Q97P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the vapor pressure of a solution of 0.39 mol of cholesterol in 5.4 mol of toluene at ${\mathbf{32}}^{\mathbf{°}}\mathit{C}$. Pure toluene has a vapor pressure of 41 torr at ${\mathbf{32}}^{\mathbf{°}}\mathit{C}$. (Assume ideal behavior.)

Step-by-Step Solution

Answer

The vapor pressure is 38 torr.

Concept

To calculate the  ${\mathbf{P}}_{\mathbf{solution}}\mathbf{,}$ use the equation  ${\mathit{P}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{=}\mathbf{ }{\mathit{X}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathit{P}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}}^{\mathbf{\circ }}$

This is known as the Raoult’s law.

Calculate mole fraction of toluene ( X toluene )

$$\begin{gathered} \mathbf{Molesoftoluene}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{mol} \\ \mathbf{Molesofcholesterol}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{39}\mathbf{mol} \end{gathered}$$

$$\begin{gathered} {\mathbf{X}}_{\mathbf{toluene}}\mathbf{=}\frac{\mathbf{molesoftoluene}}{\mathbf{molesoftoluene}\mathbf{+}\mathbf{molesofcholesterol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{4}}{\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{39}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{932}\mathbf{.} \end{gathered}$$

Vapor pressure of solution

Using $P_{solution}= X_{solvent} \times P_{solvent}^{\circ }$  calculate the vapor pressure as shown below.

${\mathbf{X}}_{\mathbf{solv}e\mathbf{nt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{932}$

$$\begin{gathered} {\mathit{P}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}}^{\mathbf{\circ }}\mathbf{=}\mathbf{ }\mathbf{41}\mathbf{ }\mathit{t}\mathit{o}\mathit{r}\mathit{r} \\ {\mathbf{P}}_{\mathbf{solution}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{932}\times \mathbf{41}\mathbf{torr} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{212}\mathbf{torr}\mathbf{.} \end{gathered}$$

Therefore, the vapor pressure of the solution is 38 torr.