Chapter 13, Problem 13.60QP

The vapor pressure of ethanol (C₂H₅OH) at 20°C is 44 mmHg, and the vapor pressure of methanol (CH₃OH) at the same temperature is 94 mmHg. A mixture of 30.0 g of methanol and 45.0 g of ethanol is prepared (and can be assumed to behave as an ideal solution). (a) Calculate the vapor pressure of methanol and ethanol above this solution at 20°C. (b) Calculate the mole fraction of methanol and ethanol in the vapor above this solution at 20°C. (c) Suggest a method for separating the two components of the solution.

Interpretation Introduction

Interpretation: For Ethanol solution,

1. (a) At $\text{20°C}$ vapor pressure to be calculated.
2. (b)  Mole fraction to be calculated.

Separating these two components to be suggested.

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.

Answer to Problem 13.60QP

Answer

Vapor pressure of Ethanol = $\text{46 mmHg}$

Vapor pressure of Methanol = $\text{22 mmHg}$

Explanation of Solution

Given data

Vapor pressure of Ethanol at $\text{20°C}$ = $\text{44 mmHg}$

Vapor pressure of Methanol at $\text{20°C}$ = $\text{94 mmHg}$

Amount of Methanol in mixture = $\text{30}\text{.0g}$

Amount of Ethanol in mixture = $\text{45}\text{.0g}$

Finding number of moles of Methanol

Molecular mass of Methanol = $\text{32}\text{.04g}$

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

By plugging in the value of amount and molecular weight of Methanol, a mole of Methanol has calculated.

Finding number of moles of Ethanol

Molecular mass of Ethanol = $\text{46}\text{.07g}$

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

By plugging in the value of amount and molecular weight of Ethanol, a mole of Ethanol has calculated.

Finding mole fraction of Methanol

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

By plugging in the value of moles of Methanol and total moles of the solution, a mole fraction of Methanol has calculated.

Finding mole fraction of Ethanol

${\text{χ}}_{\text{Ethanol}}{\text{= 1- χ}}_{\text{Methano}}{}_{\text{l}}\text{= 0}\text{.511}$

From the value of mole fraction of Methanol, mole fraction of Ethanol has calculated.

Finding vapor pressure of Methanol and Ethanol

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

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

By multiplying in the value of moles of Methanol and Ethanol with their vapor pressure at $\text{20°C}$, vapor pressure of Methanol and Ethanol has calculated.

Conclusion

Conclusion

At $\text{20°C}$ Vapor pressure of Ethanol and Methanol were calculated as $\text{46 mmHg}$ and $\text{22 mmHg}$ respectively.

Interpretation Introduction

Interpretation:

For Methanol solution, mole fraction to be calculated.

Separating these two components to be suggested

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 13.60QP

Answer

Mole fraction of Methanol = $0.68$

Explanation of Solution

Given data

Vapor pressure of Ethanol at $\text{20°C}$ = $\text{44 mmHg}$

Vapor pressure of Methanol at $\text{20°C}$ = $\text{94 mmHg}$

Amount of Methanol in mixture = $\text{30}\text{.0g}$

Amount of Ethanol in mixture = $\text{45}\text{.0g}$

Finding mole fraction of Methanol and Ethanol

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

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

By plugging in the value of vapor pressure of Methanol and total vapor pressure, a mole fraction of Methanol has calculated.

Finding mole fraction of Ethanol

${\text{χ}}_{\text{Ethanol}}{\text{= 1- χ}}_{\text{Methano}}{}_{\text{l}}\text{= 0}\text{.32}$

From the value of mole fraction of Methanol, mole fraction of Ethanol has calculated.

Conclusion

Conclusion

At $\text{20°C}$ a mole fraction of Ethanol and Methanol were calculated as $0.32$ and $0.68$ respectively.

Interpretation Introduction

Interpretation: To explain Fractional distillation

Concept introduction

Fractional distillation: Separation of individual liquid from mixture of more than two liquids by their boiling point.

Answer to Problem 13.60QP

Answer

Two components can separate by fractional distillation.

Explanation of Solution

Explanation

Fractional distillation is the process of separating one component from the mixture of more than two components according to their boiling point by using fractionating column.

Conclusion

Separating these two components has been suggested.