely proportional to the square root of the molecular masses of the gases in question. Because the relationship is to the square roots of the molecular masses, we will not observe a 2:1 ratio of effusion for neon compared to argon.
We will, however, see that more neon effuses out of container A compared to the amount of argon because neon is the lighter gas and will thus have a faster effusion rate. As a result, there will be more argon than neon in container A after the pinhole is plugged. This results in argon having a larger partial pressure than neon in container A.
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A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.
Suppose that after the pinhole is plugged, there are 100 argon atoms in container B. Approximately how many neon atoms would you predict to be in container B?
A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.
Suppose that after the pinhole is plugged, there are 100 argon atoms in container B. Approximately how many neon atoms would you predict to be in container B?
We can compare the effusion rates of these gases using the following equation.
By calling neon "gas 1" and argon "gas 2," we can compare the effusion rates of the two gases by plugging their molecular masses into the equation.
This proportion is equal to the rate of neon effusion over the rate of argon effusion, giving the ratio of neon atoms to argon atoms in container B.
As a result, 141 atoms of neon gas will effuse out of the pinhole for every 100 argon gas atoms. Keep in mind that the heavier gas will effuse at a slower rate than the lighter gas; thus, we would expect there to be more neon than argon in container B.
We can compare the effusion rates of these gases using the following equation.
By calling neon "gas 1" and argon "gas 2," we can compare the effusion rates of the two gases by plugging their molecular masses into the equation.
This proportion is equal to the rate of neon effusion over the rate of argon effusion, giving the ratio of neon atoms to argon atoms in container B.
As a result, 141 atoms of neon gas will effuse out of the pinhole for every 100 argon gas atoms. Keep in mind that the heavier gas will effuse at a slower rate than the lighter gas; thus, we would expect there to be more neon than argon in container B.
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A glass box holds equal amounts of hydrogen, nitrogen, oxygen, and bromine. The gases are allowed to exit the container through a tiny hole. Which gas will exit the hole the fastest?
A glass box holds equal amounts of hydrogen, nitrogen, oxygen, and bromine. The gases are allowed to exit the container through a tiny hole. Which gas will exit the hole the fastest?
At a particular temperature, the average kinetic energy of all gaseous molecules is equal. Since hydrogen gas has the lowest mass out of these gases, it will have the highest average velocity. This means that it will exit out of the tiny hole at a rate faster than the other gases. Conversely, bromine, which has the most mass compared to the other gases, will exit the hole the slowest.
This relationship is mathematically represented in Graham's law:
As the mass increases, the rate of effusion decreases.
At a particular temperature, the average kinetic energy of all gaseous molecules is equal. Since hydrogen gas has the lowest mass out of these gases, it will have the highest average velocity. This means that it will exit out of the tiny hole at a rate faster than the other gases. Conversely, bromine, which has the most mass compared to the other gases, will exit the hole the slowest.
This relationship is mathematically represented in Graham's law:
As the mass increases, the rate of effusion decreases.
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Which of the following gases will have the highest rate of effusion?
Which of the following gases will have the highest rate of effusion?
The rate of effusion for a gas is inversely proportional to the square-root of its molecular mass (Graham's Law).
The gas with the lowest molecular weight will effuse the fastest.
Oxygen: 
Nitrogen: 
Carbon dioxide: 
Sulfur dioxide: 
Helium: 
The lightest, and therefore fastest, gas is helium.
The rate of effusion for a gas is inversely proportional to the square-root of its molecular mass (Graham's Law).
The gas with the lowest molecular weight will effuse the fastest.
Oxygen:
Nitrogen:
Carbon dioxide:
Sulfur dioxide:
Helium:
The lightest, and therefore fastest, gas is helium.
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Which of the following gases has the highest rate of effusion?
Which of the following gases has the highest rate of effusion?
The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the gas.
The lighter a gas is, the faster it will effuse; the heavier a gas is, the slower it will effuse.
Of all the choices, helium (He) has the lowest molecular weight (atomic weight in this case), so it will have the highest rate of effusion.
The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the gas.
The lighter a gas is, the faster it will effuse; the heavier a gas is, the slower it will effuse.
Of all the choices, helium (He) has the lowest molecular weight (atomic weight in this case), so it will have the highest rate of effusion.
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Gas A has a molar mass that is 
times greater than that of Gas B. Which of these gases would be expected to effuse through a small hole faster? By how much?
Gas A has a molar mass that is
In order to answer this question, let's start by considering what effusion is and what things affect it. Effusion is the movement of a gas through a tiny hole that separates two different spaces. Because the gas particles move around in random directions with an average speed that is dependent on the temperature of the sample, lighter gas particle will move faster than heavier gas particles. This is because at a given temperature, all gas particles in a sample will have the same average kinetic energy. Consequently, we would expect gas particles with a higher molar mass to effuse more slowly than gases with a lower molar mass. This means that Gas B should effuse faster than Gas A.
The next step is to actually calculate how much greater Gas B effuses compared to Gas A. To do this, we'll need to use the following equation:
Since we know that Gas A is 
times heavier than Gas B, we can plug this into the equation to solve for the ratio of Gas B's rate of effusion to that of Gas A.
Therefore, Gas B effuses 
times faster than Gas A.
In order to answer this question, let's start by considering what effusion is and what things affect it. Effusion is the movement of a gas through a tiny hole that separates two different spaces. Because the gas particles move around in random directions with an average speed that is dependent on the temperature of the sample, lighter gas particle will move faster than heavier gas particles. This is because at a given temperature, all gas particles in a sample will have the same average kinetic energy. Consequently, we would expect gas particles with a higher molar mass to effuse more slowly than gases with a lower molar mass. This means that Gas B should effuse faster than Gas A.
The next step is to actually calculate how much greater Gas B effuses compared to Gas A. To do this, we'll need to use the following equation:
Since we know that Gas A is
Therefore, Gas B effuses
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What is the definition of the triple point?
What is the definition of the triple point?
Definition of triple point: where solid, liquid, and gas exist in equilibrium
Definition of triple point: where solid, liquid, and gas exist in equilibrium
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A solution of water is at 0.006atm and 0.01 degrees Celsius. What phase(s) are present in the sample?
A solution of water is at 0.006atm and 0.01 degrees Celsius. What phase(s) are present in the sample?
The point detailed in the question is the triple point of water on the phase diagram. At the triple point all three phases of a chemical coexist; as such the correct answer is solid, liquid, and gas.
The point detailed in the question is the triple point of water on the phase diagram. At the triple point all three phases of a chemical coexist; as such the correct answer is solid, liquid, and gas.
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Which equation accurately describes what happens to the boiling point when a solute is added to a liquid? (K = constant, M = molarity, m = molality)
Which equation accurately describes what happens to the boiling point when a solute is added to a liquid? (K = constant, M = molarity, m = molality)
The correct answer choice is the equation for boiling point elevation when solute is added to a solvent.
The correct answer choice is the equation for boiling point elevation when solute is added to a solvent.
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What is the freezing point of a solution with 
of sodium chloride in 
of water?
What is the freezing point of a solution with
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 
.
The equation for freezing point depression is
We know our constant is
The van't Hoff factor is
Using these values, we can solve for the freezing point depression.
The freezing point will be decreased by
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What is the freezing point of a 2M solution of 
in water?
What is the freezing point of a 2M solution of
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 
.
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
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What is the new freezing point of 3L of aqueous solution that contains 
of 
and 
of 
?
What is the new freezing point of 3L of aqueous solution that contains
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 
.
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
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The vapor pressure of ethanol at room temperature is 45mmHg. A nonvolatile solute is added to a vial of ethanol, resulting in a solution with a vapor pressure of 34mmHg. What is the molar fraction of the nonvolatile solute in the solution?
The vapor pressure of ethanol at room temperature is 45mmHg. A nonvolatile solute is added to a vial of ethanol, resulting in a solution with a vapor pressure of 34mmHg. What is the molar fraction of the nonvolatile solute in the solution?
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.
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.
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A volatile solute with a vapor pressure of 23mmHg is added to a solvent with a vapor pressure of 85mmHg. What is the molar fraction of the solvent if the solution has a vapor pressure of 68mmHg?
A volatile solute with a vapor pressure of 23mmHg is added to a solvent with a vapor pressure of 85mmHg. What is the molar fraction of the solvent if the solution has a vapor pressure of 68mmHg?
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 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
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What is the boiling point of a solution created when four moles of glucose are dissolved in two kilograms of water?
Assume that glucose is a nonvolatile solute.
What is the boiling point of a solution created when four moles of glucose are dissolved in two kilograms of water?
Assume that glucose is a nonvolatile solute.
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.
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.
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50g of an unknown compound are added to three kilograms of water. The compound is nonvolatile, and has a van't Hoff factor of 2. It is determined that the solution has a freezing point of 
.
What is the molar mass of the unknown compound?
50g of an unknown compound are added to three kilograms of water. The compound is nonvolatile, and has a van't Hoff factor of 2. It is determined that the solution has a freezing point of
What is the molar mass of the unknown compound?
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.
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.
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