Question 1

To calculate the freezing point of an ideal dilute solution of a single, nondissociating solute in a solvent, the minimum information one must know is 
I.The molality (of the solute).
II.The molality (of the solute) and the freezing-point-depression constant of the solvent.
III.The freezing point of the pure solvent.
IV.The molecular weight of the solute.
V.The weight of the solvent.

Accepted Answer

To calculate the freezing point depression, you need the molality of the solute and the freezing-point-depression constant of the solvent. Additionally, you need the freezing point of the pure solvent to determine the new freezing point.

Question 2

The molar mass of a solid as determined by freezing-point depression is 10% higher than the true molar mass. Which of the following experimental errors could not account for this discrepancy?

Accepted Answer

The freezing-point depression method assumes that the solute dissolves as a single particle and that the freezing point depression is proportional to the amount of solute particles present. Thus, if the solid dissociated into two particles, the measured depression would be twice that of a non-dissociated solid of the same mass, thereby leading to an incorrect determination of the molar mass. The other choices could lead to a lower mass of solid being used in the experiment, which would falsely give a higher molar mass measurement due to a greater depression of the freezing point.

Question 3

The freezing point (T_f) for t-butanol is 25.50°C, and K_f is 9.1°C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 13.9-g sample of t-butanol is measured as 24.91°C, how many grams of water are present in the sample?

Accepted Answer

The answer of The freezing point (T_f) for t-butanol is...

Question 4

When a 18.4-g sample of an unknown compound is dissolved in 522.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 5.48°C, and K_f for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

Accepted Answer

The change in freezing point (ΔTf) is 5.48°C - 3.82°C = 1.66°C. Using the formula ΔTf = Kf * m, where m is the molality, we find m = 1.66°C / 5.12°C/m = 0.3242 mol/kg. The molality is moles of solute per kg of solvent, so moles of solute = 0.3242 mol/kg * 0.522 kg = 0.1692 mol. The molar mass is mass of solute / moles of solute = 18.4 g / 0.1692 mol = 108.7 g/mol, which is approximately 1.09 x 10^2 g/mol.

Question 5

A solute added to a solvent raises the boiling point of the solution because

Accepted Answer

When a solute is added to a solvent, the solute particles mix with the solvent particles and interfere with the solvent's ability to escape from the solution as a vapor. This lowers the vapor pressure of the solvent, which means that a higher temperature is required for the solution's vapor pressure to exceed the external pressure and boil. Therefore, the boiling point of the solution is raised.

Question 6

Which of the following will cause the calculated molar mass of a compound determined by the freezing-point depression method to be greater than the true molar mass?

Accepted Answer

The answer of Which of the following will cause the...

Question 7

Adding salt to water decreases the freezing point of the water because it lowers the vapor pressure of the ice.

Accepted Answer

Adding salt to water decreases the freezing point because it disrupts the formation of the ice lattice, not because it lowers the vapor pressure of the ice. This phenomenon is known as freezing point depression.

Question 8

Using the data below, calculate the vapor pressure of benzene over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

Accepted Answer

The answer of Using the data below, calculate the vapor...

Question 9

At a given temperature, you have a mixture of benzene (P_vap = 745 torr) and toluene (P_vap = 290 torr). The mole fraction of benzene in the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

Accepted Answer

The answer of At a given temperature, you have a...

Question 10

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 50% A?

Accepted Answer

The answer of Liquid A has vapor pressure x. Liquid...

Question 11

A 5.96-g sample of a compound is dissolved in 222.0 g of benzene. The freezing point of this solution is 1.06°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.)

Accepted Answer

The freezing point depression is given by the equation:$$\Delta T_f = K_f 	imes m$$where $\Delta T_f$ is the freezing point depression, $K_f$ is the freezing point depression constant, and $m$ is the molality of the solution.We can rearrange this equation to solve for the molality:$$m = \frac{\Delta T_f}{K_f}$$Substituting the given values into this equation, we get:$$m = \frac{1.06°C}{5.12°C/m} = 0.207 m$$The molality is defined as the number of moles of solute per kilogram of solvent. We can use this to calculate the molar mass of the compound:$$M = \frac{m 	imes MW_{solvent}}{mass_{solute}}$$where $M$ is the molar mass of the compound, $MW_{solvent}$ is the molecular weight of the solvent, and $mass_{solute}$ is the mass of the solute.Substituting the given values into this equation, we get:$$M = \frac{0.207 m 	imes 78.11 g/mol}{5.96 g} = 2.85 	imes 10^2 g/mol$$Therefore, the molar mass of the compound is 2.85 x 10^2 g/mol, which corresponds to choice B.

Question 12

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 30% A?

Accepted Answer

The answer of Liquid A has vapor pressure x. Liquid...

Question 13

The molal freezing-point depression constants for benzene and water are 5.12 and 1.86, respectively. When 4.6 g of formic acid (HCOOH) is dissolved in 1.0 kg of benzene, the freezing point is depressed by 0.26°C. When the same amount of formic acid is dissolved in 1.0 kg of water, the freezing point is lowered by 0.19°C. To explain these results, we must assume that

Accepted Answer

The molal freezing-point depression constant depends on the concentration of solute molecules or ions in the solvent. The fact that the depression of the freezing point is greater when formic acid is dissolved in benzene than when it is dissolved in water suggests that there are fewer total solute particles in the former case, i.e., there is some association of formic acid molecules in the benzene solution. Conversely, in water, formic acid is more dissociated into ions, leading to the formation of more solute particles and a smaller depression of the freezing point. Therefore, choice B seems to be the best explanation.

Question 14

What is the boiling-point change for a solution containing 0.360 mol of naphthalene (a nonvolatile, nonionizing compound) in 258.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

Accepted Answer

The answer of What is the boiling-point change for a...

Question 15

For each of the following solutions, would you expect it with respect to Raoult's law) to be relatively ideal, to show a positive deviation, or to show a negative deviation? -hexane (C₆H₁₄) and octane (C₈H₁₈)

Accepted Answer

Hexane and octane are both nonpolar molecules with similar sizes and shapes, so they should mix well and exhibit relatively ideal behavior according to Raoult's law.

Question 16

A solution contains 1 mol of liquid A and 3 mol of liquid B. The vapor pressure of this solution is 314 torr at 25°C. At 25°C, the vapor pressure of liquid A is 265 torr and the vapor pressure of liquid B is 355 torr. Which of the following is true?

Accepted Answer

According to Raoult's law, the vapor pressure of a solution should be equal to the mole fraction weighted sum of the vapor pressures of the components. In this case, the expected vapor pressure would be:

(1/4) x 265 torr + (3/4) x 355 torr = 340 torr

Since the measured vapor pressure is less than the expected value, this solution exhibits a negative deviation from Raoult's law. This means that the attractive forces between unlike molecules are stronger than expected, leading to a lower vapor pressure.

Question 17

Pentane (C₅H₁₂) and hexane (C₆H₁₄) form an ideal solution. The vapor pressures of pentane and hexane at 25° C are 511 torr and 150 torr, respectively. The mole fraction of hexane in a pentane-hexane solution is 0.50. Calculate the mole fraction of pentane in the vapor that is in equilibrium at 25°C with this solution.

Accepted Answer

The answer of Pentane (C₅H₁₂) and hexane (C₆H₁₄) form an...

Question 18

Using the data below, calculate the vapor pressure of chloroform over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

Accepted Answer

The answer of Using the data below, calculate the vapor...

Question 19

When a 1.50-g sample of glutamic acid is dissolved in 100.0 g of H₂O, the resulting solution freezes at -0.190°C. K_f for H₂O is 1.86°C/m. The molar mass of glutamic acid is

Accepted Answer

First, we can use the freezing point depression formula to calculate the molality of the solution: ΔTf = Kf * molality molality = ΔTf / Kf = (-0.190°C) / (1.86°C/m) = 0.102 mol/kg Next, we can calculate the number of moles of glutamic acid in the solution: moles = mass / molar mass = 1.50 g / MM Finally, we can use the molality and number of moles to find the mass of H₂O in the solution: molality = moles of solute / mass of solvent (in kg) mass of solvent = 100.0 g / 1000 g/kg = 0.100 kg moles of solvent = molality * mass of solvent = 0.102 mol/kg * 0.100 kg = 0.0102 mol moles of solvent = moles of solute (glutamic acid) 0.0102 mol = 1.50 g / MM MM = 147 g/mol Therefore, the answer is B, 147 g/mol.

Question 20

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in the vapor above the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the solution.

Accepted Answer

The answer of At a given temperature, you have a...

Quiz 17: Properties of Solutions

The molar mass of a solid as determined by freezing-point depression is 10% higher than the true molar mass. Which of the following experimental errors could not account for this discrepancy?

The freezing point (T_f) for t-butanol is 25.50°C, and K_f is 9.1°C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 13.9-g sample of t-butanol is measured as 24.91°C, how many grams of water are present in the sample?

When a 18.4-g sample of an unknown compound is dissolved in 522.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 5.48°C, and K_f for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

A solute added to a solvent raises the boiling point of the solution because

Which of the following will cause the calculated molar mass of a compound determined by the freezing-point depression method to be greater than the true molar mass?

Adding salt to water decreases the freezing point of the water because it lowers the vapor pressure of the ice.

Using the data below, calculate the vapor pressure of benzene over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

At a given temperature, you have a mixture of benzene (P_vap = 745 torr) and toluene (P_vap = 290 torr) . The mole fraction of benzene in the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 50% A?

A 5.96-g sample of a compound is dissolved in 222.0 g of benzene. The freezing point of this solution is 1.06°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.)

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 30% A?

What is the boiling-point change for a solution containing 0.360 mol of naphthalene (a nonvolatile, nonionizing compound) in 258.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

For each of the following solutions, would you expect it with respect to Raoult's law) to be relatively ideal, to show a positive deviation, or to show a negative deviation? -hexane (C₆H₁₄) and octane (C₈H₁₈)

A solution contains 1 mol of liquid A and 3 mol of liquid B. The vapor pressure of this solution is 314 torr at 25°C. At 25°C, the vapor pressure of liquid A is 265 torr and the vapor pressure of liquid B is 355 torr. Which of the following is true?

Using the data below, calculate the vapor pressure of chloroform over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

When a 1.50-g sample of glutamic acid is dissolved in 100.0 g of H₂O, the resulting solution freezes at -0.190°C. K_f for H₂O is 1.86°C/m. The molar mass of glutamic acid is

Quiz 17: Properties of Solutions

The molar mass of a solid as determined by freezing-point depression is 10% higher than the true molar mass. Which of the following experimental errors could not account for this discrepancy?

The freezing point (Tf) for t-butanol is 25.50°C, and Kf is 9.1°C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 13.9-g sample of t-butanol is measured as 24.91°C, how many grams of water are present in the sample?

When a 18.4-g sample of an unknown compound is dissolved in 522.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 5.48°C, and Kf for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

A solute added to a solvent raises the boiling point of the solution because

Which of the following will cause the calculated molar mass of a compound determined by the freezing-point depression method to be greater than the true molar mass?

Adding salt to water decreases the freezing point of the water because it lowers the vapor pressure of the ice.

Using the data below, calculate the vapor pressure of benzene over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl3 and 50.0 g of C6H6. Assume that the solution behaves ideally.

At a given temperature, you have a mixture of benzene (Pvap = 745 torr) and toluene (Pvap = 290 torr) . The mole fraction of benzene in the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 50% A?

A 5.96-g sample of a compound is dissolved in 222.0 g of benzene. The freezing point of this solution is 1.06°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.)

Liquid A has vapor pressure x. Liquid B has vapor pressure y, and x > y. What is the mole fraction of A in the liquid mixture if the vapor above the solution is 30% A?

What is the boiling-point change for a solution containing 0.360 mol of naphthalene (a nonvolatile, nonionizing compound) in 258.0 g of liquid benzene? (Kb = 2.53°C/m for benzene)

For each of the following solutions, would you expect it with respect to Raoult's law) to be relatively ideal, to show a positive deviation, or to show a negative deviation? -hexane (C6H14) and octane (C8H18)

A solution contains 1 mol of liquid A and 3 mol of liquid B. The vapor pressure of this solution is 314 torr at 25°C. At 25°C, the vapor pressure of liquid A is 265 torr and the vapor pressure of liquid B is 355 torr. Which of the following is true?

Using the data below, calculate the vapor pressure of chloroform over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl3 and 50.0 g of C6H6. Assume that the solution behaves ideally.

When a 1.50-g sample of glutamic acid is dissolved in 100.0 g of H2O, the resulting solution freezes at -0.190°C. Kf for H2O is 1.86°C/m. The molar mass of glutamic acid is

The freezing point (T_f) for t-butanol is 25.50°C, and K_f is 9.1°C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 13.9-g sample of t-butanol is measured as 24.91°C, how many grams of water are present in the sample?

When a 18.4-g sample of an unknown compound is dissolved in 522.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 5.48°C, and K_f for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

Using the data below, calculate the vapor pressure of benzene over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

At a given temperature, you have a mixture of benzene (P_vap = 745 torr) and toluene (P_vap = 290 torr) . The mole fraction of benzene in the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

A 5.96-g sample of a compound is dissolved in 222.0 g of benzene. The freezing point of this solution is 1.06°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.)

What is the boiling-point change for a solution containing 0.360 mol of naphthalene (a nonvolatile, nonionizing compound) in 258.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

For each of the following solutions, would you expect it with respect to Raoult's law) to be relatively ideal, to show a positive deviation, or to show a negative deviation? -hexane (C₆H₁₄) and octane (C₈H₁₈)

Using the data below, calculate the vapor pressure of chloroform over a chloroform-benzene solution at 25°C, which contains 50.0 g of CHCl₃ and 50.0 g of C₆H₆. Assume that the solution behaves ideally.

When a 1.50-g sample of glutamic acid is dissolved in 100.0 g of H₂O, the resulting solution freezes at -0.190°C. K_f for H₂O is 1.86°C/m. The molar mass of glutamic acid is