Solutions - AP Chemistry
Card 1 of 1212
30 mL of 1.0 M
solution is diluted with water to a volume of 3 L. What is the new concentration of the solution?
30 mL of 1.0 M solution is diluted with water to a volume of 3 L. What is the new concentration of the solution?
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Recall that
Thus, the volume of the solution increased from 30 mL to 3000 mL, which is a factor of 100. Thus, the new concentration must be 100 times less than the original concentration, and
.
Recall that Thus, the volume of the solution increased from 30 mL to 3000 mL, which is a factor of 100. Thus, the new concentration must be 100 times less than the original concentration, and
.
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Which of the following solutions would be expected to have the highest osmotic pressure?
Which of the following solutions would be expected to have the highest osmotic pressure?
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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.
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.
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Which of the following is not a colligative property?
Which of the following is not a colligative property?
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Colligative properties are properties of solutions which depend on the number of dissolved particles in solution. The four main colligative properties are:
-
Freezing point depression: The presence of a solute lowers the freezing point of a solution as compared to that of the pure solvent.
-
Boiling point elevation: The presence of a solute increases the boiling point of a solution as compared to that of the pure solvent.
-
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.
-
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.
Colligative properties are properties of solutions which depend on the number of dissolved particles in solution. The four main colligative properties are:
-
Freezing point depression: The presence of a solute lowers the freezing point of a solution as compared to that of the pure solvent.
-
Boiling point elevation: The presence of a solute increases the boiling point of a solution as compared to that of the pure solvent.
-
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.
-
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.
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Which of the following aqueous solutions would be expected to have the greatest increase in boiling point?
Which of the following aqueous solutions would be expected to have the greatest increase in boiling point?
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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 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.
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Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?
Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?
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Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.
Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.
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What would be the best solvent would you choose to dissolve C31H64?
What would be the best solvent would you choose to dissolve C31H64?
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C31H64 is a non-polar substance and would dissolve best in a non-polar solvent like Toluene.
C31H64 is a non-polar substance and would dissolve best in a non-polar solvent like Toluene.
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Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.
Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.
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Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.
Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.
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What is the energy of a photon that has a wavelength of 540 nm?
What is the energy of a photon that has a wavelength of 540 nm?
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What transition would emit the longest wavelength of light?
What transition would emit the longest wavelength of light?
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Wavelength is inversely proportional to energy. Since the transition from n=4 to n=3 is the lowest energy transition, it would have the longest wavelength.
Wavelength is inversely proportional to energy. Since the transition from n=4 to n=3 is the lowest energy transition, it would have the longest wavelength.
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What photon wavelength can promote a transition from the n = 1 (ground state) to the n = 3 (excited state)?
What photon wavelength can promote a transition from the n = 1 (ground state) to the n = 3 (excited state)?
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A 3.0 eV light is shined on Gold ( Փ = 5.1 eV), silver (Փ =4.26 eV), Cesium (Փ = 2.1 eV), and Platinum (Փ = 6.35). If Փ is the work function of the metal, what metal would have electrons ejected?
A 3.0 eV light is shined on Gold ( Փ = 5.1 eV), silver (Փ =4.26 eV), Cesium (Փ = 2.1 eV), and Platinum (Փ = 6.35). If Փ is the work function of the metal, what metal would have electrons ejected?
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The work function is the energy required to remove an electron from a metal. Only Cs metal has a work function less than the energy of the light.
The work function is the energy required to remove an electron from a metal. Only Cs metal has a work function less than the energy of the light.
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A student has a Blue #1 dye with a molar absorptivity of 1.30 x 105 M-1 cm-1 at 631 nm. If the student runs this sample through a spectrophotometer and finds it has an absorption of 0.500 what is the dye’s molarity? Assume a path length of 1.0 cm.
A student has a Blue #1 dye with a molar absorptivity of 1.30 x 105 M-1 cm-1 at 631 nm. If the student runs this sample through a spectrophotometer and finds it has an absorption of 0.500 what is the dye’s molarity? Assume a path length of 1.0 cm.
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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)
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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 of sodium chloride in
of water?
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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 , 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
.
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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 in water?
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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 . The normal freezing point of pure water is
, which means our new freezing point is
.
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80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?
80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?
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Molarity = mol solute / L soution
mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol
L solution = 50000 mL water * 1 L/1000 mL = 50 L
2 mol / 50 L =
Molarity = mol solute / L soution
mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol
L solution = 50000 mL water * 1 L/1000 mL = 50 L
2 mol / 50 L =
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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 of
and
of
?
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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 . Since the freezing point of pure water is
, our new temperature for freezing point is
.
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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?
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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?
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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 . 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.
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