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.

${\text{P}}_{}{\text{= χ}}_{}{\text{P}}^{\text{°}}$

Where,

P- Partial pressure of each component

$\text{P°-}$ Partial pressure of its pure components

${\text{χ}}_{\text{1}}$- Mole fraction of the components

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 = $\text{100g/mol}$

Molar mass of liquid B = $\text{110g/mol}$

Vapour pressure of A = $\text{98mmHg at 55}°\text{C}$

Vapour pressure of B = $\text{42 mmHg}$ $\text{at 55°C}$

Calculation of mole fraction of each component

$\begin{array}{l}\text{1}\text{.00g A×}\frac{\text{1mol A}}{\text{100g A}}\text{ = 0}\text{.01molA}\\ \\ \text{1}\text{.00g B×}\frac{\text{1mol B}}{\text{110g B}}\text{= 0}\text{.0091molB}\end{array}$

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

$\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{molA}}{\text{molA+molB}}\text{=}\frac{\text{0}\text{.01}}{\text{(0}\text{.01+0}\text{.0091)}}\text{= 0}\text{.52}\\ \\ {\text{χ}}_{\text{B}}\text{=}\frac{\text{molB}}{\text{molA+molB}}\text{=}\frac{\text{0}\text{.0091}}{\text{(0}\text{.01+0}\text{.0091)}}\text{= 0}\text{.48}\end{array}$

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 $\text{55°C}$ 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.

${\text{P}}_{}{\text{= χ}}_{}{\text{P}}^{\text{°}}$

Where,

P- Partial pressure of each component

$\text{P°-}$ Partial pressure of its pure components

${\text{χ}}_{\text{1}}$- Mole fraction of the components

Answer to Problem 13.149QP

Partial pressure of solution A is $\text{51mmHg}$

Partial pressure of solution B is $\text{20 mmHg}$

Explanation of Solution

Given data

Molar mass of liquid A = $\text{100g/mol}$

Molar mass of liquid B = $\text{110g/mol}$

Vapour pressure of A = $\text{98mmHg at 55}°\text{C}$

Vapour pressure of B = $\text{42 mmHg}$ $\text{at 55°C}$

Calculation of partial pressure of each component

${\text{At 55°C, P}}_{\text{A}}^{\text{°}}{\text{= 98mmHg and P}}_{\text{B}}^{\text{°}}\text{= 42mmHg}$

${\text{χ}}_{\text{A}}\text{= 0}\text{.52mmHg}$ and ${\text{χ}}_B\text{= 0}\text{.48 mmHg}$

The formula for partial pressure,

${\text{P}}_1{\text{= χ}}_1{\text{P}}_1^{\text{°}}$

${\text{P}}_{\text{A}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}\text{= 0}\text{.52 × 98mmHg = 51mmHg}$

${\text{P}}_{\text{B}}{\text{= χ}}_{\text{B}}{\text{P}}_{\text{B}}^{\text{°}}\text{= 0}\text{.48× 42mmHg = 20 mmHg}$

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 $\text{51mmHg}$

Partial pressure of solution B has calculated as $\text{20 mmHg}$

(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.

${\text{P}}_{}{\text{= χ}}_{}{\text{P}}^{\text{°}}$

Where,

P- Partial pressure of each component

$\text{P°-}$ Partial pressure of its pure components

${\text{χ}}_{\text{1}}$- Mole fraction of the components

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 = $\text{100g/mol}$

Molar mass of liquid B = $\text{110g/mol}$

Vapour pressure of A = $\text{98mmHg at 55}°\text{C}$

Vapour pressure of B = $\text{42 mmHg}$ $\text{at 55°C}$

${\text{χ}}_{\text{i}}\text{=}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{total}}}$

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

${\text{P}}_{\text{total}}\text{= 51mmHg + 20mmHg = 71mmHg}$

$\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{total}}}\text{=}\frac{\text{51mmHg}}{\text{71mmHg}}\text{= 0}\text{.72}\\ \\ {\text{χ}}_{\text{B}}\text{=}\frac{{\text{P}}_{\text{B}}}{{\text{P}}_{\text{tota}}\text{l}}\text{=}\frac{\text{20mmHg}}{\text{71mmHg}}\text{= 0}\text{.28}\end{array}$

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 $\text{55°C}$ 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.

${\text{P}}_{}{\text{= χ}}_{}{\text{P}}^{\text{°}}$

Where,

P- Partial pressure of each component

$\text{P°-}$ Partial pressure of its pure components

${\text{χ}}_{\text{1}}$- Mole fraction of the components

Explanation of Solution

Partial pressure of the component A above condensed liquid at $\text{55°C}$ is $\text{71mmHg}$

Partial pressure of the component B above condensed liquid at $\text{55°C}$ is $\text{12 mmHg}$

Calculation of partial pressure of each component

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

${\text{χ}}_{\text{A}}\text{= 0}{\text{.72 and χ}}_{\text{B}}\text{= 0}\text{.28}$

${\text{P}}_{\text{A}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}\text{= 0}\text{.72 × 98mmHg = 71mmHg}$

${\text{P}}_{\text{B}}{\text{= χ}}_{\text{B}}{\text{P}}_{\text{B}}^{\text{°}}\text{= 0}\text{.28× 42mmHg = 12 mmHg}$

Conclusion

Partial pressure of the component A above condensed liquid at $\text{55°C}$ has calculated as $\text{71mmHg}$

Partial pressure of the component B above condensed liquid at $\text{55°C}$ has calculated as  $\text{12 mmHg}$