Chapter 10, Problem 34QAP

Calculate the vapor pressure of water over each of the following solutions of glycerol, C₃H₈O₃, at 28°C $\left(\text{vapor}\text{pressure}\text{of pure water}=\text{28}.\text{35 mm Hg}\right)$. Glycerol can be assumed to be nonvolatile.

(a) $\text{percent glycerol by mass}=\text{30}.6\%$

(b) 2.74 m

(c) $\text{mole fraction of glycerol}=0.\text{188}$

Interpretation Introduction

(a)

Interpretation:

To calculate vapor pressure of water if mass percent of glycerol is 30.6 %.

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 found 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 the 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 34QAP

Vapor pressure of water if mass percent of glycerol is $30.6\%$ is $26.08\text{mm}\text{Hg}$.

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mass percent of glycerol is $30.6\%$. It means in $100\text{g}$ solution mass of glycerol is $30.6\text{g}$

$\begin{array}{c}\text{Mass}\text{of}\text{solvent(water)=100}\text{g}-\text{30}\text{.6}\text{g}\\ \text{=69}\text{.4}\text{g}\end{array}$

So, mass of water is $\text{69}\text{.4}\text{g}$.

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

Mass of glycerol $=30.6\text{g}$

Molar mass of glycerol $\text{=92}\text{.09 g}{\text{mol}}^{-1}$

$\begin{array}{c}\text{Moles}\text{of glycerol=}\frac{\text{Mass of glycerol}}{\text{Molar}\text{mass of glycerol}}\\ =\frac{30.6\text{g}}{92.09\text{g}{\text{mol}}^{-1}}\\ =0.33\text{mol}\end{array}$

So, moles of glycerol is $0.33\text{mol}$.

Mass of water $=\text{69}\text{.4}\text{g}$

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

$\begin{array}{c}\text{Moles}\text{of water=}\frac{\text{Mass of water}}{\text{Molar}\text{mass of water}}\\ =\frac{69.4\text{g}}{18.02\text{g}{\text{mol}}^{-1}}\\ =3.85\text{mol}\end{array}$

So, moles of water is $3.85\text{mol}$.

Calculation of mole fraction of water

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{water}\left(X_{\text{water}}\right)\text{=}\frac{\text{Moles}\text{of water}}{\text{Moles}\text{of water}+\text{Moles}\text{of glycerol}}\\ =\frac{3.85\text{mol}}{3.85\text{mol}+0.33\text{mol}}\\ =\frac{3.85\text{mol}}{4.18\text{mol}}\\ =0.92\end{array}$

So, mole fraction of water is $0.92$.

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the 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 $X_{\text{water}}=0.92$

Vapor pressure of pure solvent $\text{=28}\text{.35}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.92\times 28.35\text{mm}\text{Hg}\\ =26.08\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $26.08\text{mm}\text{Hg}$.

Interpretation Introduction

(b)

Interpretation:

To calculate vapour pressure of water if molality of the solution is $2.74\text{m}$.

Concept introduction:

Molality is one way to define the 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 34QAP

If molality of the solution is $2.74\text{m}$, vapor pressure of water will be $26.93\text{mm}\text{Hg}$.

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given the molality is $2.74\text{m}$. So mass of solvent is $1000\text{g}\left(1\text{kg}\right)$

Moles of solute (glycerol) can be calculated as follows:

$\begin{array}{c}\text{Molality=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent(kg)}}\\ 2.74\text{m}=\frac{\text{Moles}\text{of}\text{solute}}{1\text{kg}}\\ \text{Moles}\text{of}\text{solute}=2.74\text{mol}{\text{kg}}^{-1}\text{kg}\\ \text{=}2.74\text{mol}\end{array}$

So, moles of solute (glycerol) is $2.74\text{mol}$

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 $55.5\text{mol}$.

Mole fraction of solvent can be calculated as follows:

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{solvent}\left(X_{\text{water}}\right)\text{=}\frac{\text{Moles}\text{of solvent}\left(\text{water}\right)}{\text{Moles}\text{of solvent}\left(\text{water}\right)+\text{Moles}\text{of solute}\left(\text{glycerol}\right)}\\ =\frac{55.5\text{mol}}{55.5\text{mol}+2.74\text{mol}}\\ =0.95\end{array}$

So, mole fraction of solvent (water) is $0.95$.

Raoult’s law states that vapor pressure of a solvent in a mixture is equal to the 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 $X_{\text{water}}=0.95$

Vapor pressure of pure solvent $\text{=28}\text{.35}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.95\times 28.35\text{mm}\text{Hg}\\ =26.93\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $26.93\text{mm}\text{Hg}$.

Interpretation Introduction

(c)

Interpretation:

To calculate vapor pressure of water if mole fraction of glycerol is $0.188$.

Concept introduction:

For a solution containing solute (glycerol) 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 the 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 34QAP

Vapor pressure of water if mole fraction of glycerol is $0.188$ is $23.02\text{mm}\text{Hg}$.

Explanation of Solution

Given:

Vapor pressure of pure water = 28.55 mm Hg at 28 °C.

Given mole fraction of glycerol $\left(X_{\text{glycerol}}\right)\text{is}0.188$ and vapor pressure of pure water is $28.35\text{mm}\text{Hg}$

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

$\begin{array}{c}{X}_{\text{ethylene}\text{glycol}}+X_{\text{water}}=1\\ X_{\text{water}}=1-X_{\text{ethylene}\text{glycol}}\\ =1-0.188\\ =0.812\end{array}$

So, mole fraction of water is $0.812$.

Raoult’s law states that vapor pressure of solvent in a mixture is equal to the 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 $X_{\text{water}}=0.812$

Vapor pressure of pure solvent $\text{=28}\text{.35}\text{mm}\text{Hg}$

Vapor pressure of water can be calculated as follows:

$\begin{array}{c}{P}_{\text{water}}=X_{\text{water}}\times P^0{}_{\text{water}}\\ =0.812\times 28.\text{35}\text{mm}\text{Hg}\\ =23.02\text{mm}\text{Hg}\end{array}$

Hence, vapor pressure of water is $23.02\text{mm}\text{Hg}$.