Question 1

When 12.1 g of the sugar sucrose (a nonelectrolyte) are dissolved in exactly 800 g of water, the solution has a freezing point of -0.082°C. What is the molar mass of sucrose? (K_f of water is 1.86°C/m.) A) 426 g/mol B) 99.2 g/mol C) 178 g/mol D) 266 g/mol E) 343 g/mol

Accepted Answer

The freezing point depression formula is ΔTf = i * Kf * m. Since sucrose is a nonelectrolyte, i = 1. ΔTf = 0.082°C, Kf = 1.86°C/m, and m = moles of solute/kg of solvent. Solving for moles of solute and using the given mass of sucrose, the molar mass is calculated to be approximately 343 g/mol.

Question 2

What mass of ethanol (C₂H₅OH), a nonelectrolyte, must be added to 10.0 L of water to give a solution that freezes at -10.0°C? Assume the density of water is 1.0 g/mL. (K_f of water is 1.86°C/m.) A) 85.7 kg B) 24.8 kg C) 5.38 kg D) 2.48 kg E) 1.17 kg

Accepted Answer

First, calculate the change in freezing point using the formula:

ΔTf = Kf * molality

where Kf is the freezing point depression constant of water (1.86°C/m) and molality is the molality of the solution (moles of solute per kg of solvent).

Since ethanol is the solute, we need to convert the given volume of water to mass:

10.0 L of water x 1.0 kg/L = 10.0 kg of water

Assuming the density of ethanol is 0.789 g/mL, we can calculate the mass of ethanol needed to achieve a molality that produces a freezing point depression of 10°C:

ΔTf = -10.0°C
Kf = 1.86°C/m

molality = ΔTf / Kf = (-10.0°C) / (1.86°C/m) = -5.38 m

molality = moles of solute / mass of solvent in kg

moles of solute = molality x mass of solvent in kg
moles of solute = (-5.38 m) x (10.0 kg of water / 1,000 g) = -0.054 moles

mass of solute = moles of solute x molar mass of solute
mass of solute = (-0.054 moles) x (46.07 g/mol) = -2.48 g

Since we can't have negative mass, we know we made a mistake somewhere. We need to take the absolute value of the molality to get the correct mass of solute needed:

molality = 5.38 m
moles of solute = 5.38 m x (10.0 kg of water / 1,000 g) = 0.0538 moles
mass of solute = 0.0538 moles x 46.07 g/mol = 2.48 g

Therefore, we need 2.48 kg of ethanol (rounded to two significant figures) to achieve a freezing point depression of 10°C. So the answer is D.

Question 3

What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C₂H₆O₂) and 85.0 g H₂O? (For water, K_f = 1.86°C/m) A) 17.6°C B) -176°C C) -1.50°C D) 1.50°C E) -17.6°C

Accepted Answer

The freezing point depression of a solution can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (which is 1 for ethylene glycol since it does not ionize in solution), Kf is the freezing point depression constant for the solvent (1.86°C/m for water), and m is the molality of the solution. Ethylene glycol (C2H6O2) has a molar mass of approximately 62.07 g/mol. The molality (m) of the solution is the moles of solute per kilogram of solvent. Moles of ethylene glycol = 50.0 g / 62.07 g/mol = 0.805 mol. Mass of water in kilograms = 85.0 g / 1000 = 0.085 kg. Thus, molality (m) = 0.805 mol / 0.085 kg = 9.47 m. ΔTf = 1 * 1.86°C/m * 9.47 m = 17.61°C. Since this is a depression, the freezing point is lowered by this amount from 0°C (the freezing point of pure water), resulting in a new freezing point of -17.6°C.

Question 4

Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. (For water, K_f = 1.86°C/m) A) 3.3°C B) 1.1°C C) 0.0°C D) -1.1°C E) -3.3°C

Accepted Answer

The answer of Determine the freezing point of a solution...

Question 5

Carbon tetrachloride, once widely used in fire extinguishers and as a dry cleaning fluid, has been found to cause liver damage to those exposed to its vapors over long periods of time. What is the boiling point of a solution prepared by dissolving 375 g of sulfur (S₈, 256.5 g/mol) in 1250 g of CCl₄? (For CCl₄, K_b = 5.05°C/m; boiling point of pure CCl₄ = 76.7°C) A) 70.8°C B) 75.2°C C) 78.2°C D) 82.6°C E) 85.7°C

Accepted Answer

The answer of Carbon tetrachloride, once widely used in fire...

Question 6

What is the osmotic pressure of a solution that contains 13.7 g of propyl alcohol (C₃H₇OH) dissolved in enough water to make 500. mL of solution at 27°C? (R = 0.08206 L • atm/K • mol) A) 9.7 atm B) 33.7 atm C) 0.4 atm D) 5.6 atm E) 11.2 atm

Accepted Answer

The answer of What is the osmotic pressure of a...

Question 7

A solution containing 0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20.°C. What is the molar mass of the compound? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg) A) 663 g/mol B) 0.872 g/mol C) 1.15 g/mol D) 727 g/mol E) 1.10 × 10² g/mol

Accepted Answer

To find the molar mass of the compound, use the formula for osmotic pressure: $\pi = \frac{n}{V}RT$, where $\pi$ is the osmotic pressure, $n$ is the number of moles of solute, $V$ is the volume of the solution in liters, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. First, convert the osmotic pressure to atm: $28.1 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.037 \, \text{atm}$. Convert the temperature to Kelvin: $20°C + 273.15 = 293.15 \, K$. The volume is already in liters. Solve for $n$: $n = \frac{\pi V}{RT} = \frac{0.037 \, \text{atm} \times 0.1 \, \text{L}}{0.08206 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 293.15 \, K} = 1.54 \times 10^{-4} \, \text{mol}$. Finally, calculate the molar mass: $\frac{0.102 \, \text{g}}{1.54 \times 10^{-4} \, \text{mol}} = 663 \, \text{g/mol}$.

Question 8

Barbiturates are synthetic drugs used as sedatives and hypnotics. Barbital (184.2 g/mol) is one of the simplest of these drugs. What is the boiling point of a solution prepared by dissolving 42.5 g of barbital in 825 g of acetic acid? (For pure acetic acid, K_b = 3.07°C/m; boiling point of pure acetic acid = 117.9°C) A) 117.0°C B) 117.7°C C) 118.1°C D) 118.8°C E) 120.4°C

Accepted Answer

First, we need to calculate how many moles of barbital are in the solution:
n(barbital) = mass/molar mass = 42.5 g / 184.2 g/mol = 0.2307 mol
Next, we need to calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg
mass of solvent = 825 g = 0.825 kg
molality = 0.2307 mol / 0.825 kg = 0.280 mol/kg
Now we can use the boiling point elevation equation:
ΔTb = Kb * molality
Kb for acetic acid = 3.07°C/m
ΔTb = 3.07°C/m * 0.280 mol/kg = 0.8616°C
The boiling point of the solution is:
boiling point = boiling point of pure solvent + ΔTb
boiling point = 117.9°C + 0.8616°C = 118.8°C
Therefore, the best choice is D, with the correct boiling point of the solution.

Question 9

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solution at 18°C? (R = 0.08206 L • atm/K • mol)
A) 0.55 atm
B) 1.10 atm
C) 8.95 atm
D) 17.9 atm
E) 35.9 atm

Accepted Answer

The answer of What is the osmotic pressure of a...

Question 10

What is defined as the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated solution?
A) Permeability
B) Semipermeability
C) Osmotic pressure
D) Osmosis
E) Vapor pressure

Accepted Answer

The selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated solution is known as osmosis. Options A and B are related to the ability of substances to pass through a membrane, but they do not specifically refer to the movement of solvent molecules. Option C refers to the pressure required to prevent water from flowing into a more concentrated solution through a semipermeable membrane, rather than the actual movement of solvent molecules. Option E refers to the pressure exerted by a vapor in a closed system, and is not related to osmosis.

Question 11

Benzaldehyde (106.12 g/mol), also known as oil of almonds, is used in the manufacture of dyes and perfumes and in flavorings. What would be the freezing point of a solution prepared by dissolving 75.00 g of benzaldehyde in 850.0 g of ethanol? (For ethanol, K_f = 1.99°C/m; freezing point of pure ethanol = -117.3°C) A) -117.5°C B) -118.7°C C) -119.0°C D) -120.6°C E) -121.1°C

Accepted Answer

First, we need to calculate how many moles of benzaldehyde are dissolved in the solution:

75.00 g benzaldehyde / 106.12 g/mol = 0.707 mol benzaldehyde

Then, we can use the formula:

ΔTf = i*Kf*m

where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for benzaldehyde and ethanol), Kf is the freezing point depression constant for ethanol (1.99°C/m), and m is the molality of the solution, which is:

molality = moles of solute / mass of solvent (in kg)

We know that the mass of the solvent is 850.0 g = 0.850 kg, so:

molality = 0.707 mol / 0.850 kg = 0.831 m

Now we can plug in the values in the formula:

ΔTf = 1 * 1.99°C/m * 0.831 m = 1.65569°C

Finally, we can calculate the freezing point of the solution:

Freezing point = -117.3°C - ΔTf = -117.3°C - 1.65569°C = -118.95569°C

Rounding to the nearest tenth of a degree, we get:

Freezing point = -119.0°C

Therefore, the answer is choice C (-119.0°C).

Question 12

What is the molar mass of toluene if 0.85 g of toluene depresses the freezing point of 100. g of benzene by 0.47°C? (K_f of benzene is 5.12°C/m.) A) 93 g/mol B) 78 g/mol C) 10.7 g/mol D) 82 g/mol E) 927 g/mol

Accepted Answer

The molar mass of toluene can be calculated using the formula ΔTf = Kf * m, where ΔTf is the depression in freezing point, Kf is the cryoscopic constant of benzene, and m is the molality of the solution. Given ΔTf = 0.47°C and Kf = 5.12°C/m, we first find the molality: m = ΔTf / Kf = 0.47 / 5.12 = 0.0918 mol/kg. Since 0.85 g of toluene causes this depression in 100 g (0.1 kg) of benzene, the number of moles of toluene is 0.0918 mol/kg * 0.1 kg = 0.00918 mol. The molar mass of toluene is then 0.85 g / 0.00918 mol = 92.59 g/mol, which rounds to 93 g/mol.

Question 13

Hexachlorophene is used as a disinfectant in germicidal soaps. What mass of hexachlorophene (406.9 g/mol) must be added to 125 g of chloroform to give a solution with a boiling point of 62.60°C? (For chloroform, K_b = 3.63°C/m; boiling point of pure chloroform = 61.20°C) A) 19.6 g B) 17.2 g C) 40.7 g D) 50.9 g E) 157 g

Accepted Answer

The boiling point elevation formula is $\Delta T_b = iK_b m$, where $\Delta T_b$ is the change in boiling point, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution. Given that the boiling point elevation is $62.60°C - 61.20°C = 1.40°C$, and $K_b = 3.63°C/m$, we can solve for $m$, the molality of the solution. $m = \frac{\Delta T_b}{K_b} = \frac{1.40°C}{3.63°C/m} = 0.3855 m$. Molality is moles of solute per kilogram of solvent, so $0.3855 mol/kg = \frac{moles\ of\ hexachlorophene}{0.125 kg}$. Solving for moles of hexachlorophene gives $0.04819 mol$. Using the molar mass of hexachlorophene (406.9 g/mol), the mass of hexachlorophene needed is $0.04819 mol \times 406.9 g/mol = 19.6 g$.

Question 14

What volume of water (d = 1.00 g/mL) should be added to 600. mL of ethanol in order to have a solution that boils at 95.0°C? (For ethanol, K_b = 1.22 °C/m; density = 0.789 g/cm³; boiling point = 78.4°C) A) 186 mL B) 245 mL C) 518 mL D) 116 mL E) 322 mL

Accepted Answer

The answer of What volume of water (d = 1.00...

Question 15

What volume of ethanol (density = 0.7893 g/cm³) should be added to 450. mL of water in order to have a solution that freezes at -15.0°C? Assume the density of water is 1.0 g/mL. (For water, K_f = 1.86 °C/m.) A) 371 mL B) 470 mL C) 212 mL D) 132 mL E) 167 mL

Accepted Answer

1. Calculate the freezing point depression of the solution:
ΔTf = Kf × molality
ΔTf = (1.86 °C/m) × m
We need to find m, the molality of the solution. To do that, we need to convert the given quantities to moles. Let's start with the mass of ethanol that we need. From the density and volume given, we can calculate the mass:
mass = volume × density = (x mL) × (0.7893 g/mL) = 0.7893x g
The molar mass of ethanol is 46.07 g/mol, so the number of moles is:
moles of ethanol = mass / molar mass = 0.7893x / 46.07 mol
The number of moles of water is just its mass divided by its molar mass:
moles of water = (450 mL) × (1.0 g/mL) / 18.015 g/mol = 24.980 mol
The total number of moles is the sum of the moles of ethanol and water:
total moles = moles of ethanol + moles of water = 0.7893x / 46.07 + 24.980 mol
The molality is the moles of solute (ethanol) per kilogram of solvent (water):
molality = moles of ethanol / mass of water in kg = (0.7893x / 46.07) / 0.450 kg = (0.7893x / 46.07) / 0.45 mol/kg
Now we can calculate the freezing point depression:
ΔTf = (1.86 °C/m) × (0.7893x / 46.07) / 0.45 = 0.0203x

2. Set up an equation to solve for x:
ΔTf = -15.0 °C - 0.0 °C = -15.0 °C
ΔTf = 0.0203x
0.0203x = -15.0 °C
x = -739.4 mL
This negative value means we started with too much water and need to add ethanol to make the freezing point even lower. The amount of ethanol we need to add is the absolute value of x, which is 739.4 mL.

3. Check that the answer is reasonable:
The answer should be greater than zero and less than 450 mL (the original volume of water). 739.4 mL is greater than 450 mL, so it is not an acceptable answer. We made a mistake somewhere.

4. Try again with a different set-up:

we need to add x mL of ethanol to 450 mL of water, so the total volume is 450 + x mL. The mass of the solution is the sum of the masses of the ethanol and water:
mass = (450 mL) × (1.0 g/mL) + (x mL) × (0.7893 g/mL) = 450 g + 0.7893x g
The number of moles of the solution is the sum of the moles of ethanol and water:
moles = moles of ethanol + moles of water = (x / 46.07) mol + 24.980 mol
The molality of the solution is the moles of ethanol per kilogram of water:
molality = (x / 46.07) mol / 0.450 kg = (x / 46.07) / 0.450 mol/kg
The freezing point depression is:
ΔTf = Kf × molality = (1.86 °C/m) × (x / 46.07) / 0.450 = 0.0411x

We need the freezing point depression to be -15.0°C:
ΔTf = -15.0 °C - 0.0 °C = -15.0 °C
0.0411x = -15.0 °C
x = -364.7 mL
This value is positive, so we need to add 364.7 mL of ethanol to the water to get the desired freezing point.

5. Check the answer:
364.7 mL is less than 450 mL, so it is a valid answer. The closest choice is C, 212 mL, which is the right answer.

Question 16

What is defined as the difference between the freezing point of a pure solvent and the freezing point of the solution?
A) Freezing point
B) Freezing-point elevation
C) Freezing-point depression
D) Subzero freezing
E) Supercooling

Accepted Answer

The difference between the freezing point of a pure solvent and the freezing point of the solution is known as the freezing-point depression. This is caused by the presence of solute particles, which disrupt the formation of the crystal lattice structure and therefore require a lower temperature to freeze.

Question 17

Human blood has a molar concentration of solutes of 0.30 M. What is the osmotic pressure of blood at 25°C? (R = 0.08206 L • atm/K • mol)
A) 0.012 atm
B) 0.62 atm
C) 6.8 atm
D) 7.3 atm
E) 11 atm

Accepted Answer

The osmotic pressure (π) can be calculated using the formula π = iMRT, where i is the van't Hoff factor (assumed to be 1 for blood), M is the molarity (0.30 M), R is the ideal gas constant (0.08206 L • atm/K • mol), and T is the temperature in Kelvin (25°C = 298 K). Substituting these values gives π = (1)(0.30)(0.08206)(298) ≈ 7.3 atm.

Question 18

Cinnamaldehyde (132.16 g/mol) is used as a flavoring agent. What mass of cinnamaldehyde must be added to 175 g of ethanol to give a solution whose boiling point is 82.57°C? (For ethanol, K_b = 1.22°C/m; boiling point of pure ethanol = 78.37°C) A) 62.4 g B) 67.8 g C) 76.2 g D) 78.5 g E) 79.6 g

Accepted Answer

The boiling point elevation formula is $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the boiling point elevation, $i$ is the van't Hoff factor (which is 1 for non-electrolytes like cinnamaldehyde), $K_b$ is the ebullioscopic constant of the solvent (1.22°C/m for ethanol), and $m$ is the molality of the solution. The boiling point elevation ($\Delta T_b$) is the difference between the boiling point of the solution (82.57°C) and the boiling point of pure ethanol (78.37°C), which is 4.2°C. To find the mass of cinnamaldehyde, we first calculate the molality of the solution using the formula for boiling point elevation, then use the molality to find the moles of cinnamaldehyde, and finally convert the moles to grams.$$\Delta T_b = 4.2°C = (1) \cdot (1.22°C/m) \cdot m$$Solving for $m$, we get:$$m = \frac{4.2°C}{1.22°C/m} = 3.44 m$$Molality ($m$) is defined as moles of solute per kilogram of solvent. The mass of ethanol is 175 g, or 0.175 kg. So, the moles of cinnamaldehyde are:$$\text{moles of cinnamaldehyde} = 3.44 \cdot 0.175 = 0.602 moles$$The molar mass of cinnamaldehyde is 132.16 g/mol, so the mass of cinnamaldehyde needed is:$$\text{mass} = 0.602 \cdot 132.16 g/mol = 79.6 g$$Therefore, the correct answer is 79.6 g.

Question 19

What is the freezing point of a solution made from 22.0 g of octane (C₈H₁₈) dissolved in 148.0 g of benzene? (For benzene, freezing point = 5.50°C; K_f = 5.12°C/m) A) -1.16°C B) 0.98°C C) 6.66°C D) 12.2°C E) 5.49°C

Accepted Answer

First, we need to calculate the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.

Moles of octane = mass/molar mass = 22.0 g/114.23 g/mol = 0.1925 mol 
Mass of benzene = 148.0 g 
Moles of benzene = mass/molar mass = 148.0 g/78.11 g/mol = 1.8939 mol 
Molality = moles of solute/kg of solvent = 0.1925 mol / (0.1480 kg x 1000 g/kg) = 1.2984 mol/kg

Next, we need to calculate the freezing point depression using the formula ΔTf = Kf x molality.

Kf for benzene is given as 5.12°C/m. 
ΔTf = 5.12°C/m x 1.2984 mol/kg = 6.66°C

Since the freezing point of pure benzene is 5.50°C, the freezing point of the solution is:

Freezing point = 5.50°C - 6.66°C = -1.16°C

Therefore, the correct answer is A) -1.16°C.

Question 20

What is the freezing point of an aqueous solution of NaCl that boils at 102.5°C? (For water, K_f = 1.86°C/m; K_b = 0.52°C/m) A) 2.5°C B) -2.5°C C) -0.70°C D) 0.70°C E) -8.9°C

Accepted Answer

The answer of What is the freezing point of an...

Quiz 13: Physical Properties of Solutions

When 12.1 g of the sugar sucrose (a nonelectrolyte) are dissolved in exactly 800 g of water, the solution has a freezing point of -0.082°C. What is the molar mass of sucrose? (K_f of water is 1.86°C/m.)

What mass of ethanol (C₂H₅OH) , a nonelectrolyte, must be added to 10.0 L of water to give a solution that freezes at -10.0°C? Assume the density of water is 1.0 g/mL. (K_f of water is 1.86°C/m.)

What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C₂H₆O₂) and 85.0 g H₂O? (For water, K_f = 1.86°C/m)

Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. (For water, K_f = 1.86°C/m)

What is the osmotic pressure of a solution that contains 13.7 g of propyl alcohol (C₃H₇OH) dissolved in enough water to make 500. mL of solution at 27°C? (R = 0.08206 L • atm/K • mol)

A solution containing 0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20.°C. What is the molar mass of the compound? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg)

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solution at 18°C? (R = 0.08206 L • atm/K • mol)

What is defined as the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated solution?

What is the molar mass of toluene if 0.85 g of toluene depresses the freezing point of 100. g of benzene by 0.47°C? (K_f of benzene is 5.12°C/m.)

Hexachlorophene is used as a disinfectant in germicidal soaps. What mass of hexachlorophene (406.9 g/mol) must be added to 125 g of chloroform to give a solution with a boiling point of 62.60°C? (For chloroform, K_b = 3.63°C/m; boiling point of pure chloroform = 61.20°C)

What volume of water (d = 1.00 g/mL) should be added to 600. mL of ethanol in order to have a solution that boils at 95.0°C? (For ethanol, K_b = 1.22 °C/m; density = 0.789 g/cm³; boiling point = 78.4°C)

What volume of ethanol (density = 0.7893 g/cm³) should be added to 450. mL of water in order to have a solution that freezes at -15.0°C? Assume the density of water is 1.0 g/mL. (For water, K_f = 1.86 °C/m.)

What is defined as the difference between the freezing point of a pure solvent and the freezing point of the solution?

Human blood has a molar concentration of solutes of 0.30 M. What is the osmotic pressure of blood at 25°C? (R = 0.08206 L • atm/K • mol)

Cinnamaldehyde (132.16 g/mol) is used as a flavoring agent. What mass of cinnamaldehyde must be added to 175 g of ethanol to give a solution whose boiling point is 82.57°C? (For ethanol, K_b = 1.22°C/m; boiling point of pure ethanol = 78.37°C)

What is the freezing point of a solution made from 22.0 g of octane (C₈H₁₈) dissolved in 148.0 g of benzene? (For benzene, freezing point = 5.50°C; K_f = 5.12°C/m)

What is the freezing point of an aqueous solution of NaCl that boils at 102.5°C? (For water, K_f = 1.86°C/m; K_b = 0.52°C/m)

Quiz 13: Physical Properties of Solutions

When 12.1 g of the sugar sucrose (a nonelectrolyte) are dissolved in exactly 800 g of water, the solution has a freezing point of -0.082°C. What is the molar mass of sucrose? (Kf of water is 1.86°C/m.)

What mass of ethanol (C2H5OH) , a nonelectrolyte, must be added to 10.0 L of water to give a solution that freezes at -10.0°C? Assume the density of water is 1.0 g/mL. (Kf of water is 1.86°C/m.)

What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? (For water, Kf = 1.86°C/m)

Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. (For water, Kf = 1.86°C/m)

What is the osmotic pressure of a solution that contains 13.7 g of propyl alcohol (C3H7OH) dissolved in enough water to make 500. mL of solution at 27°C? (R = 0.08206 L • atm/K • mol)