The freezing points of wine and vodka have to be calculated. Concept Introduction: Freezing point depression: The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below. Δ T f = T f ∘ − T f = K f m c Where, Δ T f is freezing point depression, T f is the freezing point of solution, T f ∘ is the freezing point of pure solvent, K f is the freezing point depression constant and m c is the colligative molality.

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Answer 1

Question

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

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Answer 2

Question

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

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Answer 3

Question

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

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Answer 4

Question

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

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Answer 5

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

Answer 6

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(2.34 mc)0∘C−(Tf)=4.4 KTf=-4.4oC

For vodka:

ΔTf=Tf∘−Tf=Kfmc=(1.86 K.mc−1)×(11.43 mc)0∘C−(Tf)=21 KTf=-21oC

The value of ΔTf in kelvins is equal to the value of Δtf in degrees Celsius.

Therefore, the freezing points of wine and vodka are -4.4oC and -21oC respectively.

Answer 7

Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent. This effect is called as the freezing point depression. The mathematical relationship is given below.

ΔTf=Tf∘−Tf=Kfmc

Where, ΔTf is freezing point depression, Tf is the freezing point of solution, Tf∘ is the freezing point of pure solvent, Kf is the freezing point depression constant and mc is the colligative molality.

Answer 8

Expert Solution & Answer

Answer to Problem 16.70P

The freezing points of wine and vodka are -4.4oC and -21oC respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The value of Kf for water is 1.86 K.mc−1 and its freezing point is 0∘C.

Wine and vodka are about 12% and 40% of ethanol by volume. That means, wine contains 12 mL of ethanol and 88 mL of water. Similarly, vodka contains 40 mL of ethanol and 60 mL of water.

Given that, the densities of ethanol and water are 0.79 g.mL−1 and 1.00 g.mL−1 respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

Mass=Density×VolumeMass of ethanol in wine=0.79 g.mL−1×12 mL=9.48 gMass of water in wine=1.00 g.mL−1×88 mL=88 gMass of ethanol in vodka=0.79 g.mL−1×40 mL=31.6 gMass of water in vodka=1.00 g.mL−1×60 mL=60 g

The colligative molality of wine and vodka can be calculated as given below.

Molality=Moleskg of solventMolality of wine=Mass of ethanolMolar mass of ethanolkg of solventMolality of wine=9.48 g46.07 g.mL−188 g×(1 kg1000 g)Molality of wine=2.34 mMolality of vodka=Mass of ethanolMolar mass of ethanolkg of solventMolality of vodka= 31.6g46.07 g.mL−160 g×(1 kg1000 g)Molality of vodka=11.43 m