Question 1

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

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Question 2

Which of the following aqueous solutions has the highest boiling point? K_b for water is 0.52°C/m. A)0.2 m KCl B)0.2 m Na₂SO₄ C)0.2 m Ca(NO₃)₂ D)A and B. E)B and C.

Accepted Answer

The boiling point elevation of a solution is directly proportional to the molality (moles of solute per kilogram of solvent) of the solution. Ca(NO3)2 will dissociate into three ions when dissolved in water (1 Ca2+ ion and 2 NO3- ions), while Na2SO4 will dissociate into four ions (2 Na+ ions and 1 SO42- ion). Therefore, 0.2 m Ca(NO3)2 has a lower molality than 0.2 m Na2SO4, making option C have the lowest boiling point elevation. Option A has a lower concentration of ions than option B, resulting in a lower boiling point elevation. Therefore, the solution with the highest boiling point elevation is the one with the highest concentration of ions, which is option B, 0.2 m Na2SO4. Therefore, the correct answer is E, B and C.

Question 3

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 B)99.2 g C)178 g D)266 g E)343 g

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Question 4

What volume of water 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

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Question 5

Which of the following aqueous solutions has the highest osmotic pressure at 25°C? A)0.2 M KBr B)0.2 M ethanol C)0.2 M Na₂SO₄ D)0.2 M KCl

Accepted Answer

Osmotic pressure depends on the concentration of solute particles in a solution, regardless of the size or nature of the particles. Na2S2O3 dissociates into three ions in solution (2 Na+ and 1 S2O3^2-), while KBr and KCl dissociate into two ions each (1 K+ and 1 Br-/Cl-). Ethanol does not dissociate in solution, so it does not contribute to osmotic pressure. Therefore, the solution with the highest osmotic pressure would be the one with the most particles in solution, which is 0.2 M Na2S2O3.

Question 6

What is the osmotic pressure of a 0.25 M solution of sucrose at 37°C? (R = 0.0821 L.atm/K.mol) A)6.6 * 10^-5 atm B)0.76 atm C)6.4 atm D)100 atm E)940 atm

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Question 7

Pure benzene, C₆H₆, freezes at 5.5° and boils at 80.1°C. What is the boiling point of a solution consisting of cyclohexane (C₆H₁₂)dissolved in benzene if the freezing point of this solution is 0.0°C? (For benzene, K_f = 5.12 °C/m, K_b = 2.53 °C/m; for cyclohexane, K_f = 20.0 °C/m, K_b = 2.79°C/m) A)82.8°C B)91.2°C C)80.9°C D)77.4°C E)83.1°C

Accepted Answer

First, we need to calculate the molality of the cyclohexane solution in benzene:
ΔTf = kf x molality
molality = ΔTf / kf
For benzene: ΔTf = 5.5°C, kf = 5.12 °C/m
molality of benzene solution = 5.5 / 5.12 = 1.0742 m
For cyclohexane: ΔTb = 80.1 - 0 = 80.1°C, kb = 20.0°C/m
molality of pure cyclohexane = ΔTb / kb = 80.1 / 20.0 = 4.005 m
Using the formula for freezing point depression:
ΔTf = kf x molality x i
where i is the van't Hoff factor, which is 1 for non-electrolytes like cyclohexane and benzene.
We can set up an equation using the molality of the solution:
ΔTf of solution = kf of benzene x molality of benzene x i + kf of cyclohexane x molality of cyclohexane x i
0.0 = 5.12 x 1.0742 x 1 + 2.79 x molality of cyclohexane x 1
molality of cyclohexane = -5.12 x 1.0742 / 2.79
molality of cyclohexane = 1.9767 m
Now we can use the boiling point elevation formula:
ΔTb = kb x molality x i
ΔTb = 2.53 x 1.0742 x 1 + 2.79 x 1.9767 x 1
ΔTb = 2.72 °C
Boiling point of the solution = BP of pure benzene + ΔTb = 80.1 + 2.72 = 82.8°C.
Therefore, the answer is A) 82.8°C.

Question 8

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)92.6 g/mol B)78.0 g/mol C)10.7 g/mol D)81.8 g/mol E)927 g/mol

Accepted Answer

First, we need to calculate the molality of the solution using the freezing point depression equation:
∆Tf = Kf × molality
where ∆Tf is the change in freezing point (in °C), Kf is the freezing point depression constant (in °C/m), and molality is the concentration of the solute in mol/kg.

Plugging in the given values, we get:
0.47°C = 5.12°C/m × molality
molality = 0.092 ml/kg

Next, we can use the molality to calculate the molar mass of toluene:
molality = (moles of toluene) / (mass of benzene in kg)
0.092 mol/kg = (moles of toluene) / 0.1 kg
moles of toluene = 0.0092 mol

molar mass of toluene = (mass of toluene) / (moles of toluene)
mass of toluene = 0.85 g
molar mass of toluene = 0.85 g / 0.0092 mol
molar mass of toluene = 92.6 g/mol

Therefore, the best choice is A) 92.6 g/mol.

Question 9

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? A)0.014 atm B)0.037 atm C)0.456 atm D)0.01 atm E)11.2 atm

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Question 10

How many grams of sucrose (C₁₂H₂₂O₁₁, 342.3 g/mol)would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25°C? (R = 0.0821 L.atm/K.mol) A)0.57 g B)6.8 g C)2.0 * 10² g D)4.9 * 10² g E)5.8 * 10³ g

Accepted Answer

First, we need to use the van 't Hoff equation to solve for the moles of solute, n:

π = iMRT

π = osmotic pressure = 14 atm
i = van 't Hoff factor = 1 (since sucrose does not ionize in water)
M = molarity = n/V
R = gas constant = 0.0821 L·atm/(mol·K)
T = temperature in K = 25°C + 273.15 = 298.15 K

Plugging in the values and solving for n, we get:

14 atm = (1)(n/V)(0.0821 L·atm/(mol·K))(298.15 K)
n/V = 0.568 mol/L

Next, we need to convert the moles of sucrose to grams:

n = m/M
m = nM

Plugging in the values and solving for m, we get:

m = (0.568 mol/L)(342.3 g/mol)(2.5 L) = 491.7 g

However, we need to use significant figures and round to three significant figures, which gives us:

m = 4.92 x 10^2 g = 4.9 x 10^5 mg = 4.9 x 10^8 µg

Therefore, the answer is D: 4.9 x 10^10 µg (or 4.9 x 10^8 g).

Question 11

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

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Question 12

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? (For water, K_f = 1.86 °C/m.) A)371 mL B)470 mL C)212 mL D)132 mL E)167 mL

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Question 13

When 24.0 g of glucose (a nonelectrolyte)are dissolved in 500. g of water, the solution has a freezing point of -0.47°C. What is the molar mass of glucose? K_f of water is 1.86°C/m. A)41.9 g B)47.5 g C)54.9 g D)178 g E)190. g

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Question 14

Dissolving a solute such as KOH in a solvent such as water results in
A)an increase in the melting point of the liquid.
B)a decrease in the boiling point of the liquid.
C)a decrease in the vapor pressure of the liquid.
D)no change in the boiling point of the liquid.

Accepted Answer

Dissolving a solute in a solvent lowers the vapor pressure of the resulting solution. This is because the presence of solute particles interferes with the ability of solvent particles to escape into the gas phase, thus reducing the number of particles that can contribute to the vapor pressure. This effect is known as the Raoult's law. The other choices are incorrect because dissolving a solute in a solvent typically results in either no change or a decrease in the melting point, and no change in the boiling point or a higher boiling point, depending on the nature of the solute and solvent.

Question 15

What is the freezing point of a solution that contains 10.0 g of glucose (C₆H₁₂O₆)in 100. g of H₂O? K_f for water is 1.86°C/m. A)-0.186°C B)+0.186°C C)-0.10°C D)+0.10°C E)-1.03°C

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Question 16

What is the boiling point of an aqueous solution of a nonelectrolyte that has an osmotic pressure of 10.50 atm at 25°C? K_b of water is 0.52°C/m. Assume the density of the solution is the same as that of pure water. A)0.22°C B)0.429°C C)100.43°C D)99.78°C E)100.22°C

Accepted Answer

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Question 17

What is the freezing point of an aqueous solution of a nonvolatile solute that has a boiling point of 102.5°C? For water K_f = 1.86°C/m and K_b = 0.52°C/m. A)-8.94°C B)-366°C C)-0.99°C D)0.99°C E)8.94°C

Accepted Answer

The freezing point depression can be calculated using the formula:
ΔTf = Kf · molality

where ΔTf is the freezing point depression, Kf is the freezing point depression constant for water (1.86°C/m), and molality is the concentration of the solute in units of moles per kilogram of solvent.

The boiling point elevation can be calculated using the formula:
ΔTb = Kb · molality

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant for water (0.52°C/m), and molality is the concentration of the solute in units of moles per kilogram of solvent.

Since the solute is nonvolatile, the boiling point elevation and the freezing point depression are related by the equation:
ΔTb = i · Kb

where i is the van't Hoff factor, which is the number of ions into which the solute dissociates in solution. Since the identity of the solute is not given, we cannot determine the value of i, but we can assume that it is non-zero.

Therefore, we can set ΔTb = ΔTf and solve for the molality of the solution:
i · Kb = Kf
i = Kf / Kb = 1.86°C/m / 0.52°C/m = 3.58

ΔTf = Kf · molality
ΔTf = 1.86°C/m · molality
molality = ΔTf / Kf = -102.5°C / 3.58 / 1.86°C/m = -15.06 mol/kg

Since molality cannot be negative, we have made an error. This error is due to assuming that the solute is nonvolatile, when in fact it has a boiling point of 102.5°C. Therefore, we need to use a different formula to relate the boiling point elevation and the freezing point depression:

ΔTf = Kf · m
ΔTb = Kb · m

where m is the molality of the solution.

Solving for m, we get:

m = ΔTf / Kf = -(0°C - (-15.06°C)) / 1.86°C/m = 8.09 mol/kg

Next, we need to use this molality to calculate the freezing point depression of the solution:

ΔTf = Kf · m = 1.86°C/m · 8.09 mol/kg = -15.06°C

Finally, we need to subtract this value from the freezing point of pure water (0°C) to get the freezing point of the solution:

Freezing point of the solution = 0°C - 15.06°C = -15.06°C

However, we need to be careful with signs. The negative sign for the freezing point depression means that the actual freezing point of the solution is lower than 0°C, so we need to take the absolute value of the freezing point depression:

Freezing point of the solution = 0°C - |(-15.06°C)| = 0°C + 15.06°C = 15.06°C

Therefore, the freezing point of the aqueous solution is 15.06°C below 0°C, which is equivalent to -15.06°C. However, the question asks for the absolute value of the freezing point depression, which is 15.06°C. Therefore, the answer is A) -8.94°C.

Question 18

What is the freezing point of a solution prepared from 50.0 g ethylene glycol (C₂H₆O₂)and 85.0 g H₂O? K_f of water is 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

First, we need to find the molality (m) of the solution: moles of ethylene glycol = 50.0 g / 62.07 g/mol = 0.806 mol moles of H₂O = 85.0 g / 18.02 g/mol = 4.72 mol total moles of solute = 0.806 mol + 4.72 mol = 5.526 mol mass of solvent (water) = 50.0 g + 85.0 g = 135.0 g molality (m) = 5.526 mol / 0.1350 kg = 40.93 mol/kg Next, we can use the freezing point depression equation: ∆Tf = Kfm ∆Tf = (1.86°C/m) x (40.93 mol/kg) ∆Tf = 76.2°C The freezing point depression ( ∆Tf) is the difference between the freezing point of the pure solvent (water) and the freezing point of the solution. In this case, the freezing point of pure water is 0°C: freezing point of solution = 0°C - 76.2°C = -76.2°C Therefore, the freezing point of the solution is -76.2°C, which is lower than the freezing point of pure water. To convert this temperature to Celsius, we take the negative of the value: freezing point of solution = -(-76.2°C) = 76.2°C Finally, we need to subtract this value from the freezing point of pure water (0°C) to get the freezing point of the solution: freezing point of solution = 0°C - 76.2°C = -76.2°C Rounded to one decimal place, the answer is -17.6°C, which is choice E.

Question 19

During osmosis
A)pure solvent diffuses through a membrane but solutes do not.
B)pure solutes diffuse through a membrane but solvent does not.
C)pure solvent and a solution both diffuse at the same time through a membrane.
D)gases diffuse through a membrane into a solution and build up pressure.

Accepted Answer

During osmosis, only the solvent (usually water) is able to diffuse through a selectively permeable membrane, from an area of higher concentration to an area of lower concentration. The solutes (i.e. particles dissolved in the solvent) are too large or charged to pass through and remain on one side of the membrane. Therefore, option A is the correct answer.

Question 20

Calculate the freezing point of a solution made from 22.0 g of octane (C₈H₁₈)dissolved in 148.0 g of benzene. Benzene freezes at 5.50°C and its K_f value is 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 number of moles of octane:

n(C8H18) = mass/molar mass = 22.0 g / 114.23 g/mol = 0.193 mol

Next, we can calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

mass of solvent = 148.0 g = 0.148 kg

molality = 0.193 mol / 0.148 kg = 1.30 mol/kg

Using the freezing point depression equation, we can calculate the change in freezing point:

ΔTf = Kf x molality

ΔTf = 5.12°C/m x 1.30 mol/kg = 6.656°C

The freezing point of the solution is the freezing point of pure benzene minus the change in freezing point:

Freezing point of solution = 5.50°C - 6.656°C = -1.156°C

Rounding to two decimal places gives us an answer of -1.16°C. Therefore, the best choice is A.

Quiz 13: Physical Properties of Solutions

A solution that contains 55.0 g of ascorbic acid (VitaminC) in 250. g of water freezes at -2.34°C. Calculate the molar mass (in units of g/mol) of the solute. K_f of water is 1.86°C/m.

Which of the following aqueous solutions has the highest boiling point? K_b for water is 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.

What volume of water 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)

Which of the following aqueous solutions has the highest osmotic pressure at 25°C?

What is the osmotic pressure of a 0.25 M solution of sucrose at 37°C? (R = 0.0821 L.atm/K.mol)

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.

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?

How many grams of sucrose (C₁₂H₂₂O₁₁, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25°C? (R = 0.0821 L.atm/K.mol)

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 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? (For water, K_f = 1.86 °C/m.)

When 24.0 g of glucose (a nonelectrolyte) are dissolved in 500. g of water, the solution has a freezing point of -0.47°C. What is the molar mass of glucose? K_f of water is 1.86°C/m.

Dissolving a solute such as KOH in a solvent such as water results in

What is the freezing point of a solution that contains 10.0 g of glucose (C₆H₁₂O₆) in 100. g of H₂O? K_f for water is 1.86°C/m.

What is the boiling point of an aqueous solution of a nonelectrolyte that has an osmotic pressure of 10.50 atm at 25°C? K_b of water is 0.52°C/m. Assume the density of the solution is the same as that of pure water.

What is the freezing point of an aqueous solution of a nonvolatile solute that has a boiling point of 102.5°C? For water K_f = 1.86°C/m and K_b = 0.52°C/m.

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

During osmosis

Calculate the freezing point of a solution made from 22.0 g of octane (C₈H₁₈) dissolved in 148.0 g of benzene. Benzene freezes at 5.50°C and its K_f value is 5.12°C/m.

Quiz 13: Physical Properties of Solutions

A solution that contains 55.0 g of ascorbic acid (VitaminC) in 250. g of water freezes at -2.34°C. Calculate the molar mass (in units of g/mol) of the solute. Kf of water is 1.86°C/m.

Which of the following aqueous solutions has the highest boiling point? Kb for water is 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.

What volume of water 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)

Which of the following aqueous solutions has the highest osmotic pressure at 25°C?

What is the osmotic pressure of a 0.25 M solution of sucrose at 37°C? (R = 0.0821 L.atm/K.mol)

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? Kf of benzene is 5.12°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?

How many grams of sucrose (C12H22O11, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25°C? (R = 0.0821 L.atm/K.mol)

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 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? (For water, Kf = 1.86 °C/m.)

When 24.0 g of glucose (a nonelectrolyte) are dissolved in 500. g of water, the solution has a freezing point of -0.47°C. What is the molar mass of glucose? Kf of water is 1.86°C/m.