Q96P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the vapor pressure of a solution of 34.0 g of glycerol (C₃H₈O₃) in 500.0 g of water at ${\mathbf{25}}^{\mathbf{°}}\mathit{C}$. The vapor pressure of water at ${\mathbf{25}}^{\mathbf{°}}\mathit{C}$ is 23.76 torr. (Assume ideal behavior.)

Step-by-Step Solution

Answer

The vapor pressure of the solution is 23.42 torr.

Concept

To calculate the ${\mathit{P}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}$  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 water X H 2 O

First, we have to calculate the number of moles of glycerol.

$$\begin{gathered} {\mathbf{n}}_{\mathbf{glycerol}}\mathbf{=}\frac{\mathbf{givenmassofglycerol}}{\mathbf{molecularmassofglycerol}} \\ \mathbf{=}\left[\frac{\mathbf{34}\mathbf{g}}{\mathbf{92}\mathbf{.}\mathbf{09}\mathbf{g}/\mathbf{mol}}\right] \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{369}\mathbf{mol}. \end{gathered}$$

$$\begin{gathered} {\mathbf{n}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\left[\frac{\mathbf{500}\mathbf{g}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}}\right] \\ \mathbf{=}\mathbf{27}\mathbf{.}\mathbf{746}{\mathbf{molH}}_{\mathbf{2}}\mathbf{O}. \end{gathered}$$

 

 $$\begin{gathered} {\mathbf{X}}_{\mathbf{H}\mathbf{2}\mathbf{O}}\mathbf{=}\frac{{\mathbf{n}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{{\mathbf{n}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{+}{\mathbf{n}}_{\mathbf{glycerol}}} \\ \mathbf{=}\frac{\mathbf{27}\mathbf{.}\mathbf{74} \mathbf{mol}}{\left(\mathbf{0}\mathbf{.}\mathbf{369}\mathbf{+}\mathbf{27}\mathbf{.}\mathbf{74}\right) \mathbf{mol}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{986}. \end{gathered}$$

    

   

calculate vapor pressure

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

$$\begin{gathered} {\mathbf{X}}_{\mathbf{solv}e\mathbf{nt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{986} \\ {\mathit{P}}_{\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}}^{\mathbf{\circ }}\mathbf{=}\mathbf{ }\mathbf{23}\mathbf{.}\mathbf{76}\mathbf{ }\mathit{t}\mathit{o}\mathit{r}\mathit{r} \\ {\mathbf{P}}_{\mathbf{solution}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{986}\mathbf{\times }\mathbf{23}\mathbf{.}\mathbf{76}\mathbf{torr} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{42}\mathbf{torr}\mathbf{.} \end{gathered}$$

Therefore, the required vapor pressure is 23.42 torr.