Calculate the vapor pressure of water over each of the following solutions of glycerol, C 3 H 8 O 3 , at 28°C ( vapor pressure of pure water = 28 . 35 mm Hg ) . Glycerol can be assumed to be nonvolatile. (a) percent glycerol by mass = 30 .6 % (b) 2.74 m (c) mole fraction of glycerol = 0. 188

Question

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Answer 1

Textbook Question

Chapter 10, Problem 34QAP

Calculate the vapor pressure of water over each of the following solutions of glycerol, C₃H₈O₃, at 28°C $( vapor pressure of pure water = 28 . 35 mm Hg )$ . Glycerol can be assumed to be nonvolatile.

(a) $percent glycerol by mass = 30 .6 %$

(b) 2.74 m

(c) $mole fraction of glycerol = 0. 188$

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

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Answer 2

Textbook Question

Chapter 10, Problem 34QAP

Calculate the vapor pressure of water over each of the following solutions of glycerol, C₃H₈O₃, at 28°C $( vapor pressure of pure water = 28 . 35 mm Hg )$ . Glycerol can be assumed to be nonvolatile.

(a) $percent glycerol by mass = 30 .6 %$

(b) 2.74 m

(c) $mole fraction of glycerol = 0. 188$

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

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Answer 3

Textbook Question

Chapter 10, Problem 34QAP

Calculate the vapor pressure of water over each of the following solutions of glycerol, C₃H₈O₃, at 28°C $( vapor pressure of pure water = 28 . 35 mm Hg )$ . Glycerol can be assumed to be nonvolatile.

(a) $percent glycerol by mass = 30 .6 %$

(b) 2.74 m

(c) $mole fraction of glycerol = 0. 188$

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

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Answer 4

Textbook Question

Chapter 10, Problem 34QAP

Calculate the vapor pressure of water over each of the following solutions of glycerol, C₃H₈O₃, at 28°C $( vapor pressure of pure water = 28 . 35 mm Hg )$ . Glycerol can be assumed to be nonvolatile.

(a) $percent glycerol by mass = 30 .6 %$

(b) 2.74 m

(c) $mole fraction of glycerol = 0. 188$

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

Answer 8

Expert Solution

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

Concept introduction:

Formula to calculate moles of a component is −

$Moles=MassMolar mass$

Mole fraction of a component in a solution can be found out by dividing moles of the component by summation of mole of all components in the solution.

$Mole fraction (X)=Moles of componentSummation of moles of all components$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent, and $P0solvent$ is vapor pressure of pure solvent.

Answer 13

Answer to Problem 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6%$ is $26.08 mm Hg$ .

Answer 14

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6%$ . It means in $100 g$ solution mass of glycerol is $30.6 g$

$Mass of solvent(water)=100 g−30.6 g=69.4 g$

So, mass of water is $69.4 g$ .

Moles of each component of the solution are to be calculated as follows:

Mass of glycerol $=30.6 g$

Molar mass of glycerol $=92.09 g mol−1$

$Moles of glycerol=Mass of glycerolMolar mass of glycerol=30.6 g92.09 g mol−1=0.33 mol$

So, moles of glycerol is $0.33 mol$ .

Mass of water $=69.4 g$

Molar mass of water $=18.02 g mol−1$

$Moles of water=Mass of waterMolar mass of water=69.4 g18.02 g mol−1=3.85 mol$

So, moles of water is $3.85 mol$ .

Calculation of mole fraction of water

$Mole fraction of water(Xwater)=Moles of waterMoles of water+Moles of glycerol=3.85 mol3.85 mol+0.33 mol=3.85 mol4.18 mol=0.92$

So, mole fraction of water is $0.92$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.92$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.92×28.35 mm Hg=26.08 mm Hg$

Hence, vapor pressure of water is $26.08 mm Hg$ .

Answer 15

Expert Solution

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74 m$ .

Concept introduction:

Molality is one way to define the concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$Molality=Moles of soluteMass of solvent(kg)$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer 20

Answer to Problem 34QAP

If molality of the solution is $2.74 m$ , vapor pressure of water will be $26.93 mm Hg$ .

Answer 21

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74 m$ . So mass of solvent is $1000 g(1 kg)$

Moles of solute (glycerol) can be calculated as follows:

$Molality=Moles of soluteMass of solvent(kg)2.74 m=Moles of solute1 kgMoles of solute=2.74 mol kg−1 kg=2.74 mol$

So, moles of solute (glycerol) is $2.74 mol$

Moles of solvent can be calculated as follows:

Molar mass of water is $=18.02 g mol−1$

Mass of solvent (water) =1000 g

$Moles of solvent(water)=1000 g18.02 g mol−1=55.5 mol$

So, moles of solvent is $55.5 mol$ .

Mole fraction of solvent can be calculated as follows:

$Mole fraction of solvent(Xwater)=Moles of solvent (water)Moles of solvent (water)+Moles of solute (glycerol)=55.5 mol55.5 mol+2.74 mol=0.95$

So, mole fraction of solvent (water) is $0.95$ .

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.95$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.95×28.35 mm Hg=26.93 mm Hg$

Hence, vapor pressure of water is $26.93 mm Hg$ .

Answer 22

Expert Solution

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.

Mole fraction of water $Xwater=0.812$

Vapor pressure of pure solvent $=28.35 mm Hg$

Vapor pressure of water can be calculated as follows:

$Pwater=Xwater×P0water=0.812×28.35 mm Hg=23.02 mm Hg$

Hence, vapor pressure of water is $23.02 mm Hg$ .

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$ .

Concept introduction:

For a solution containing solute (glycerol) and solvent (water), summation of mole fraction is equal to one.

$Xethylene glycol+Xwater=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent

Answer 27

Answer to Problem 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02 mm Hg$ .

Answer 28

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $(X glycerol) is 0.188$ and vapor pressure of pure water is $28.35 mm Hg$

Addition of mole fraction of all components of a solution is equal to one.

$Xethylene glycol+Xwater=1Xwater=1−Xethylene glycol=1−0.188=0.812$

So, mole fraction of water is $0.812$ .

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the product of mole fraction of solvent and vapor pressure of pure solvent.

$Psolvent=Xsolvent×P0solvent$

Where, $Psolvent$ is vapor pressure of solvent in a mixture, $Xsolvent$ is mole fraction of solvent and $P0solvent$ is vapor pressure of pure solvent.