Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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Chapter 13, Problem 13.111P

(a)

Interpretation Introduction

Interpretation:

The molality and van’t Hoff factor for the aqueous KCl solution are to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

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

Molality=amount(mol)ofsolutemass(kg)ofsolvent (1)

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

ΔTf=ikfm (2)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

The van’t Hoff factor is a measure of the effect of solute on the colligative properties. It is represented by i. It is defined as the ratio of measured values for the electrolytic solution to the expected values for non-electrolytic solutions. It is a unitless quantity.

(a)

Expert Solution
Check Mark

Answer to Problem 13.111P

The molality and van’t Hoff factor for the aqueous KCl solution are 0.000068m and 2.

Explanation of Solution

Consider the mass of the solution to be. 100g.

The formula to calculate the mass of the compound is as follows:

Mass of compound=(Mass of solution)(mass %of compound100%) (3)

Substitute 100g for the mass of solution and 0.500 % for the mass percent of the compound in equation (3) to calculate the mass of KCl.

Mass of KCl=(100 g)(0.500 %100%)=0.500g

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

Moles of compound=Given massMolar mass (4)

Substitute 0.500g for the given mass and 74.55g/mol for the molar mass in equation (4) to calculate the moles of KCl.

Moles of KCl=(0.500g)(1mol74.55g)=0.0067069081mol

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

Mass of solution=Mass of solute+Mass of solvent (5)

Rearrange equation (5) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solutionMass of solute (6)

Substitute 100g for the mass of solution and 0.500g for the mass of solute in equation (6) to calculate the mass of water.

Mass of water=100g0.500 g=99.500 g

Substitute 0.0067069081mol for the amount of solute and 99.500 g for the mass of solvent in equation (1) to calculate the molality of KCl.

Molality of KCl=(0.0067069081mol99.500g)(103g1kg)=0.0000674061m=0.000068m

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

ΔTf=Tf(solvent)Tf(solution) (7)

Substitute 0°C for Tf(solvent) and 0.234°C for Tf(solution) in equation (7).

ΔTf=0°C(0.234°C)=0.234°C

Rearrange equation (2) to calculate the van’t Hoff factor of the solution as follows:

i=ΔTfmkf (8)

Substitute 0.234°C for ΔTf, 0.0000674061m for m and 1.86°C/m for kf in equation (8).

i=0.593°C(0.0000674061m)(1.86°C/m)=1.866398=1.87

KCl dissociates to give one K+ and one Cl so its van’t Hoff factor should be close to 2.

Conclusion

The molality and van’t Hoff factor for the aqueous KCl solution are 0.000068m and 2.

(b)

Interpretation Introduction

Interpretation:

The molality and van’t Hoff factor for the aqueous H2SO4 solution are to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

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

Molality=amount(mol)ofsolutemass(kg)ofsolvent (1)

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

ΔTf=ikfm (2)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

The van’t Hoff factor is a measure of the effect of solute on the colligative properties. It is represented by i. It is defined as the ratio of measured values for the electrolytic solution to the expected values for non-electrolytic solutions. It is a unitless quantity.

(b)

Expert Solution
Check Mark

Answer to Problem 13.111P

The molality and van’t Hoff factor for the aqueous H2SO4 solution are 0.10298746m and 3.

Explanation of Solution

Consider the mass of the solution to be. 100g.

Substitute 100g for the mass of solution and % for the mass percent of the compound in equation (3) to calculate the mass of H2SO4.

Mass of H2SO4=(100 g)(%100%)=1g

Substitute 1g for the given mass and 98.08g/mol for the molar mass in equation (4) to calculate the moles of H2SO4.

Moles of H2SO4=(1g)(1mol98.08g)=0.0101957586mol

Substitute 100g for the mass of solution and 1g for the mass of solute in equation (6) to calculate the mass of water.

Mass of water=100g1 g=99 g

Substitute 0.0101957586mol for the amount of solute and 99 g for the mass of solvent in equation (1) to calculate the molality of H2SO4.

Molality of H2SO4=(0.0101957586mol99g)(103g1kg)=0.10298746m=0.1023m

Substitute 0°C for Tf(solvent) and 0.423°C for Tf(solution) in equation (7).

ΔTf=0°C(0.423°C)=0.423°C

Substitute 0.423°C for ΔTf, 0.10298746m for m and 1.86°C/m for kf in equation (8).

i=0.423°C(0.10298746m)(1.86°C/m)=2.20842=2.21

H2SO4 is a strong acid so it completely dissociates to give two H+ and one SO42 and therefore its van’t Hoff factor is 3.

Conclusion

The molality and van’t Hoff factor for the aqueous CH3COOH solution are 0.10298746m and 3.

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Chapter 13 Solutions

Chemistry: The Molecular Nature of Matter and Change

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