Question 1

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 2

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 freezing-point depression is defined as the difference between the freezing point of a pure solvent and the freezing point of the solution.

Question 3

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 freezing point depression can be calculated using the formula: ΔTf = i*Kf*m, where i is the van't Hoff factor (assume 1 for sucrose), Kf is the freezing point depression constant for water (1.86°C/m), and m is the molality of the solution.

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg 
mass of solvent = 175 g = 0.175 kg

molality = 0.31 mol / 0.175 kg 
molality = 1.77 m

Now we can use the freezing point depression formula:

ΔTf = i*Kf*m 
ΔTf = 1 * 1.86°C/m * 1.77 m 
ΔTf = 3.29 ≈ 3.3°C

The freezing point of pure water is 0°C, so the freezing point of the solution would be:

Freezing point = 0°C - ΔTf 
Freezing point = 0°C - 3.3°C = -3.3°C

Therefore, the correct answer is E (-3.3°C).

Question 4

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, calculate the molality of the solution:

moles of benzaldehyde = 75.00 g / 106.12 g/mol = 0.707 mol
mass of ethanol = 850.0 g
molality = 0.707 mol / 0.850 kg = 0.831 m

Next, calculate the freezing point depression:

ΔTf = Kf × molality
ΔTf = 1.99°C/m × 0.831 m = 1.654°C

Finally, calculate the freezing point of the solution:

freezing point of solution = freezing point of pure solvent - ΔTf
freezing point of solution = -117.3°C - 1.654°C = -119.0°C

Therefore, the freezing point of the solution prepared by dissolving 75.00 g of benzaldehyde in 850.0 g of ethanol is -119.0°C. The correct answer is choice C.

Question 5

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

First, we need to find the molality of the NaCl solution using its boiling point elevation.

ΔTb = Kb * molality

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant of water (0.52°C/m), and molality is the molality of the NaCl solution.

Thus, molality = ΔTb / Kb = (102.5 - 100) / 0.52 = 4.81 mol/kg

Next, we can use the molality to find the freezing point depression of the NaCl solution.

ΔTf = Kf * molality

where ΔTf is the freezing point depression, Kf is the freezing point depression constant of water (1.86°C/m).

Thus, ΔTf = Kf * molality = 1.86 * 4.81 = 8.94°C

Finally, we can find the freezing point of the NaCl solution by subtracting the freezing point depression from the freezing point of pure water (0°C).

Freezing point = 0°C - ΔTf = 0°C - 8.94°C = -8.94°C ≈ -8.9°C

Therefore, the answer is E (-8.9°C).

Question 6

Barbiturates are synthetic drugs used as sedatives and hypnotics.Barbital (184.2g/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 7

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 answer of When 12.1 g of the sugar sucrose...

Question 8

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 answer of Human blood has a molar concentration of...

Question 9

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

First, we need to calculate the boiling point elevation caused by the addition of hexachlorophene to chloroform.
ΔT = i*K*m
where i=1 (since hexachlorophene doesn't ionize in solution)
K = 3.63°C/m (given)
m = molality of the solution (mol solute/kg solvent)
We need to find the amount of hexachlorophene that gives a solution with a boiling point of 62.60°C when added to 125 g of chloroform. Since we're given the molecular weight of hexachlorophene, we can calculate the number of moles of hexachlorophene needed.
n(hexachlorophene) = m(hexachlorophene)/M(hexachlorophene)
We don't know m(hexachlorophene) yet, but we can calculate the amount of chloroform in moles:
n(chloroform) = m(chloroform)/M(chloroform) = 125 g / 119.38 g/mol = 1.048 mol
We also know that the molality of the solution is:
molality = moles of solute / kg of solvent
We can solve for m(hexachlorophene) using this relationship:
m(hexachlorophene) = molality * kg of solvent
We know that 125 g of chloroform is equivalent to 0.125 kg, so we can rearrange the above equation to solve for molality:
molality = moles of solute / (kg of solvent - mass of solute)
If x is the mass of hexachlorophene we need to add, then:
molality = n(hexachlorophene) / (0.125-x) kg
Setting ΔT = 62.60 - 61.20 = 1.40°C, we can solve for x:
1.40 = (1)(3.63)(n(hexachlorophene) / (0.125-x))
x = 19.6 g (rounded to one decimal place)

Question 10

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 11

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 12

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 called osmosis. Options A and E are related to the movement of molecules through a substance, but not specifically through a membrane. Option B, semipermeability, refers to the property of a membrane that allows certain substances to pass through while blocking others. Option C, osmotic pressure, is related to osmosis but specifically refers to the pressure required to prevent the movement of solvent molecules through a membrane in order to maintain equilibrium between the two solutions.

Question 13

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 1.0g/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

First, we need to calculate the molality (m) of the solution:
ΔTf = Kfm
-15.0°C - 0°C = (-1) * 1.86°C/m * m
m = 8.06 mol/kg

Next, we need to convert the volume of water to mass:
mass of water = 450 mL * 1.0 g/mL = 450 g

Then, we can use the molality to calculate the mass of ethanol needed:
m = moles of solute / mass of solvent (in kg)
moles of solute = m * mass of solvent (in kg)
moles of solute = 8.06 mol/kg * 0.45 kg = 3.627 mol

mass of ethanol = moles of solute * molar mass of ethanol
mass of ethanol = 3.627 mol * 46.07 g/mol = 166.8 g

Finally, we can convert the mass of ethanol to volume:
volume of ethanol = mass of ethanol / density of ethanol
volume of ethanol = 166.8 g / 0.7893 g/mL = 211.3 mL

Therefore, the best choice is C, 212 mL (rounded to the nearest mL).

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 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 16

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

Explanation:ΔT_f = K_f x m0.47°C = 5.12°C/m x mm = 0.092 mol/kgM = (mass of solute/moles of solute) x 1000 g/kgM = (9.2 g/0.092 mol) x 1000 g/kgM = 100 g/mol

Question 17

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

The answer of What mass of ethanol (C₂H₅OH),a nonelectrolyte,must be...

Question 18

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 19

A solution that contains 55.0 g of ascorbic acid (vitamin C)in 250.g of water freezes at -2.34°C.Calculate the molar mass (in units of g/mol)of the solute. (For water,K_f = 1.86°C/m) A)1.26 g/mol B)10.9 g/mol C)43.6 g/mol D)175 g/mol E)277 g/mol

Accepted Answer

The freezing point depression formula is ΔTf = i * Kf * m. Here, ΔTf = 2.34°C, Kf = 1.86°C/m, and i = 1 (since ascorbic acid is a non-electrolyte). Solving for molality (m), we get m = 2.34 / 1.86 = 1.258 m. The molality is moles of solute per kg of solvent, so moles of solute = 1.258 m * 0.250 kg = 0.3145 moles. The molar mass is mass of solute / moles of solute = 55.0 g / 0.3145 moles = 175 g/mol.

Question 20

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:
molality (m) = moles of solute / mass of solvent in kg
Molar mass of octane = 114.23 g/mol
Number of moles of octane = 22.0 g / 114.23 g/mol = 0.1927 mol
Mass of benzene = 148.0 g = 0.1480 kg
molality = 0.1927 mol / 0.1480 kg = 1.301 mol/kg
Next, we can use the freezing point depression equation:
ΔTf = Kf × m
where ΔTf is the change in freezing point, Kf is the freezing point depression constant, and m is the molality.
Plugging in the values we have:
ΔTf = 5.12°C/m × 1.301 mol/kg
ΔTf = 6.67°C
Since we want to find the freezing point of the solution, we need to subtract ΔTf from the freezing point of pure benzene:
Freezing point of solution = 5.50°C - 6.67°C = -1.17°C, which rounds to -1.16°C.
Therefore, the answer is A) -1.16°C.

Quiz 13: Physical Properties of Solutions

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)

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

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 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)

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.)

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)

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)

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

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 1.0g/mL.(For water,K_f = 1.86 °C/m.)

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 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)

g of benzene by 0.47°C? (K_f of benzene is 5.12°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 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 solution that contains 55.0 g of ascorbic acid (vitamin C) in 250.g of water freezes at -2.34°C.Calculate the molar mass (in units of g/mol) of the solute. (For water,K_f = 1.86°C/m)

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)

Quiz 13: Physical Properties of Solutions

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)

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

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 freezing point of an aqueous solution of NaCl that boils at 102.5°C? (For water,Kf = 1.86°C/m; Kb = 0.52°C/m)

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.)

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)

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,Kb = 3.63°C/m; boiling point of pure chloroform = 61.20°C)

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,Kb = 1.22°C/m; boiling point of pure ethanol = 78.37°C)

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

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 1.0g/mL.(For water,Kf = 1.86 °C/m.)

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,Kb = 1.22 °C/m; density = 0.789 g/cm3; boiling point = 78.4°C)

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)

g of benzene by 0.47°C? (Kf of benzene is 5.12°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.)