Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.133P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the solute is to be calculated.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

The formula to calculate the change in boiling point is as follows:

  ΔTb=ikbm        (1)

Here,

ΔTb is the change in boiling point.

i is van’t Hoff factor.

kb is the boiling point elevation constant.

m is the molality of the solution.

(a)

Expert Solution
Check Mark

Answer to Problem 13.133P

68g/mol is the molar mass of the solute.

Explanation of Solution

The formula to calculate the change in boiling point is as follows:

  ΔTb=Tb(solution)Tb(solvent)        (2)

Substitute 100.45°C for Tb(solution) and 100°C for Tb(solvent) in equation (2).

  ΔTb=100.45°C100°C=0.45°C

The solute is a nonvolatile non-electrolyte so its van’t Hoff factor is 1.

Rearrange equation (1) to calculate the molarity of the solution as follows:

  m=ΔTbikb        (3)

Substitute 1 for i, 0.45°C for ΔTb, 0.512°C/m for kb in equation (3).

  m=0.45°C(1)(0.512°C/m)=0.878906m

The density of the solution is calculated as follows:

  Density of solution(ρ)=Mass (M)of solutionVolume (V)of solution        (4)

Rearrange equation (4) to calculate the mass of the solution as follows:

  Mass of solution=(Density of solution)(Volume of solution)        (5)

Substitute 25mL for the mass of solution and 0.997g/mL for the density of the solution in equation (5) to calculate the mass of water.

  Mass of water=(25mL)(0.997g1mL)(1kg103g)=0.0249250kg

The formula to calculate the molality of the solution is as follows:

  Molality=amount(mol)ofsolutemass(kg)ofsolvent        (6)

Rearrange equation (6) to calculate the moles of solute as follows:

  Amountofsolute=(Molality)(massofsolvent)        (7)

Substitute 0.878906m for the molality and 0.0249250kg for the mass of solvent in equation (7) to calculate the amount of solute.

  Amount of solute=(0.878906m)(0.0249250kg)=0.0219067mol

The formula to calculate the number of moles is as follows:

  Number of moles=Given massMolar mass        (8)

Rearrange equation (8) to calculate the molar mass as follows:

  Molar mass=Given massNumber of moles        (9)

Substitute 1.50g for the given mass and 0.0219067mol for the number of moles in equation (9).

  Molar mass=1.50g0.0219067mol=68.4722g/mol68g/mol.

Conclusion

Boiling point elevation is a colligative property as it depends on the number of moles of solute.

(b)

Interpretation Introduction

Interpretation:

The molar mass of the solute is to be calculated.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

The formula to calculate the change in boiling point is as follows:

  ΔTb=ikbm        (1)

Here,

ΔTb is the change in boiling point.

i is van’t Hoff factor.

kb is the boiling point elevation constant.

m is the molality of the solution.

(b)

Expert Solution
Check Mark

Answer to Problem 13.133P

2.1×102g/mol is the  molar mass of the solute.

Explanation of Solution

The solute is ionic with general formula AB2 or A2B so it will dissociate into three ions and therefore its van’t Hoff factor is 3.

Substitute 3 for i, 0.45°C for ΔTb, 0.512°C/m for kb in equation (3).

  m=0.45°C(3)(0.512°C/m)=0.29296875m

Substitute 0.29296875m for the molality and 0.0249250kg for the mass of solvent in equation (7) to calculate the amount of solute.

  Amount of solute=(0.29296875m)(0.0249250kg)=0.00730225mol

Substitute 1.50g for the given mass and 0.00730225mol for the number of moles in equation (9).

  Molar mass=1.50g0.00730225mol=205.416g/mol2.1×102g/mol.

Conclusion

Boiling point elevation depends on the moles of solute and therefore it is a colligative property.

(c)

Interpretation Introduction

Interpretation:

The difference between the actual formula mass and that calculated from the boiling point elevation is to be explained.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

(c)

Expert Solution
Check Mark

Answer to Problem 13.133P

The actual molar mass of CaN2O6 is 164.10g/mol that lies in between that calculated when the compound is nonelectrolyte and that calculated when it is an electrolyte. This indicates the compound is ionic but dissociation is incomplete.

Explanation of Solution

The molar mass of CaN2O6 is 164.10g/mol. This molar mass is less than that calculated when the compound is considered to be ionic (2.1×102g/mol) but it is more than that calculated when the compound is considered as a nonelectrolyte (68g/mol). This indicates that the compound dissociates into its ions in the solution and therefore it is an electrolyte. But the ions do not completely dissociate in the solution.

Conclusion

Boiling point elevation is a colligative property as it depends on the number of moles of solute.

(d)

Interpretation Introduction

Interpretation:

The van’t Hoff factor for the solution is to be calculated.

Concept introduction:

The formula to relate the elevation in boiling point and van’t Hoff factor is as follows:

  ΔTb=ikbm        (1)

Here,

ΔTb is the change in boiling point.

i is van’t Hoff factor.

kb is the boiling point elevation constant.

m is the molality of the solution.

(d)

Expert Solution
Check Mark

Answer to Problem 13.133P

2.4 is the van’t Hoff factor.

Explanation of Solution

Substitute 1.50g for the given mass and 164.10g/mol for the molar mass in equation (8) to calculate the moles of CaN2O6.

  Moles of CaN2O6=1.50g164.10g/mol=0.0091408mol

Substitute 0.0091408mol for the moles of solute and 0.0249250kg for the mass of solvent in equation (6) to calculate the molality of CaN2O6.

  Molality of CaN2O6=0.0091408mol0.0249250kg=0.3667314 m

Rearrange equation (1) to calculate the van’t Hoff factor is as follows:

  i=ΔTbkbm        (10)

Substitute 0.45°C for ΔTb, 0.512°C/m for kb, 0.3667314 m for m in equation (10).

  i=0.45°C(0.512°C/m)(0.3667314 m)=2.39659 2.4.

Conclusion

The van’t Hoff factor for the solution is 2.4.

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Chapter 13 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY