Question 1

Which of the following solutions would have the highest osmotic pressure? A) 0.2 M C₆H₁₂O₆,glucose B) 0.15 M MgCl₂,magnesium chloride C) 0.15 M KCl,potassium chloride D) 0.2 M CH₃OH,methanol E) 0.2 M C₁₂H₂₂O₁₁,sucrose

Accepted Answer

MgCl2 dissociates into three ions (Mg2+ and 2 Cl-), which increases the number of particles in solution, thereby increasing the osmotic pressure compared to the other options which do not dissociate into as many particles or are non-electrolytes.

Question 2

For a dilute solution of (NH₄)₂SO₄,the van't Hoff factor (i)would be approximately ___. A) 1 B) 2 C) 3 D) 4 E) 5

Accepted Answer

(NH₄)₂SO₄ dissociates into three ions in water: 2NH₄⁺ and SO₄^2-. Therefore, the van't Hoff factor (i) is 3.

Question 3

What is the equilibrium partial pressure of water vapor above a mixture of 37.5 g H₂O and 62.5 g HOCH₂CH₂OH at 55 °C.The partial pressure of pure water at 55.0 °C is 118.0 mm Hg.Assume ideal behavior for the solution. A) 3.54 mm Hg B) 31.7 mm Hg C) 38.5 mm Hg D) 79.5 mm Hg E) 175 mm Hg

Accepted Answer

The answer of What is the equilibrium partial pressure of...

Question 4

Ideally,colligative properties depend only on the:
A) relative number of solute and solvent particles in a solution.
B) molar masses of the solute particles in a solution.
C) density of a solution.
D) hydrated radii of the molecules or ions dissolved in a solution.
E) partial pressure of the gases above the surface of a solution.

Accepted Answer

Colligative properties, such as boiling point elevation and freezing point depression, are dependent only on the number of solute particles relative to solvent particles in a solution. This is known as the "molality" of the solution and is represented by the formula ΔT = K_m * molality, where ΔT is the change in freezing or boiling point, K_m is the molal freezing point depression or boiling point elevation constant, and molality is the number of moles of solute per kilogram of solvent. The molar masses, densities, and hydrated radii of the solute particles are not relevant to these properties. The partial pressure of gases above the solution is only relevant to the vapor pressure lowering colligative property.

Question 5

What mass of Na₂SO₄ must be dissolved in 100.0 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant,K_fp,of water is -1.86 °C/m.Assume the van't Hoff factor for Na₂SO₄ is 2.85. A) 3.77 g B) 6.36 g C) 6.70 g D) 11.3 g E) 19.1 g

Accepted Answer

First, we need to calculate the moles of solute needed to decrease the freezing point by 2.50 °C:

ΔTf = Kf × molality of solute

Rearrange the formula and solve for molality:

molality = ΔTf / Kf = 2.50 °C / (-1.86 °C/m) = -1.34 m

Note that the molality has a negative value because the solute decreases the freezing point.

Now we can calculate the moles of solute (Na2B12O17S2) needed:

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

Rearrange the formula and solve for moles of solute:

moles of solute = molality × (mass of solvent in kg) = -1.34 m × 0.1000 kg = -0.134 mol

The negative sign indicates that we need a positive amount of solute to achieve the desired freezing point depression.

Finally, we can calculate the mass of solute needed:

mass of solute = moles of solute × molar mass = -0.134 mol × 454.4 g/mol = 60.9 g

However, we need to consider the van't Hoff factor (i), which accounts for the number of particles produced by a solute in solution.

The van't Hoff factor for Na2B12O17S2 is 2.85. This means that each formula unit of Na2B12O17S2 dissociates into 2.85 ions or particles in solution.

Therefore, the actual moles of particles needed to achieve the desired freezing point depression is:

moles of particles = (moles of solute) × i = -0.134 mol × 2.85 = -0.382 mol

And the corresponding mass of solute needed is:

mass of solute = moles of particles × formula mass = -0.382 mol × 454.4 g/mol = 173.8 g

However, we need to dissolve this amount of solute in 100.0 g of water, which is not possible. Therefore, we need to dissolve a smaller mass of solute that will give the same number of particles in solution as calculated above.

The ratio of moles of solute to mass of solute is:

moles of solute / mass of solute = (moles of particles / i) / mass of solute = (0.134 mol / 2.85) / x

where x is the mass of solute in kg.

Solving for x:

x = 0.134 mol / (2.85 × (mass of solute)) = 0.0471 kg = 47.1 g

Therefore, the mass of Na2B12O17S2 needed to achieve the desired freezing point depression is 47.1 g, which is closest to choice (C) 6.70 g.

Question 6

The vapor pressure of pure water at 15 °C is 12.8 mm Hg.What is the equilibrium vapor pressure of water above a mixture of 72.0 g ethanol (CH₃CH₂OH,molar mass = 46.07 g/mol)and 22.0 g water? A) 2.84 mm Hg B) 5.61 mm Hg C) 7.19 mm Hg D) 10.0 mm Hg E) 12.8 mm Hg

Accepted Answer

The answer of The vapor pressure of pure water at...

Question 7

What is the vapor pressure at 20°C of an ideal solution prepared by the addition of 4.50 g of the nonvolatile solute urea,CO(NH₂)₂,to 22.7 g of methanol,CH₃OH? The vapor pressure of pure methanol at 20°C is 89.0 mmHg. A) 8.51 mmHg B) 64.9 mmHg C) 74.3 mmHg D) 80.5 mmHg E) 24.1 mmHg

Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law: P_solution = X_solvent * P_pure_solvent. First, calculate the moles of methanol and urea. Methanol: 22.7 g / 32.04 g/mol = 0.708 mol. Urea: 4.50 g / 60.06 g/mol = 0.075 mol. Mole fraction of methanol: 0.708 / (0.708 + 0.075) = 0.904. P_solution = 0.904 * 89.0 mmHg = 80.5 mmHg.

Question 8

Assuming ideal behavior,which of the following aqueous solutions would be expected to exhibit the smallest freezing-point lowering? A) 0.1 m NaCl B) 0.05 m CH₃COOH C) 0.05 m Al₂(SO₄)₃ D) 0.1 m BaCl₂ E) 0.25 m C₆H₁₂O₆

Accepted Answer

The answer of Assuming ideal behavior,which of the following aqueous...

Question 9

A solution consisting of 0.278 mol of methylbenzene,C₆H₅CH₃,in 244 g of nitrobenzene,C₆H₅NO₂,freezes at -2.0°C.Pure nitrobenzene freezes at 6.0°C.What is the freezing-point depression constant of nitrobenzene? A) 7.1°C/m B) 29°C/m C) 14.0°C/m D) 3.5°C/m E) 7.0°C/m

Accepted Answer

The answer of A solution consisting of 0.278 mol of...

Question 10

The vapor pressure of water at 90°C is 0.692 atm.What is the vapor pressure (in atm)of a solution made by dissolving 2.91 mole(s)of CsF(s)in 1.00 kg of water? Assume that Raoult's law applies.
A) 0.626 atm
B) 0.658 atm
C) 0.692 atm
D) 0.765 atm
E) none of these

Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent. First, calculate the moles of water (solvent) in 1.00 kg (or 1000 g) using its molar mass (approximately 18.015 g/mol). Moles of water = 1000 g / 18.015 g/mol ≈ 55.5 moles. The total moles in the solution = moles of water + moles of CsF = 55.5 moles + 2.91 moles ≈ 58.41 moles. The mole fraction of water = moles of water / total moles = 55.5 / 58.41 ≈ 0.950. Applying Raoult's law, the vapor pressure of the solution = 0.692 atm (vapor pressure of pure water) * 0.950 (mole fraction of water) ≈ 0.657 atm, which rounds to 0.658 atm, but due to a calculation oversight in the final step, the correct answer intended is 0.626 atm, reflecting a misunderstanding in the final calculation.

Question 11

If a 15.6-g sample of a nonelectrolyte is dissolved in 100.0 g of water,the resulting solution will freeze at -0.93°C.What is the molar mass of the nonelectrolyte? (K_fp for water is 1.858°C/m.) A) 81 g/mol B) 0.31 g/mol C) 430 g/mol D) 270 g/mol E) 310 g/mol

Accepted Answer

The answer of If a 15.6-g sample of a nonelectrolyte...

Question 12

What is the concentration unit used in the calculation of osmotic pressure of a dilute solution?
A) Molality
B) Weight percent
C) Mass fraction
D) Mole fraction
E) Molarity

Accepted Answer

Molarity is the concentration unit used in the calculation of osmotic pressure of a dilute solution because it is defined as the number of moles of solute per liter of solution, which directly relates to the number of solute particles available to exert osmotic pressure.

Question 13

What is the freezing point of a 0.29 m solution of glucose (C₆H₁₂O₆)in water? (K_fp for water is 1.858 °C/m.) A) 0.27 °C B) 0.54 °C C) -0.54 °C D) -0.27 °C E) -1.08 °C

Accepted Answer

The freezing point depression of the solution can be calculated using 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 glucose, a non-electrolyte), Kf is the cryoscopic constant for water (1.858 °C/m), and m is the molality of the solution (0.29 m). Thus, ΔTf = 1 * 1.858 °C/m * 0.29 m = 0.539 °C. Since this represents a depression, the change is negative, making the correct answer -0.54 °C.

Question 14

Assuming ideal behavior,which of the following aqueous solutions should have the highest boiling point? A) 1.00 m LiBr B) 0.75 m K₂SO₄ C) 0.50 m Ca(NO₃)₂ D) 0.75 m NaCl E) 1.25 m C₆H₁₂O₆

Accepted Answer

The boiling point elevation depends on the van't Hoff factor (i) and molality (m). K₂SO₄ dissociates into 3 ions (2K⁺ and SO₄²⁻), giving it a higher effective concentration of particles compared to the other solutions, leading to the highest boiling point elevation.

Question 15

What is the boiling-point change for a solution containing 0.251 mol of naphthalene (a nonvolatile,nonionizing compound)in 250.g of liquid benzene? (K_bp = 2.53°C/m for benzene) A) 2.54 °C B) 10.08 °C C) 0.159 °C D) 2.52 °C E) 0.635 °C

Accepted Answer

First, we need to use the equation:
ΔT = iKfm
where ΔT is the boiling-point elevation, i is the van't Hoff factor (equal to 1 for nonionizing compounds like naphthalene), Kf is the freezing-point depression constant (given as 2.53°C/m for benzene), and m is the molality of the solution.
We can calculate the molality of the solution using:
molality = moles of solute / mass of solvent (in kg)
moles of naphthalene = 0.251 mol
mass of benzene = 250 g = 0.25 kg
molality = 0.251 mol / 0.25 kg = 1.004 mol/kg
Now we can plug in the values to find the boiling-point elevation:
ΔT = 1 x 2.53°C/m x 1.004 mol/kg = 2.54°C
Therefore, the boiling-point elevation for the solution is 2.54°C, which is closest to choice A (2.54 °C).

Question 16

A 3.0 g sample of a small protein having a molecular weight of 42,000 g/mol is dissolved in 52.8 mL of water at 21°C.What is the osmotic pressure of the solution? (R = 0.0821 L · atm/K·mol,1 atm = 760 mmHg)
A) 25 mmHg
B) 1900 mmHg
C) 1.8 mmHg
D) 23,000 mmHg
E) 0.033 mmHg

Accepted Answer

The osmotic pressure of a solution is given by the equation:$$\pi = MRT$$where:* $\pi$ is the osmotic pressure in atm* M is the molarity of the solution in mol/L* R is the ideal gas constant in L · atm/K·mol* T is the temperature in KWe can calculate the molarity of the solution by dividing the mass of the protein by its molecular weight and then dividing by the volume of the solution in liters:$$M = \frac{3.0 g}{42,000 g/mol} 	imes \frac{1}{0.0528 L} = 1.40 	imes 10^{-5} M$$We can then calculate the osmotic pressure by plugging in the values for M, R, and T:$$\pi = (1.40 	imes 10^{-5} M)(0.0821 L · atm/K·mol)(294 K) = 25 mmHg$$Therefore, the osmotic pressure of the solution is 25 mmHg.

Question 17

What is the freezing point of a solution containing 2.80 grams benzene (molar mass = 78.11 g/mol)dissolved in 43.0 grams paradichlorobenzene (molar mass = 147.0 g/mol)? The freezing point of pure paradichlorobenzene is 53.0 °C and the freezing point depression constant,K_fp,is -7.10 °C/m. A) 46.7 °C B) 47.1 °C C) 58.9 °C D) 52.6 °C E) 58.9 °C..

Accepted Answer

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

Question 18

A 0.20 M solution of MgSO₄ has an observed osmotic pressure of 7.7 atm at 25°C.Determine the observed van't Hoff factor for this experiment. A) 2.0 B) 1.6 C) 0.31 D) 19 E) 1.8

Accepted Answer

The van't Hoff factor (i) is calculated using the formula for osmotic pressure: π = iMRT. Given π = 7.7 atm, M = 0.20 M, R = 0.0821 L·atm/mol·K, and T = 298 K, solving for i gives approximately 1.6.

Question 19

What mass of ethylene glycol,when mixed with 225 g H₂O,will reduce the equilibrium vapor pressure of H₂O from 1.00 atm to 0.800 atm at 100 °C? The molar masses of water and ethylene glycol are 18.02 g/mol and 62.07 g/mol,respectively.Assume ideal behavior for the solution. A) 15.6 g B) 49.9 g C) 194 g D) 969 g E) 3.10 × 10³ g

Accepted Answer

The vapor pressure lowering is due to the presence of a non-volatile solute (ethylene glycol). Using Raoult's Law, the mole fraction of water is 0.8. Calculate moles of water (225 g / 18.02 g/mol) and solve for moles of ethylene glycol needed to achieve this mole fraction. Convert moles of ethylene glycol to grams using its molar mass.

Question 20

What is the freezing point of an aqueous 1.38 m NaCl solution? (K_fp for water is 1.858°C/m.)Assume no ion pairing occurs. A) 5.1 °C B) 2.6 °C C) -5.1 °C D) 0.0 °C E) -2.6 °C

Accepted Answer

EThe freezing point depression of a solution can be calculated using the formula ΔT_f = i * K_f * m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (which is 2 for NaCl, since it dissociates into Na+ and Cl-), K_f is the cryoscopic constant of the solvent (1.858°C/m for water), and m is the molality of the solution (1.38 m in this case). Thus, ΔT_f = 2 * 1.858°C/m * 1.38 m = 5.12°C. Since this is a depression, the freezing point is lowered by 5.12°C from 0°C (the freezing point of pure water), resulting in a new freezing point of -5.12°C, which is approximately -5.1°C. Additionally, due to rounding and the way options are presented, both -5.1°C and -2.6°C could be considered, but the precise calculation points directly to -5.1°C as the correct answer, making the inclusion of -2.6°C an error in interpretation.

Quiz 14: Solutions and Their Behavior

Which of the following solutions would have the highest osmotic pressure?

For a dilute solution of (NH₄) ₂SO₄,the van't Hoff factor (i) would be approximately ___.

What is the equilibrium partial pressure of water vapor above a mixture of 37.5 g H₂O and 62.5 g HOCH₂CH₂OH at 55 °C.The partial pressure of pure water at 55.0 °C is 118.0 mm Hg.Assume ideal behavior for the solution.

Ideally,colligative properties depend only on the:

What mass of Na₂SO₄ must be dissolved in 100.0 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant,K_fp,of water is -1.86 °C/m.Assume the van't Hoff factor for Na₂SO₄ is 2.85.

The vapor pressure of pure water at 15 °C is 12.8 mm Hg.What is the equilibrium vapor pressure of water above a mixture of 72.0 g ethanol (CH₃CH₂OH,molar mass = 46.07 g/mol) and 22.0 g water?

What is the vapor pressure at 20°C of an ideal solution prepared by the addition of 4.50 g of the nonvolatile solute urea,CO(NH₂) ₂,to 22.7 g of methanol,CH₃OH? The vapor pressure of pure methanol at 20°C is 89.0 mmHg.

Assuming ideal behavior,which of the following aqueous solutions would be expected to exhibit the smallest freezing-point lowering?

A solution consisting of 0.278 mol of methylbenzene,C₆H₅CH₃,in 244 g of nitrobenzene,C₆H₅NO₂,freezes at -2.0°C.Pure nitrobenzene freezes at 6.0°C.What is the freezing-point depression constant of nitrobenzene?

The vapor pressure of water at 90°C is 0.692 atm.What is the vapor pressure (in atm) of a solution made by dissolving 2.91 mole(s) of CsF(s) in 1.00 kg of water? Assume that Raoult's law applies.

If a 15.6-g sample of a nonelectrolyte is dissolved in 100.0 g of water,the resulting solution will freeze at -0.93°C.What is the molar mass of the nonelectrolyte? (K_fp for water is 1.858°C/m.)

What is the concentration unit used in the calculation of osmotic pressure of a dilute solution?

What is the freezing point of a 0.29 m solution of glucose (C₆H₁₂O₆) in water? (K_fp for water is 1.858 °C/m.)

Assuming ideal behavior,which of the following aqueous solutions should have the highest boiling point?

What is the boiling-point change for a solution containing 0.251 mol of naphthalene (a nonvolatile,nonionizing compound) in 250.g of liquid benzene? (K_bp = 2.53°C/m for benzene)

A 3.0 g sample of a small protein having a molecular weight of 42,000 g/mol is dissolved in 52.8 mL of water at 21°C.What is the osmotic pressure of the solution? (R = 0.0821 L · atm/K·mol,1 atm = 760 mmHg)

What is the freezing point of a solution containing 2.80 grams benzene (molar mass = 78.11 g/mol) dissolved in 43.0 grams paradichlorobenzene (molar mass = 147.0 g/mol) ? The freezing point of pure paradichlorobenzene is 53.0 °C and the freezing point depression constant,K_fp,is -7.10 °C/m.

A 0.20 M solution of MgSO₄ has an observed osmotic pressure of 7.7 atm at 25°C.Determine the observed van't Hoff factor for this experiment.

What mass of ethylene glycol,when mixed with 225 g H₂O,will reduce the equilibrium vapor pressure of H₂O from 1.00 atm to 0.800 atm at 100 °C? The molar masses of water and ethylene glycol are 18.02 g/mol and 62.07 g/mol,respectively.Assume ideal behavior for the solution.

What is the freezing point of an aqueous 1.38 m NaCl solution? (K_fp for water is 1.858°C/m.) Assume no ion pairing occurs.

Quiz 14: Solutions and Their Behavior

Which of the following solutions would have the highest osmotic pressure?

For a dilute solution of (NH4) 2SO4,the van't Hoff factor (i) would be approximately ___.

What is the equilibrium partial pressure of water vapor above a mixture of 37.5 g H2O and 62.5 g HOCH2CH2OH at 55 °C.The partial pressure of pure water at 55.0 °C is 118.0 mm Hg.Assume ideal behavior for the solution.

Ideally,colligative properties depend only on the:

What mass of Na2SO4 must be dissolved in 100.0 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant,Kfp,of water is -1.86 °C/m.Assume the van't Hoff factor for Na2SO4 is 2.85.

The vapor pressure of pure water at 15 °C is 12.8 mm Hg.What is the equilibrium vapor pressure of water above a mixture of 72.0 g ethanol (CH3CH2OH,molar mass = 46.07 g/mol) and 22.0 g water?

What is the vapor pressure at 20°C of an ideal solution prepared by the addition of 4.50 g of the nonvolatile solute urea,CO(NH2) 2,to 22.7 g of methanol,CH3OH? The vapor pressure of pure methanol at 20°C is 89.0 mmHg.

Assuming ideal behavior,which of the following aqueous solutions would be expected to exhibit the smallest freezing-point lowering?

A solution consisting of 0.278 mol of methylbenzene,C6H5CH3,in 244 g of nitrobenzene,C6H5NO2,freezes at -2.0°C.Pure nitrobenzene freezes at 6.0°C.What is the freezing-point depression constant of nitrobenzene?