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At\[\text{ 90 }\!\!{}^\circ\!\!\text{ C }\] , the vapour pressure of toluene is $\text{ 400 mm }$ and that of xylene is \[\text{ 150 mm }\] . What is the composition of liquid mixture with respect to xylene (write answer after multiplying with 100) that will boil at \[\text{ 90 }\!\!{}^\circ\!\!\text{ C }\]when the pressure of mixture is $\text{ 0}\text{.5 atm }$ ?

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Last updated date: 13th Jun 2024
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Answer
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Hint: The ideal binary solution is a mixture of two components. The vapour pressure on the solution can be expressed in terms of molar ratio. The relation between the pressure of the mixture and the partial pressure of components and molar ratio is expressed as follows;
$\text{ }{{\text{P}}_{\text{mixture}}}\text{ = }{{\text{X}}_{\text{A}}}\text{ p}_{\text{A}}^{\text{0}}\text{ + }{{\text{X}}_{\text{B}}}\text{ p}_{\text{B}}^{\text{0}}\text{ }$
Where ${{x}_{A}}$ is the mole fraction of component A and ${{x}_{B}}$ is the mole fraction of B.

Complete step by step answer:
According to Raoult's law, the partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.
For a binary solution let’s make of the $\text{ }{{\text{n}}_{\text{A}}}\text{ }$moles of volatile liquid A and $\text{ }{{\text{n}}_{\text{B}}}\text{ }$moles of a volatile liquid B. If $\text{ }{{P}_{A}}\text{ }$and $\text{ }{{P}_{B}}\text{ }$are the partial pressure of the two liquid components, then, according to the Raoult’s law,
$\text{ }{{P}_{A}}\text{ = }{{x}_{A}}\text{ }p_{A}^{0}\text{ }$ And $\text{ }{{P}_{B}}\text{ = }{{x}_{B}}\text{ }p_{B}^{0}\text{ }$
Where ${{x}_{A}}$ is the mole fraction of component A and ${{x}_{B}}$ is the mole fraction of B.
For the binary mixture of A and B the, according to Dalton’s law of partial pressure, the total vapour pressure P is given as,
$\text{ }{{\text{P}}_{\text{mixture}}}\text{ = }{{\text{X}}_{\text{A}}}\text{ p}_{\text{A}}^{\text{0}}\text{ + }{{\text{X}}_{\text{B}}}\text{ p}_{\text{B}}^{\text{0}}\text{ }$
Where ${{x}_{A}}$ is the mole fraction of component A and ${{x}_{B}}$ is the mole fraction of B.

We have given the following data:
Vapour pressure of xylene is,$\text{ p}_{\text{Xylene}}^{\text{0}}\text{ = 150 mm }$
Vapour pressure of toluene is, $\text{ p}_{\text{Toluene}}^{\text{0}}\text{ }=\text{400 mm }$
The total vapour pressure of the mixture is, $\text{ 0}\text{.5 atm }$
To find: the molar ratios of the xylene and toluene
Let’s first convert the pressure of the mixture from atmospheric pressure (atm). We have,
$\begin{align}
 & \text{ P = 0}\text{.5 atm = }\frac{76}{2}\text{ mm of Hg } \\
 & \therefore \text{ P = 380 mm of Hg } \\
\end{align}$
And we know that, the sum of the mole fraction of the components A and B is always equal to unity. Thus we have,
$\begin{align}
 & \text{ }{{\text{X}}_{\text{Xylene}}}\text{ + }{{\text{X}}_{\text{Toluene}}}\text{ = 1} \\
 & \therefore \text{ }{{\text{X}}_{\text{Xylene}}}\text{ = 1}-{{\text{X}}_{\text{Toluene}}}\text{ } \\
\end{align}$
Let's rewrite the equation of the pressure for the mixture of xylene and toluene. We have,
$\text{ }{{\text{P}}_{\text{mixture}}}\text{ = }{{\text{X}}_{\text{Xylene}}}\text{ p}_{\text{Xylene}}^{\text{0}}\text{ + }{{\text{X}}_{\text{Toluene}}}\text{ p}_{\text{Toluene}}^{\text{0}}\text{ }$
Let’s substitute the values, we have,
$\begin{align}
  & \text{ }{{\text{P}}_{\text{mixture}}}\text{ = p}_{\text{Xylene}}^{\text{0}}\text{(1}-{{\text{X}}_{\text{Toluene}}}\text{) + }{{\text{X}}_{\text{Toluene}}}\text{ p}_{\text{Toluene}}^{\text{0}} \\
& \Rightarrow \text{380 = 150 (1}-{{\text{X}}_{\text{Toluene}}}\text{) + 400 (}{{\text{X}}_{\text{Toluene}}}) \\
 & \Rightarrow 380\text{ = 150 }-\text{150 }{{\text{X}}_{\text{Toluene}}}\text{ + 400 }{{\text{X}}_{\text{Toluene}}} \\
 & \Rightarrow 230\text{ = 250}{{\text{X}}_{\text{Toluene}}} \\
 & \therefore {{\text{X}}_{\text{Toluene}}}=\frac{230\text{ }}{\text{250}}\text{ = 0}\text{.92} \\
\end{align}$

Therefore, the mole fraction of xylene is as follows,
$\begin{align}
 & {{\text{X}}_{\text{Xylene}}}\text{ = 1}-{{\text{X}}_{\text{Toluene}}} \\
 & \therefore {{\text{X}}_{\text{Xylene}}}=1-\text{0}\text{.92 = 0}\text{.08} \\
\end{align}$
Thus, the mole fraction of toluene is equal to $\text{0}\text{.92}$ and the mole fraction of xylene is equal to$\text{0}\text{.08}$.

Note: The Raoult’s law is valid for the ideal or the dilute solutions. Applied to the non-ideal of the concentrated solution does not cause the change in the order of the magnitude of the calculated vapour pressure. Note that, when the solution contains i component then the pressure of the mixture is given as,
$\text{ }{{\text{p}}_{\text{i}}}\text{ = }{{\text{x}}_{\text{i}}}\text{ p}_{\text{i}}^{\text{0}}\text{ }$
The total vapour pressure may be expressed as,
$\text{P = }\sum{{{p}_{i}}}$