Animated Solution for Chemistry - Solutions: A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290\text{ mm}$ at $300\text{ K}$. The vapour pressure of propyl alcohol is $200\text{ mm}$. If the mole fraction of ethyl alcohol is $0.6$, its vapour pressure (in mm) at the same temperature will be

Imagine you are standing in a laboratory, observing a sealed glass flask. Inside this flask is a perfectly blended liquid mixture of two closely related compounds: ethyl alcohol and propyl alcohol. To the naked eye, it looks like a still, calm pool of liquid. But at the microscopic level, it is a chaotic, bustling dance floor.

Molecules of both alcohols are constantly gaining enough kinetic energy to break free from the liquid's surface and leap into the empty space above. This invisible gas phase exerts a pressure on the walls of the flask, which we call the vapour pressure. When the rate of molecules escaping equals the rate of molecules returning to the liquid, a state of dynamic equilibrium is reached.

In our specific problem, we are told that the total vapour pressure exerted by this mixture at $300\text{ K}$ is $290\text{ mm}$ of mercury. We are also given a crucial piece of data: if the flask contained only pure propyl alcohol, its vapour pressure would be $200\text{ mm}$. Finally, we know the recipe of our mixture: the mole fraction of ethyl alcohol ($X_{\text{ethyl}}$) is $0.6$. Our ultimate quest is to determine the vapour pressure of pure ethyl alcohol ($P_{\text{ethyl}}^{\circ }$).

Raoult's Law

To solve this mystery, we must call upon the genius of the 19th-century French chemist François-Marie Raoult. He discovered a beautifully simple relationship for ideal solutions. Raoult's Law states that the partial vapour pressure of any volatile component in a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.

Mathematically, for a binary mixture of components A and B, the total pressure ($P_{\text{total}}$) is the sum of their partial pressures:

Expanding this using Raoult's Law, we get our master equation:

This equation is the bridge that connects the macroscopic pressure we measure to the microscopic composition of the liquid.

Finding the Missing Piece

Before we can plug our numbers into the master equation, we realize we are missing the mole fraction of propyl alcohol ($X_{\text{propyl}}$). However, nature demands balance. In any mixture, the sum of all mole fractions must equal exactly $1$ (or $100\%$).

Therefore, finding $X_{\text{propyl}}$ is a simple subtraction:

Now, we have all the pieces of the puzzle. Let's substitute our known values into Raoult's Law:

We begin by calculating the contribution of propyl alcohol to the total pressure. Multiplying its pure pressure by its mole fraction:

So, out of the total $290\text{ mm}$ of pressure, $80\text{ mm}$ is provided by the propyl alcohol molecules. Our equation simplifies to:

To isolate our unknown variable, we subtract $80$ from both sides:

We are at the final step. To find the pure vapour pressure of ethyl alcohol, we divide $210$ by $0.6$:

And there we have it! The vapour pressure of pure ethyl alcohol at $300\text{ K}$ is $350\text{ mm}$.

Does this answer make physical sense? Absolutely. Ethyl alcohol has a shorter carbon chain than propyl alcohol, meaning its intermolecular London dispersion forces are weaker. Weaker forces mean it is easier for the molecules to escape into the vapour phase, resulting in a higher pure vapour pressure ($350\text{ mm}$ vs $200\text{ mm}$).

While we assumed this mixture behaves as an ideal solution, always remember that in the real world, molecules have preferences. If they repel each other, the total pressure might be higher than Raoult's Law predicts (a positive deviation). If they attract each other strongly, the pressure might be lower (a negative deviation). But for this problem, the ideal assumption leads us straight to the perfect answer!