Chapter 10, Problem 33QAP

Calculate the vapor pressure of water over each of the following ethylene glycol (C₂H₆O₂) solutions at 22°C $\left(\text{vp pure water}=\text{19}.\text{83 mm Hg}\right)$. Ethylene glycol can be assumed to be nonvolatile.

(a) $X_{\text{ethylene glycol}}=0.\text{288}$(b) $\text{percent ethylene glycol by mass}=\text{39}.0\%$(c) 2.42 m ethylene glycol

Interpretation Introduction

(a)

Interpretation:

Vapor pressure of water is to be calculated if mole fraction of ethylene glycol is $0.228$.

Concept introduction:

For a solution containing solute (ethylene glycol) and solvent (water) summation of mole fraction is equal to one.

$X_{\text{ethylene}\text{glycol}}+X_{\text{water}}=1$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent

Answer to Problem 33QAP

Vapor pressure of water if mole fraction of ethylene glycol is $0.228$ is $14.12\text{mm}\text{Hg}$.

Explanation of Solution

Given:

Mole fraction of ethylene glycol $\left(X_{\text{ethylene}\text{glycol}}\right)\text{is}0.228$ and vapor pressure of pure water is $\text{19}\text{.83}\text{mm}\text{Hg}$.

Addition of mole fraction of all components of a solution is equal to one.

$\begin{array}{c}{X}_{\text{ethylene}\text{glycol}}+X_{\text{water}}=1\\ X_{\text{water}}=1-X_{\text{ethylene}\text{glycol}}\\ =1-0.288\\ =0.712\end{array}$

So, mole fraction of water(solvent) is $0.712$.

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent.

Mole fraction of water $X_{\text{water}}=0.712$

Vapor pressure of pure solvent $\text{=19}\text{.83}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.712\times \text{19}\text{.83}\text{mm}\text{Hg}\\ =14.12\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $14.12\text{mm}\text{Hg}$.

Interpretation Introduction

(b)

Interpretation:

The vapor pressure of water is to be calculated if mass percent of ethylene glycol is 39.0 %.

Concept introduction:

Formula to calculate moles of a component is −

$\text{Moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Mole fraction of a component in a solution can be find out by dividing moles of the component by summation of mole of all components in the solution.

$\text{Mole}\text{fraction}\left(X\right)\text{=}\frac{\text{Moles}\text{of component}}{\text{Summation}\text{of}\text{moles}\text{of}\text{all}\text{components}}$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent.

Answer to Problem 33QAP

Vapor pressure of water if mass percent of ethylene glycol is $39.0\%$ is $16.66\text{mm}\text{Hg}$

Explanation of Solution

Given:

Vapor pressure of pure water is $\text{19}\text{.83}\text{mm}\text{Hg}$

Mass percent of ethylene glycol is $39.0\%$.

Mass percent of ethylene glycol is $39.0\%$.

It means in $100\text{g}$ solution mass of ethylene glycol is $39\text{g}$.

$\begin{array}{c}\text{Mass}\text{of}\text{solvent(water)=100}\text{g}-\text{39}\text{g}\\ \text{=61}\text{g}\end{array}$

So, mass of water is $61\text{g}$

Moles of each component of the solution are to be calculated as follows:

Mass of ethylene glycol $=39\text{g}$

Molar mass of ethylene glycol $\text{=62}\text{.07 g}{\text{mol}}^{-1}$

$\begin{array}{c}\text{Moles}\text{of ethylene glycol=}\frac{\text{Mass of ethylene glycol}}{\text{Molar}\text{mass of ethylene glycol}}\\ =\frac{39\text{g}}{62.07\text{g}{\text{mol}}^{-1}}\\ =0.63\text{mol}\end{array}$

So, moles of ethylene glycol is $0.63\text{mol}$.

Mass of water $=61\text{g}$

Molar mass of water $\text{=18}\text{.02 g}{\text{mol}}^{-1}$

$\begin{array}{c}\text{Moles}\text{of water=}\frac{\text{Mass of water}}{\text{Molar}\text{mass of water}}\\ =\frac{61\text{g}}{18.02\text{g}{\text{mol}}^{-1}}\\ =3.39\text{mol}\end{array}$

So, moles of water is $3.39\text{mol}$.

Calculation of mole fraction of water

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{water}\left(X_{\text{water}}\right)\text{=}\frac{\text{Moles}\text{of water}}{\text{Moles}\text{of water}+\text{Moles}\text{of ethylene glycol}}\\ =\frac{3.39\text{mol}}{3.39\text{mol}+0.63\text{mol}}\\ =\frac{3.39\text{mol}}{4.02\text{mol}}\\ =0.84\end{array}$

So, mole fraction of water is $0.84$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent.

Mole fraction of water $X_{\text{water}}=0.84$

vapor pressure of pure solvent $\text{=19}\text{.83}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.84\times \text{19}\text{.83}\text{mm}\text{Hg}\\ =16.66\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $16.66\text{mm}\text{Hg}$.

Interpretation Introduction

(c)

Interpretation:

The vapor pressure of water is to be calculated if molality of the solution is $2.24\text{m}$.

Concept introduction:

Molality is one way to define concentration of solution. It is the ratio of moles of solute to mass of solvent in kilogram.

Formula of molality is-

$\text{Molality=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent(kg)}}$

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent

Answer to Problem 33QAP

If molality of the solution is $2.24\text{m}$, vapor pressure of water will be $19.03\text{mm}\text{Hg}$.

Explanation of Solution

Given:

Vapor pressure of pure water is $\text{19}\text{.83}\text{mm}\text{Hg}$

The molality is $2.24\text{m}$.

Since, the molality is $2.24\text{m}$. So mass of solvent is $1000\text{g}\left(1\text{kg}\right)$

Moles of solute (ethylene glycol) can be calculated as follows:

$\begin{array}{c}\text{Molality=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent(kg)}}\\ 2.24\text{m}=\frac{\text{Moles}\text{of}\text{solute}}{1\text{kg}}\\ \text{Moles}\text{of}\text{solute}=2.24\text{mol}{\text{kg}}^{-1}\text{kg}\\ \text{=}2.24\text{mol}\end{array}$

So, moles of solute (ethylene glycol) is $2.24\text{mol}$

Moles of solvent can be calculated as follows:

Molar mass of water is $\text{=18}\text{.02 g}{\text{mol}}^{-1}$

Mass of solvent (water) =1000 g

$\begin{array}{c}\text{Moles}\text{of}\text{solvent}\left(\text{water}\right)\text{=}\frac{1000\text{g}}{18.02\text{g}{\text{mol}}^{-1}}\\ =55.5\text{mol}\end{array}$

So, moles of solvent is $55.5\text{mol}$.

Calculation of mole fraction of solvent is as follows:

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{solvent}\left(X_{\text{water}}\right)\text{=}\frac{\text{Moles}\text{of solvent}\left(\text{water}\right)}{\text{Moles}\text{of solvent}\left(\text{water}\right)+\text{Moles}\text{of solute}\left(\text{ethylene glycol}\right)}\\ =\frac{55.5\text{mol}}{55.5\text{mol}+2.24\text{mol}}\\ =0.96\end{array}$

So, mole fraction of solvent(water) is $0.96$.

Raoult’s law states that vapor pressure of solvent in a mixture is equal to product of mole fraction of solvent and vapor pressure of pure solvent.

$P_{\text{solvent}}=X_{\text{solvent}}\times P^0{}_{\text{solvent}}$

Where, $P_{\text{solvent}}$ is vapor pressure of solvent in a mixture, $X_{\text{solvent}}$ is mole fraction of solvent and $P^0{}_{\text{solvent}}$ is vapor pressure of pure solvent.

Mole fraction of water $X_{\text{water}}=0.96$

vapor pressure of pure solvent $\text{=19}\text{.83}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.96\times \text{19}\text{.83}\text{mm}\text{Hg}\\ =19.03\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $19.03\text{mm}\text{Hg}$.