Question

At ${{100}^{\circ }}C$, benzene and toluene have vapor pressure of $1375\And 558$Torr respectively. Assuming they form an ideal binary solution, calculate the composition of the solution that boils at $1\,atm\And {{100}^{\circ }}C$. What is the composition of vapor issuing in these conditions?
A. ${{X}_{b}}=0.2472,\,{{Y}_{b}}=0.4473$
B. ${{X}_{b}}=0.4473,\,\,{{Y}_{b}}=0.2472$
C. ${{X}_{b}}=0.362,\,{{Y}_{b}}=0.321$
D. None of these

Answer 1

In this question, we have to find out the mole fraction of the benzene in solution and also in vapor phase. For finding the mole fraction in solution we will use Raoult's law which states that partial pressure of each component in the solution is directly proportional to its mole fraction.

Complete step-by-step answer: Let us assume that ${{P}_{T}}$= partial pressure of toluene
${{P}_{B}}$= partial pressure of benzene
${{X}_{B}}$= mole fraction of benzene
${{X}_{T}}$= mole fraction of toluene
Given, ${{P}_{total}}=1atm=760torr$
We know that According to Dalton’s law of partial pressure, the total pressure $\left( {{P}_{total}} \right)$over the solution phase is equal to the sum of partial pressures of the components of the solution, that are benzene and toluene.
${{P}_{Total}}={{P}_{B}}+{{P}_{T}}$ …. (equation1)
Also, by Raoult’s law we can deduce that
${{P}_{B}}={{P}_{B}}^{\circ }{{X}_{A}}$ and ${{P}_{T}}={{P}_{T}}^{\circ }{{X}_{T}}$
Where ${{P}_{B}}^{\circ }$= vapor pressure of pure benzene = $1375$ and
${{P}_{T}}^{\circ }$= vapor pressure of pure toluene =$558$
Substituting, the values of ${{P}_{B}}\And {{P}_{T}}$in equation 1, we get
${{P}_{Total}}={{P}_{B}}^{\circ }{{X}_{B}}+{{P}_{T}}^{\circ }{{X}_{T}}$ … (equation 2)
Now we know that in any solution, sum of mole fraction of all components is always equal to 1
$\begin{aligned} & \therefore {{X}_{B}}+{{X}_{T}}=1 \\\ & \therefore {{X}_{T}}=1-{{X}_{B}} \\\ \end{aligned}$
Substituting the value of ${{X}_{T}}$in equation2, we get
$\begin{aligned} & \Rightarrow {{P}_{total}}={{P}_{B}}^{\circ }{{X}_{B}}+{{P}_{T}}^{\circ }\left( 1-{{X}_{B}} \right) \\\ & \Rightarrow {{P}_{total}}={{P}_{T}}^{\circ }+{{X}_{B}}\left( {{P}_{B}}^{\circ }-{{P}_{T}}^{\circ } \right) \\\ \end{aligned}$
On substituting the values of ${{P}_{total}},\,{{P}_{T}}^{\circ }\And {{P}_{B}}^{\circ }$in the above expression, we can find the value of ${{X}_{B}}$
$\begin{aligned} & \Rightarrow 760=558+{{X}_{B}}\left( 1375-558 \right) \\\ & \Rightarrow 760=558+{{X}_{B}}\left( 817 \right) \\\ & \Rightarrow 817{{X}_{B}}=760-558 \\\ & \Rightarrow 817{{X}_{B}}=202 \\\ & \Rightarrow {{X}_{B}}=\dfrac{202}{817}=\,0.2472 \\\ & \therefore {{X}_{B}}=0.2472 \\\ \end{aligned}$
Now we know that the mole fraction of a constituent in vapor phase is given by the formula;
${{Y}_{Constituent}}=\dfrac{{{P}_{constituent}}}{{{P}_{Total}}}$
Therefore, mole fraction of benzene in vapor phase will be,
${{Y}_{B}}=\dfrac{{{P}_{B}}}{{{P}_{total}}}$
$\begin{aligned} & \Rightarrow {{Y}_{B}}=\dfrac{1375\times 0.2472}{760}=0.4473 \\\ & \therefore {{Y}_{B}}=0.4473 \\\ \end{aligned}$

Hence, the correct option is option A.

Note: It is essential to indicate the temperature while stating a vapor pressure since vapor pressure increases with the temperature. Also, note that there are different units of pressure so while solving the problem convert all the units in the same form.