Chapter 13, Problem 13.106SP

(a)

Interpretation Introduction

Interpretation:

Mole fraction of each component present in the given solution 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,

$\text{P-}$ Partial pressure of each component

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

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

Mole fraction: Concentration of the solution can also expressed by mole fraction. Mole fraction is equal to moles of the component divided by total moles of the mixture.

$\text{Mole fraction =}\frac{\text{moles of a component}}{\text{total moles}}$

Answer to Problem 13.106SP

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 = $\text{95mmHg at 55}°\text{C}$

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

Calculate mole fraction of each component:

Assume $1g$ of each A and B mixed then, the mole fractions are calculated as follows,

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

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

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

Substituting the values of moles of each component and total moles of the component, the mole fraction of each component has calculated.

(b)

Interpretation Introduction

Interpretation:

The partial pressure of the components over the given 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,

$\text{P-}$ Partial pressure of each component

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

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

Answer to Problem 13.106SP

Partial pressure of solution A is $\text{50mmHg}$

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

Explanation of Solution

Calculate partial pressure of each component:

${\text{At 55°C, P}}_{\text{A}}^{\text{°}}{\text{= 95mmHg and P}}_{\text{B}}^{\text{°}}\text{= 42mmHg}$

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

The formula for partial pressure,

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

${\text{P}}_{\text{A}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}\text{= 0}\text{.52 × 95}\text{mmHg = 49}\text{.4}\text{mmHg}$

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

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 been calculated.

(c)

Interpretation Introduction

Interpretation:

The mole fraction of each component in the condensed liquid and the vapour pressure of the components above the condensed liquid at $\text{55°C}$ should 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,

$\text{P-}$ Partial pressure of each component

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

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

Answer to Problem 13.106SP

Mole fraction of component A in condensed liquid is $0.71$

Mole fraction of component B in condensed liquid is $0.29$

Partial pressure of the component A above condensed liquid at $\text{55°C}$ is $\text{67}\text{.45mmHg}$

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

Explanation of Solution

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

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

${\text{P}}_{\text{total}}\text{= 49}\text{.4 mmHg + 20mmHg = 69}\text{.4mmHg}$

$\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{total}}}\text{=}\frac{\text{49}\text{.4}\text{mmHg}}{\text{69}\text{.4}\text{mmHg}}\text{= 0}\text{.71}\\ \\ {\text{χ}}_{\text{B}}\text{=1-}\text{0}\text{.71= 0}\text{.29}\end{array}$

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

Calculation of partial pressure of each component

The mole fraction of each component in condensed liquid is,

${\text{χ}}_{\text{A}}\text{= 0}{\text{.71 and χ}}_{\text{B}}\text{= 0}\text{.29}$

${\text{P}}_{\text{A}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}\text{= 0}\text{.71 × 95mmHg = 67}\text{.45mmHg}$

${\text{P}}_{\text{B}}{\text{= χ}}_{\text{B}}{\text{P}}_{\text{B}}^{\text{°}}\text{= 0}\text{.29× 42mmHg = 12 mmHg}$