Chapter 13, Problem 13.51QP

(a)

Interpretation Introduction

Interpretation:

Vapor pressure of methanol and ethanol has to be calculated.

Concept introduction

Vapor pressure is the pressure which is expend by a gas above its condensed phase (either solid phase or liquid phase) and it does not depends on the system’s volume and surface of the or liquid.

The vapor pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

$\begin{array}{l}{P}_A=X_A{P}_{total}\\ where{X}_{A}isthemolefraction\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.

Equation for mole fraction of a molecule in a mixture of two molecules can be given as:

$\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}\text{) }}$

Answer to Problem 13.51QP

Vapor pressure of ethanol is $\text{46 mmHg}$

Vapor pressure of methanol is $\text{22 mmHg}$

Explanation of Solution

Given data follows as:

Vapor pressure of ethanol at $\text{20°C}$ is $\text{44 mmHg}$

Vapor pressure of methanol at $\text{20°C}$ is $\text{94 mmHg}$

Amount of methanol in mixture is $\text{30}\text{.0g}$

Amount of ethanol in mixture is $\text{45}\text{.0g}$

From its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Number of moles of methanol is given as:

Molecular mass of methanol is $\text{32}\text{.04g}$

$\begin{array}{c}\text{mole of Methanol = 30}\text{.0g ×}\frac{\text{1mol}}{\text{32}\text{.04g}}\\ \\ \text{= 0}\text{.936 mol}\end{array}$

Number of moles of ethanol is given as:

Molecular mass of ethanol is $\text{46}\text{.07g}$

$\begin{array}{c}\text{mole of Ethanol= 45}\text{.0g ×}\frac{\text{1mol}}{\text{46}\text{.07g}}\\ \\ \text{= 0}\text{.977 mol}\end{array}$

Equation for mole fraction of a molecule in a mixture of two molecules can be given as:

$\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}\text{) }}$

Mole fraction of methanol is given as:

$\begin{array}{c}{\text{χ}}_{\text{Methanol}}\text{ = }\frac{\text{0}\text{.936 mol}}{\text{0}\text{.936 mol + 0}\text{.977mol}}\\ \\ \text{= 0}\text{.489}\end{array}$

Mole fraction of ethanol is given as:

$\begin{array}{c}{\text{χ}}_{\text{Ethanol}}\text{= 1}-{\text{ χ}}_{\text{Methano}}{}_{\text{l}}\\ \\ \text{=}\left(1-\text{0}\text{.489}\right)\\ \\ \text{= 0}\text{.511}\end{array}$

From the value of mole fraction of methanol and mole fraction of ethanol can be calculated.

Equation for the vapor pressure is given as:

$\begin{array}{l}{P}_A=X_A{P}_{total}\\ where{X}_{A}isthemolefraction\end{array}$

Vapor pressure of methanol and ethanol is calculated as:

$\begin{array}{c}{\text{P}}_{\text{Methanol}}\text{= (0}\text{.489)(94mmHg) }\\ \\ \text{= 46mmHg}\end{array}$

Vapor pressure of ethanol is $\text{46 mmHg}$

$\begin{array}{c}{\text{P}}_{\text{Ethanol}}\text{= (0}\text{.511)(44mmHg) }\\ \\ \text{= 22 mmHg}\end{array}$

Vapor pressure of methanol is $\text{22 mmHg}$

(b)

Interpretation Introduction

Interpretation:

Mole fraction of methanol and ethanol has to be calculated.

Concept introduction:

Mole fraction $\text{(χ)}$ is mole of individual component divided by total moles of the solution.

$\text{χ =}\frac{\text{moles of one component}}{\text{total moles of the solution}}$

Answer to Problem 13.51QP

Mole fraction of methanol is $0.68$

Mole fraction of ethanol is $0.32$

Explanation of Solution

Given data follows as:

Vapor pressure of ethanol at $\text{20°C}$ is $\text{44 mmHg}$

Vapor pressure of methanol at $\text{20°C}$ is $\text{94 mmHg}$

Amount of methanol in mixture is $\text{30}\text{.0g}$

Amount of ethanol in mixture is $\text{45}\text{.0g}$

Mole fraction of methanol and ethanol is calculated as:

$\text{n = PV/RT}$ and T and V are same for both vapors. Therefore number of moles of each substance is proportional to partial pressure. Then the equation for the mole fractions become as follows:

$\begin{array}{c}{\text{χ}}_{\text{Methanol}}\text{ = }\frac{{\text{P}}_{\text{Methanol}}}{{\text{ P}}_{\text{Methanol}}{\text{ + P}}_{\text{Ethanol}}}\\ \\ \text{= }\frac{\text{46mmHg}}{\text{46mmHg + 22mmHg}}\\ \\ \text{= 0}\text{.68}\end{array}$

Mole fraction of methanol is $0.68$

From the value of mole fraction of methanol, mole fraction of ethanol has calculated.

Mole fraction of ethanol can be given as:

$\begin{array}{c}{\text{χ}}_{\text{Ethanol}}{\text{= 1- χ}}_{\text{Methano}}{}_{\text{l}}\\ \\ \text{=}\left(1-0.68\right)\\ \\ \text{= 0}\text{.32}\end{array}$

Mole fraction of ethanol is $0.32$