The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure. Concept Introduction: The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as P 1 = χ 1 P 1 ∘ Here, P 1 ∘ is the vapor pressure of the pure solvent, P 1 is the pressure exerted by a solvent over a solution, and χ 1 is the mole fraction of the solvent. The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as P t o t a l = P A + P B or P t o t a l = χ A P ∘ A + χ B P ∘ B The mole fraction of compound can be calculated as χ A = n A n A + n B

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Answer 1

Question

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

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Answer 2

Question

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

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Answer 3

Question

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

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Answer 4

Question

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

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Answer 5

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

Answer 6

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

Answer 7

Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P 1=χ1P1∘$

Here, $P1∘$ is the vapor pressure of the pure solvent, $P1$ is the pressure exerted by a solvent over a solution, and $χ1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$

The mole fraction of compound can be calculated as

$χA=nAnA+nB$

Answer 8

Expert Solution & Answer

Answer to Problem 124AP

Solution: $1.9×102 mmHg$ and $4.0×102mmHg$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information: Moles of A: $1.2 mole$ .

Moles of B: $2.3 mole$ .

Total vapor pressure is $331 mmHg$ .

Total vapor pressure upon addition of another mole of B is $347 mmHg$ .

Calculate the mole fraction of A by using the expression given below:

$χA=nAnA+nB$

Here, $nA$ is the number of moles of A and $nB$ is the number of moles of B.

Substitute $1.2 mol$ for $nA$ and $2.3 mol$ for $nB$

$χA = 1.2 mol1.2 mol + 2.3 mol = 0.3429$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$χB=1−0.3429 =0.6571$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$Ptotal=PA+PB$

or

$Ptotal=χAP∘A+χBP∘B$ …...…... (1)

Substituting $Ptotal$ and the mole fractions calculated gives the equation below:

$333 mmHg=0.3429P∘A+0.6571P∘B$

Now, solving for $P∘A$ , we get

$P∘A=331 mmHg−0.6571 P∘B0.3429=965.3 mmHg−1.916P∘B$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$χA = 1.2 mol1.2 mol + 3.3 mol = 0.2667$

$χB=1−0.2667 =0.7333$

Substituting $Ptotal$ and the mole fractions in equation (1) gives the equation as follows:

$347 mmHg=0.2669P∘A+0.7333P∘B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$347 mmHg=0.2667(965.3mmHg−1.916P∘B)+0.7333P∘B0.2223P∘B=89.55 mmHgP∘B=402.8 mmHg=4.0×102mmHg$

Substitute the value of $P∘B$ into Equation (1) to solve for $PA∘$

$P∘A=965.3 mmHg − 1.916(402.8 mmHg)=193.5 mmHg=1.9×102 mmHg$

Answer 12

Conclusion

The vapor pressure of pure A and B is $1.9×102 mmHg$ and $4.0×102mmHg$ , respectively.

Answer 13

Ch. 13.1 - Prob. 1PPA Ch. 13.1 - Prob. 1PPB Ch. 13.1 - Prob. 1PPC Ch. 13.2 - Prob. 1CP Ch. 13.2 - Prob. 2CP Ch. 13.2 - Practice Problem ATTEMPT Determine (a) the...Ch. 13.2 - Practice Problem BUILD Determine the molality of...Ch. 13.2 - Prob. 1PPC Ch. 13.3 - Practice Problem ATTEMPT An aqueous solution that...Ch. 13.3 - Practice Problem BUILD Determine the percent...

Answer to Problem 124AP

Explanation of Solution

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Chapter 13 Solutions