Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
Question
Chapter 13, Problem 13.149QP

(a)

Interpretation Introduction

Interpretation:

For mixture of two volatile liquids,

Mole fraction of each component has to be calculated.

Concept introduction:

Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.

P= χP°

Where,

P- Partial pressure of each component

P°- Partial pressure of its pure components

χ1 - Mole fraction of the components

(a)

Expert Solution
Check Mark

Answer to Problem 13.149QP

Mole fraction of component A is 0.52

Mole fraction of component B is 0.48

Explanation of Solution

Given data

Molar mass of liquid A = 100g/mol

Molar mass of liquid B = 110g/mol

Vapour pressure of A = 98mmHg at 55°C

Vapour pressure of B = 42 mmHg at 55°C

Calculation of mole fraction of each component

1.00g A×1mol A100g A = 0.01molA1.00g B×1mol B110g B= 0.0091molB

The mole fraction of the component is calculated by moles of the component is divided by the total number of moles in the mixture.

χA=molAmolA+molB=0.01(0.01+0.0091)= 0.52χB=molBmolA+molB=0.0091(0.01+0.0091)= 0.48

By plugging in the values of moles of each component and total moles of the component, the mole fraction of each component has calculated.

Conclusion

Mole fraction of component A has calculated as 0.52

Mole fraction of component B has calculated as 0.48

(b)

Interpretation Introduction

Interpretation:

For mixture of two volatile liquids,

Partial pressure of A and B over the solution at 55°C has to be calculated.

Concept introduction:

Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.

P= χP°

Where,

P- Partial pressure of each component

P°- Partial pressure of its pure components

χ1 - Mole fraction of the components

(b)

Expert Solution
Check Mark

Answer to Problem 13.149QP

Partial pressure of solution A is 51mmHg

Partial pressure of solution B is 20 mmHg

Explanation of Solution

Given data

Molar mass of liquid A = 100g/mol

Molar mass of liquid B = 110g/mol

Vapour pressure of A = 98mmHg at 55°C

Vapour pressure of B = 42 mmHg at 55°C

Calculation of partial pressure of each component

At 55°C, PA°= 98mmHg and PB°= 42mmHg

χA= 0.52mmHg and χB= 0.48 mmHg

The formula for partial pressure,

P1= χ1P1°

PA= χAPA°= 0.52 × 98mmHg = 51mmHg

PB= χBPB°= 0.48× 42mmHg = 20 mmHg

According to Raoult’s law, the vapour pressure of the solution is sum of the individual partial pressure exerted by the solution and then using partial pressure equation, partial pressure of each component has calculated.

Conclusion

Partial pressure of solution A has calculated as 51mmHg

Partial pressure of solution B has calculated as 20 mmHg

(c)

Interpretation Introduction

Interpretation:

For mixture of two volatile liquids,

Mole fraction of each component in the condensed liquid to be calculated.

Concept introduction:

Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.

P= χP°

Where,

P- Partial pressure of each component

P°- Partial pressure of its pure components

χ1 - Mole fraction of the components

(c)

Expert Solution
Check Mark

Answer to Problem 13.149QP

Mole fraction of component A in condensed liquid is 0.72

Mole fraction of component B in condensed liquid is 0.28

Explanation of Solution

Given data

Molar mass of liquid A = 100g/mol

Molar mass of liquid B = 110g/mol

Vapour pressure of A = 98mmHg at 55°C

Vapour pressure of B = 42 mmHg at 55°C

χi=PiPtotal

The mole fraction is equal to partial pressure of the component divided by the total pressure.

Ptotal= 51mmHg + 20mmHg = 71mmHg

χA=PAPtotal=51mmHg71mmHg= 0.72χB=PBPtotal=20mmHg71mmHg= 0.28

By plugging in the value of partial pressure of each component and total pressure, the mole fraction of each component at condensed liquid has calculated.

Conclusion

Mole fraction of component A in condensed liquid has calculated as 0.72

Mole fraction of component B in condensed liquid has calculated as  0.28

(d)

Interpretation Introduction

Interpretation:

For mixture of two volatile liquids,

Partial pressure of the components above the condensed liquid at 55°C to be calculated.

Concept introduction:

Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.

P= χP°

Where,

P- Partial pressure of each component

P°- Partial pressure of its pure components

χ1 - Mole fraction of the components

(d)

Expert Solution
Check Mark

Explanation of Solution

Partial pressure of the component A above condensed liquid at 55°C is 71mmHg

Partial pressure of the component B above condensed liquid at 55°C is 12 mmHg

Calculation of partial pressure of each component

The mole fraction of each component in condensed liquid was calculated I part (c) is,

χA= 0.72 and  χB= 0.28

PA= χAPA°= 0.72 × 98mmHg = 71mmHg

PB= χBPB°= 0.28× 42mmHg = 12 mmHg

Conclusion

Partial pressure of the component A above condensed liquid at 55°C has calculated as 71mmHg

Partial pressure of the component B above condensed liquid at 55°C has calculated as  12 mmHg

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Chapter 13 Solutions

Chemistry: Atoms First

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