Question 1

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 2

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

Raoult's law states that the partial pressure of a component in a solution is equal to the product of its mole fraction in the solution and its vapor pressure in the pure state. For a two-component solution, this can be expressed as:P_A = X_A * P_A^0P_B = X_B * P_B^0where P_A and P_B are the partial pressures of components A and B, respectively, X_A and X_B are their mole fractions in the solution, and P_A^0 and P_B^0 are their vapor pressures in the pure state.In this problem, we are given the vapor pressures of pentane and hexane at 25° C:P_pentane^0 = 511 torrP_hexane^0 = 150 torrWe are also given the mole fraction of hexane in the solution:X_hexane = 0.50We can use Raoult's law to calculate the partial pressure of hexane in the solution:P_hexane = X_hexane * P_hexane^0 = 0.50 * 150 torr = 75 torrThe total pressure of the solution is the sum of the partial pressures of the two components:P_total = P_pentane + P_hexaneWe can use Dalton's law to calculate the mole fraction of pentane in the vapor:Y_pentane = P_pentane / P_totalwhere P_pentane is the partial pressure of pentane in the vapor.We can rearrange this equation to solve for P_pentane:P_pentane = Y_pentane * P_totalWe can substitute the expression for P_total into this equation:P_pentane = Y_pentane * (P_pentane + P_hexane)We can solve this equation for Y_pentane:Y_pentane = P_pentane / (P_pentane + P_hexane)We can substitute the given values into this equation:Y_pentane = (P_total - P_hexane) / (P_total - P_hexane + P_hexane) = (P_total - 75 torr) / P_totalWe do not know the value of P_total, but we can use the fact that the mole fraction of hexane in the solution is 0.50 to find a relationship between P_total and P_pentane:X_hexane = P_hexane / P_total = 0.50We can rearrange this equation to solve for P_total:P_total = P_hexane / 0.50 = 75 torr / 0.50 = 150 torrWe can substitute this value into the equation for Y_pentane:Y_pentane = (P_total - 75 torr) / P_total = (150 torr - 75 torr) / 150 torr = 0.77Therefore, the mole fraction of pentane in the vapor that is in equilibrium at 25°C with this solution is 0.77.

Question 3

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 4

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...

Question 5

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 13.7-g sample of t-butanol is measured as 24.81°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 6

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

The actual vapor pressure of the solution (314 torr) is lower than the vapor pressure predicted by Raoult's law based on the composition of the solution [(1/4) x 265 torr + (3/4) x 355 torr = 342 torr]. Therefore, the solution exhibits a negative deviation from Raoult's law.

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 of the water because it disrupts the formation of the ice crystal lattice, not because it lowers the vapor pressure of the ice.

Question 8

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 need to calculate the molality of the solution using the freezing point depression formula: ΔTf = Kf · molality ΔTf = -0.190°C Kf = 1.86°C/m molality = ΔTf / Kf = -0.190°C / (1.86°C/m) = -0.102 mol/kg Now, we can use the molality formula to calculate the number of moles of glutamic acid in the solution: molality = moles of solute / mass of solvent in kg moles of solute = molality × mass of solvent in kg = -0.102 mol/kg × 0.1000 kg = -0.0102 mol However, we can see that we obtained a negative value for the number of moles, which does not make sense. This means that we made an error somewhere. The error is actually in the assumption that the freezing point depression is solely due to the presence of the solute (glutamic acid) in the solution. In fact, H₂O also contributes to the freezing point depression. We need to use the van't Hoff factor (i) to correct our calculation: ΔTf = Kf · molality · i H₂O is a non-electrolyte, so i = 1. Glutamic acid, on the other hand, is an amino acid that dissociates into ions in water, so i is greater than 1. The actual value of i depends on the extent of dissociation, which we do not know. However, we can assume that it is close to 1 for the sake of this problem. Let's assume i = 1, and redo the calculation: ΔTf = Kf · molality · i = 1.86°C/m × 0.102 mol/kg × 1 = 0.190°C Now, we obtain a positive value for ΔTf, which is consistent with our intuition. The corrected number of moles of glutamic acid is: molality = moles of solute / mass of solvent in kg moles of solute = molality × mass of solvent in kg = 0.102 mol/kg × 0.1000 kg = 0.0102 mol Finally, we can calculate the molar mass of glutamic acid: molar mass = mass / moles = 1.50 g / 0.0102 mol = 147 g/mol Therefore, the correct answer is B.

Question 9

What is the boiling-point change for a solution containing 0.308 mol of naphthalene (a nonvolatile, nonionizing compound) in 266.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

Accepted Answer

We can use the equation ΔTb = Kb · m · i, where ΔTb is the boiling point elevation, Kb is the molal boiling-point elevation constant, m is the molality of the solution (moles of solute per kilogram of solvent), and i is the van 't Hoff factor, which is 1 for nonionizing solutes like naphthalene. First, we need to calculate the molality of the solution using the formula

molality = moles of solute / mass of solvent in kg

moles of naphthalene = 0.308 mol
mass of benzene = 266.0 g = 0.2660 kg

molality = 0.308 mol / 0.2660 kg ≈ 1.16 mol/kg

Now we can plug in the values to find ΔTb:

ΔTb = (2.53°C/m) · (1.16 mol/kg) · 1
ΔTb ≈ 2.93°C

Therefore, the boiling-point change for the solution is 2.93°C, which corresponds to answer choice (E).

Question 10

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

Accepted Answer

When a solute is added to a solvent, it forms a solution with a lower vapor pressure than the pure solvent. This is because the solute particles occupy some of the surface area of the liquid, reducing the number of solvent molecules that can escape into the gas phase. Thus, to reach the same vapor pressure as the pure solvent, the solution must be heated to a higher temperature, which raises its boiling point. Option B is incorrect because the boiling point is determined by the vapor pressure of the liquid, not the presence of the solute. Option C does not accurately reflect the relationship between volume and boiling point. Option D is also incorrect because the solute particles lower the vapor pressure of the solvent, not raise it.

Question 11

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 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 50% A?

Accepted Answer

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

Question 13

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 14

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 for benzene is greater than that of water, meaning that the freezing point of benzene is more sensitive to changes in solute concentration. Since the depression of the freezing point of benzene is greater than that of water with the same amount of solute, this suggests that the formic acid molecules are closely associated in benzene (resulting in a larger reduction in the freezing point) but are mostly in a monomeric state in water.

Question 15

When a 19.4-g sample of an unknown compound is dissolved in 458.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 3.86°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 0.04°C. Using the formula ΔTf = Kf * m, where m is the molality, we find m = 0.04°C / 5.12°C/m = 0.0078125 mol/kg. The molality is moles of solute per kg of solvent, so moles of solute = 0.0078125 mol/kg * 0.458 kg = 0.003578 mol. The molar mass is mass of solute / moles of solute = 19.4 g / 0.003578 mol = 1.34 × 10^2 g/mol.

Question 16

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 (II), as well as the freezing point of the pure solvent (III) to determine the new freezing point.

Question 17

A 5.22-g sample of a compound is dissolved in 200.0 g of benzene. The freezing point of this solution is 1.08°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

Explanation:ΔT_f = K_fm1.08°C = 5.12°C/m * mm = 0.211 mM = 5.22 g / 0.211 mol = 24.7 g/mol

Question 18

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

Dissociation would result in a lower apparent molar mass, not a higher one. The other errors would result in a higher apparent molar mass.

Question 19

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 20

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 hydrocarbons with similar molecular sizes and shapes, leading to similar intermolecular forces between them. This similarity results in a mixture that behaves relatively ideally according to Raoult's law.

Quiz 17: Properties of Solutions

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?

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.

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 13.7-g sample of t-butanol is measured as 24.81°C, how many grams of water are present in the sample?

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?

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

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

What is the boiling-point change for a solution containing 0.308 mol of naphthalene (a nonvolatile, nonionizing compound) in 266.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

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

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?

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 19.4-g sample of an unknown compound is dissolved in 458.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 3.86°C, and K_f for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

A 5.22-g sample of a compound is dissolved in 200.0 g of benzene. The freezing point of this solution is 1.08°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.)

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?

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?

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)

Quiz 17: Properties of Solutions

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?

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.

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 13.7-g sample of t-butanol is measured as 24.81°C, how many grams of water are present in the sample?

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?

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

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

What is the boiling-point change for a solution containing 0.308 mol of naphthalene (a nonvolatile, nonionizing compound) in 266.0 g of liquid benzene? (Kb = 2.53°C/m for benzene)

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

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?

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 19.4-g sample of an unknown compound is dissolved in 458.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 3.86°C, and Kf for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

A 5.22-g sample of a compound is dissolved in 200.0 g of benzene. The freezing point of this solution is 1.08°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.)

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?

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?

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 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.

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 13.7-g sample of t-butanol is measured as 24.81°C, how many grams of water are present in the sample?

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

What is the boiling-point change for a solution containing 0.308 mol of naphthalene (a nonvolatile, nonionizing compound) in 266.0 g of liquid benzene? (K_b = 2.53°C/m for benzene)

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.

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 19.4-g sample of an unknown compound is dissolved in 458.0 g of benzene, the freezing point of the resulting solution of 3.82°C. The freezing point of pure benzene is 3.86°C, and K_f for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

A 5.22-g sample of a compound is dissolved in 200.0 g of benzene. The freezing point of this solution is 1.08°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.)

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)