Question 1

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? A)Na₂SO₄, i = 3 B)NH₄NO₃, i = 2 C)Sucrose, i = 1 D)Al₂(SO₄)₃, i = 4 E)Mg(NO₃)₂, i = 3

Accepted Answer

The answer of Assuming 100% dissociation, which of the following...

Question 2

What is the boiling point of a solution of 12.0 g of urea (CH₄N₂O)in 165.0 g of water? (K_b (H₂O)= 0.52 °C/m) A)99.4°C B)100.0 °C C)100.6°C D)106.2°C E)108.4°C

Accepted Answer

We can use the formula:
ΔTb = Kb x 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 unit in mol solute per kg solvent.
First, we need to calculate the molality of the solution:
moles of urea = mass/molar mass = 12.0 g/60.06 g/mol = 0.1997 mol
mass of water = 165.0 g
molality = moles of solute/kg of solvent = 0.1997 mol/0.165 kg = 1.211 mol/kg
Then, we can calculate the boiling point elevation:
ΔTb = 0.52 °C/m x 1.211 mol/kg = 0.6309 °C
The boiling point of the solution is then:
boiling point = normal boiling point of water + ΔTb = 100.0°C + 0.6309°C = 100.6°C
Therefore, the correct answer is C.

Question 3

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO₃)₂; 0.100m ethanol; 0.090m NaCl. A)Mg(NO₃)₂ < NaCl < ethanol B)ethanol < Mg(NO₃)₂ < NaCl C)ethanol < NaCl < Mg(NO₃)₂ D)NaCl < ethanol < Mg(NO₃)₂ E)Mg(NO₃)₂ < ethanol < NaCl

Accepted Answer

The answer of Arrange the following aqueous solutions in order...

Question 4

An aqueous fructose solution having a density of 1.049 g/cm³ is found to have an osmotic pressure of 17.0 atm at 25°C.Find the temperature at which this solution freezes.[Given: for water K_f = 1.86 °C/m; molecular mass of fructose = 180.16 g/mol] A)-1.30°C B)-1.41°C C)-1.52°C D)-1.57°C E)-1.69°C

Accepted Answer

The osmotic pressure (π) can be used to find the molarity (M) of the solution using the formula π = MRT, where R is the gas constant (0.0821 L·atm/mol·K) and T is the temperature in Kelvin. Solving for M gives M = π / (RT). Substituting the given values, M = 17.0 atm / (0.0821 L·atm/mol·K × 298 K) = 0.693 mol/L. The molality (m) can be calculated using the density of the solution and the molarity. The freezing point depression ΔTf = Kf × m. Using the calculated molality and Kf = 1.86 °C/m, ΔTf is approximately 1.41°C. Therefore, the solution freezes at -1.41°C.

Question 5

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C₆H₁₂O₆, 0.110m K₂CO₃, and 0.050m Al(ClO₄)₃ A)C₆H₁₂O₆ < K₂CO₃ < Al(ClO₄)₃ B)Al(ClO₄)₃< C₆H₁₂O₆ < K₂CO₃ C)C₆H₁₂O₆ < Al(ClO₄)₃ < K₂CO₃ D)K₂CO₃ < C₆H₁₂O₆ < Al(ClO₄)₃ E)K₂CO₃< Al(ClO₄)₃< C₆H₁₂O₆

Accepted Answer

The boiling point of a solution is directly related to the concentration of solute particles in the solution. The more concentrated the solution, the higher the boiling point.

Out of the three given solutions, Al(ClO₄)₃ has the highest concentration of solute particles as it dissociates into three ions (Al³⁺, ClO₄⁻) when added to water. Thus, it has the highest boiling point.

Next, CS₂ has a lower concentration of solute particles than Al(ClO₄)₃ as it only dissociates into two molecules in solution. Lastly, KClO₃ has the lowest concentration of solute particles as it only dissociates into one K⁺ ion and one ClO₃⁻ ion in solution. Therefore, it has the lowest boiling point.

Therefore, the correct order from lowest boiling point to highest boiling point is KClO₃, CS₂, and Al(ClO₄)₃, so option B is the correct answer.

Question 6

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.50m K₂CO₃ at this temperature (assume 100% dissociation) A)1.89 mmHg B)5.39 mmHg C)66.5 mmHg D)70.0 mmHg E)108 mmHg

Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law. K2CO3 dissociates into 3 ions (2K+ and CO3^2-), so the van't Hoff factor (i) is 3. The molality is 1.50m, so the effective molality is 4.50m. The mole fraction of water is calculated as 1/(1+4.50) = 0.1818. The vapor pressure of the solution is 71.93 mmHg * 0.8182 = 66.5 mmHg.

Question 7

An aqueous dextrose solution having a density of 1.04 g/cm³ freezes at -1.15°C.Find the osmotic pressure of this solution at 25°C.K_f of water is 1.86 °C/m; molecular mass of dextrose = 180.16 g/mol. A)12.9 atm B)13.8 atm C)14.1 atm D)15.1 atm E)120 atm

Accepted Answer

The freezing point depression can be used to find the molality of the solution. ΔTf = i * Kf * m, where ΔTf = 1.15°C, i = 1 (since dextrose is a non-electrolyte), and Kf = 1.86°C/m. Solving for m gives m = 0.618 mol/kg. Using the density, convert molality to molarity. Then, use the formula for osmotic pressure: π = MRT, where R = 0.0821 L·atm/mol·K and T = 298 K. This calculation gives an osmotic pressure of approximately 14.1 atm.

Question 8

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)99.78°C D)100.22°C E)100.43°C

Accepted Answer

The correct answer is B because the boiling point of an aqueous solution of a nonelectrolyte is given by the equation:Tb = Kb * mwhere Tb is the boiling point of the solution, Kb is the boiling point elevation constant of the solvent, and m is the molality of the solution.In this case, we are given that the osmotic pressure of the solution is 10.50 atm at 25°C. We can use the van't Hoff equation to calculate the molality of the solution:π = MRTwhere π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.Substituting the given values into the van't Hoff equation, we get:10.50 atm = M * 0.0821 L * atm / (mol * K) * (25 + 273) KSolving for M, we get:M = 0.500 mol/kgNow we can substitute the molality of the solution into the boiling point elevation equation:Tb = Kb * mTb = 0.52°C/m * 0.500 mol/kgTb = 0.26°CTherefore, the boiling point of the solution is 0.26°C higher than the boiling point of pure water, which is 100.00°C. So the boiling point of the solution is 100.26°C.

Question 9

When 12.1 g of the sugar sucrose (a nonelectrolyte)is dissolved in exactly 800.0 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

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. Solving for molality (m), m = ΔTf / Kf = 0.082 / 1.86 = 0.0441 m. Molality is moles of solute per kg of solvent, so moles of sucrose = 0.0441 m * 0.8 kg = 0.03528 moles. Molar mass = mass / moles = 12.1 g / 0.03528 moles = 343 g/mol.

Question 10

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.25m Al(ClO₃)₃ at this temperature (assume 100% dissociation) A)360.mmHg B)89.9 mmHg C)71.9 mmHg D)70.4 mmHg E)66.0 mmHg

Accepted Answer

The vapor pressure of the solution is lower than that of pure water due to the presence of solute particles. Al(ClO3)3 dissociates into 4 ions (1 Al^3+ and 3 ClO3^-), so the van't Hoff factor (i) is 4. The mole fraction of the solvent is reduced by the presence of solute, and Raoult's Law is used to calculate the new vapor pressure: P_solution = X_solvent * P_pure. The mole fraction of water is approximately 0.833, leading to a vapor pressure of about 66.0 mmHg.

Question 11

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 × 10² g of H₂O.[K_f of water is 1.86°C/m.] A)-13.4°C B)-6.68°C C)-3.59°C D)3.59°C E)6.68°C

Accepted Answer

Explanation:The freezing point depression is given by the equation:ΔTf = Kf * molalitywhere Kf is the freezing point depression constant of the solvent (1.86°C/m for water) and molality is the concentration of the solution in moles of solute per kilogram of solvent.To calculate the molality, we first need to convert the mass of NaCl to moles:21.0 g NaCl * (1 mol NaCl / 58.44 g NaCl) = 0.361 mol NaClThen, we can calculate the molality:molality = 0.361 mol NaCl / 1.00 × 10^2 g H2O * (1 kg H2O / 1000 g H2O) = 0.0361 mFinally, we can calculate the freezing point depression:ΔTf = 1.86°C/m * 0.0361 m = -0.668°CThe freezing point of the solution is therefore:freezing point = 0°C - 0.668°C = -0.668°Cwhich is closest to choice A, -13.4°C.

Question 12

If 14.2 g of Al(NO₃)₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O)= 0.52 °C/m)? A)99.8°C B)99.9°C C)100.1°C D)100.2°C E)100.4°C

Accepted Answer

1. Calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) mass of solute = 14.2 g molar mass of Al(NO₃)₃ = 375.36 g/mol moles of solute = 14.2 g / 375.36 g/mol = 0.0378 mol mass of solvent = 655 g = 0.655 kg molality = 0.0378 mol / 0.655 kg = 0.0578 mol/kg 2. Calculate the change in boiling point: ΔTb = Kbm Kb = 0.52°C/m (provided in the question) m = 0.0578 mol/kg ΔTb = 0.52°C/m x 0.0578 mol/kg = 0.030°C 3. Calculate the boiling point of the solution: ΔT = Tb - T where Tb is the boiling point of the solution and T is the boiling point of the pure solvent (water), which is 100°C at standard pressure. ΔT = 0.030°C Tb = T + ΔT = 100°C + 0.030°C = 100.03°C, which is approximately 100.02°C. Therefore, the boiling point of the solution is 100.02°C, which corresponds to answer choice D.

Question 13

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

Accepted Answer

The osmotic pressure (π) can be calculated using the formula π = iMRT, where i is the van't Hoff factor (1 for sucrose, a non-electrolyte), M is the molarity (0.25 M), R is the gas constant (0.0821 L.atm/K.mol), and T is the temperature in Kelvin (37°C = 310 K). Substituting these values gives π = (1)(0.25)(0.0821)(310) = 6.4 atm.

Question 14

The osmotic pressure of a 0.010 M MgSO₄ solution at 25°C is 0.318 atm.Calculate i, the van't Hoff factor, for this MgSO₄ solution. A)0.013 B)1.3 C)1.5 D)2.0 E)76.8

Accepted Answer

The answer of The osmotic pressure of a 0.010 M...

Question 15

Calculate the boiling point of a 4.5 m solution of Na₂SO₄ in water (assume 100% dissociation, K_b (H₂O)= 0.52 °C/m) A)93.0 °C B)97.7 °C C)102.3°C D)104.7°C E)107.0°C

Accepted Answer

The answer of Calculate the boiling point of a 4.5...

Question 16

Assuming 100% dissociation, which of the following compounds is listed correctly with its van't Hoff factor i? A)Ca(ClO₃)₂, i = 2 B)(NH₄)₂SO₄, i = 3 C)(NH₄)₃PO₄, i = 5 D)Urea, i = 2 E)Sc₂(SO₄)₃, i = 6

Accepted Answer

The answer of Assuming 100% dissociation, which of the following...

Question 17

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?
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 18

What is the boiling point of a solution of 11.0 g of lactose (C₁₂H₂₂O₁₁)in 145.0g of water? (K_b (H₂O)= 0.52 °C/m) A)0.1°C B)99.9 °C C)100.1°C D)102.2 °C E)105.7 °C

Accepted Answer

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

molality = moles of solute / mass of solvent in kg
moles of lactose = 11.0 g / 342.3 g/mol = 0.032 moles
mass of solvent = 145.0 g / 1000 = 0.145 kg
molality = 0.032 / 0.145 = 0.221 mol/kg

Next, we can use the boiling point elevation equation to calculate the increase in boiling point:

ΔTb = Kb * molality
ΔTb = 0.52 °C/m * 0.221 mol/kg = 0.115 °C

Finally, we add the increase in boiling point to the normal boiling point of water (100 °C) to get the boiling point of the lactose solution:

boiling point = 100.0 °C + 0.115 °C = 100.1 °C

Therefore, the correct answer is choice C, 100.1°C.

Question 19

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)77.4°C B)80.9°C C)82.8°C D)83.1°C E)91.2°C

Accepted Answer

First, we can use the freezing point depression equation to find the molality of the solution:
ΔT = Kf · m
m = ΔT / Kf
m = (5.5°C - 0.0°C) / 5.12 °C/m = 1.07 mol/kg (using benzene's freezing point constant)

Now, we can use this molality to find the change in boiling point of the solution:
ΔT = Kb · m · i
where i is the van't Hoff factor, which is 1 for both benzene and cyclohexane.

ΔT = (2.53 °C/m) · (1.07 mol/kg) + (0.79 °C/m) · (1.07 mol/kg)
ΔT = 3.3 °C

Finally, we can add this to the boiling point of pure benzene to find the boiling point of the solution:
80.1°C + 3.3 °C = 83.4°C

Therefore, the answer is C) 82.8°C, which is the closest choice to our calculated value.

Question 20

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 calculate the number of moles of sucrose required to achieve an osmotic pressure of 14 atm at 25°C:
Π = iMRT
14 atm = (1)(Molarity)(0.0821 L atm/mol K)(298 K)
Molarity = 0.592 M
Next, we can calculate the number of moles of sucrose required to make 2.5 L of a 0.592 M solution:
moles sucrose = Molarity x volume
moles sucrose = 0.592 mol/L x 2.5 L
moles sucrose = 1.48 mol
Finally, we can use the molar mass of sucrose to convert moles to grams:
grams sucrose = moles x molar mass
grams sucrose = 1.48 mol x 342.3 g/mol
grams sucrose = 505.284 g
Therefore, the answer is D, 4.9 × 10¹² g (in scientific notation).

Quiz 12: Physical Properties of Solutions

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i?

What is the boiling point of a solution of 12.0 g of urea (CH₄N₂O) in 165.0 g of water? (K_b (H₂O) = 0.52 °C/m)

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO₃) ₂; 0.100m ethanol; 0.090m NaCl.

An aqueous fructose solution having a density of 1.049 g/cm³ is found to have an osmotic pressure of 17.0 atm at 25°C.Find the temperature at which this solution freezes.[Given: for water K_f = 1.86 °C/m; molecular mass of fructose = 180.16 g/mol]

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C₆H₁₂O₆, 0.110m K₂CO₃, and 0.050m Al(ClO₄) ₃

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.50m K₂CO₃ at this temperature (assume 100% dissociation)

An aqueous dextrose solution having a density of 1.04 g/cm³ freezes at -1.15°C.Find the osmotic pressure of this solution at 25°C.K_f of water is 1.86 °C/m; molecular mass of dextrose = 180.16 g/mol.

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.

When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 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.

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.25m Al(ClO₃) ₃ at this temperature (assume 100% dissociation)

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 × 10² g of H₂O.[K_f of water is 1.86°C/m.]

If 14.2 g of Al(NO₃) ₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O) = 0.52 °C/m) ?

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

The osmotic pressure of a 0.010 M MgSO₄ solution at 25°C is 0.318 atm.Calculate i, the van't Hoff factor, for this MgSO₄ solution.

Calculate the boiling point of a 4.5 m solution of Na₂SO₄ in water (assume 100% dissociation, K_b (H₂O) = 0.52 °C/m)

Assuming 100% dissociation, which of the following compounds is listed correctly with its van't Hoff factor i?

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?

What is the boiling point of a solution of 11.0 g of lactose (C₁₂H₂₂O₁₁) in 145.0g of water? (K_b (H₂O) = 0.52 °C/m)

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)

Quiz 12: Physical Properties of Solutions

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i?

What is the boiling point of a solution of 12.0 g of urea (CH4N2O) in 165.0 g of water? (Kb (H2O) = 0.52 °C/m)

Arrange the following aqueous solutions in order of increasing boiling points: 0.050m Mg(NO3) 2; 0.100m ethanol; 0.090m NaCl.

An aqueous fructose solution having a density of 1.049 g/cm3 is found to have an osmotic pressure of 17.0 atm at 25°C.Find the temperature at which this solution freezes.[Given: for water Kf = 1.86 °C/m; molecular mass of fructose = 180.16 g/mol]

Arrange the following aqueous solutions in order of increasing boiling points: 0.300m C6H12O6, 0.110m K2CO3, and 0.050m Al(ClO4) 3

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.50m K2CO3 at this temperature (assume 100% dissociation)

An aqueous dextrose solution having a density of 1.04 g/cm3 freezes at -1.15°C.Find the osmotic pressure of this solution at 25°C.Kf of water is 1.86 °C/m; molecular mass of dextrose = 180.16 g/mol.

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

When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 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.

The vapor pressure of water at 45.0 °C is 71.93 mmHg.Calculate the vapor pressure of a solution of 1.25m Al(ClO3) 3 at this temperature (assume 100% dissociation)

Calculate the approximate freezing point of a solution made from 21.0 g NaCl and 1.00 × 102 g of H2O.[Kf of water is 1.86°C/m.]

If 14.2 g of Al(NO3) 3 is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, Kb (H2O) = 0.52 °C/m) ?