Animated Solution for Chemistry - Solutions: Comprehension Passage

Imagine you are looking at a closed container. Inside, a liquid mixture of two volatile components, A and B, sits quietly. But at the microscopic level, it's a chaotic dance! Molecules of A and B are constantly escaping into the space above, creating a vapour phase, while others condense back into the liquid. This dynamic equilibrium is the heart of our problem.

We are given a 5 molal solution of B in A. The total vapour pressure of this system is $100\text{ mm Hg}$, and the pure vapour pressure of A is $105\text{ mm Hg}$. Our mission is to uncover the hidden properties of component B, specifically its pure vapour pressure, its molar volumes, and its concentration in the vapour phase.

Raoult's Law

To find anything about B, we first need to know how it behaves on its own—we need its pure vapour pressure, $P_B^{\circ }$. This is where Raoult's Law comes to the rescue. It tells us that the total vapour pressure of an ideal solution is the sum of the partial pressures of its components:

But wait, we don't have the mole fractions ($X_A$ and $X_B$) yet! We only have the molality. Let's decode that. A 5 molal solution means there are $5\text{ moles}$ of solute B in exactly $1\text{ kg}$ ($1000\text{ g}$) of solvent A.

Since the molar mass of A is $50\text{ g/mol}$, $1000\text{ g}$ of A corresponds to:

Now, the total number of moles in our liquid is $20+5=25\text{ moles}$. The mole fractions are simply:

With the mole fractions in hand, we can plug everything back into Raoult's Law:

We've found it! The pure vapour pressure of B is $80\text{ mm Hg}$.

The first question asks for the ratio of the molar volume of pure B in the vapour phase to its molar volume in the liquid phase. Let's tackle the vapour phase first. Since the vapour behaves ideally, we can use the Ideal Gas Equation:

We must be careful with units here. The gas constant $R$ is $0.08{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$, so our pressure must be in atmospheres. We convert $80\text{ mm Hg}$ to atm by dividing by $760$:

Now, what about the liquid phase? The molar volume of a liquid is simply its molar mass divided by its density:

To find the ratio, we must align our units. Let's convert the vapour volume to milliliters:

The ratio is:

This massive number beautifully illustrates how much more space a substance occupies as a gas compared to a liquid!

The second question asks for the mole fraction of B in the vapour phase, denoted as $y_B$. To find this, we turn to Dalton's Law of Partial Pressures, which states that the partial pressure of a gas is its mole fraction times the total pressure:

We already know the total pressure is $100\text{ mm Hg}$. The partial pressure of B, $P_B$, is simply its contribution to the total pressure, which we calculated earlier using Raoult's Law:

Plugging this in:

And there we have it! The mole fraction of B in the vapour phase is 0.16. Notice how the mole fraction of B dropped from $0.2$ in the liquid to $0.16$ in the vapour. This makes perfect physical sense because B has a lower pure vapour pressure ($80\text{ mm Hg}$) compared to A ($105\text{ mm Hg}$), meaning B is less volatile and therefore less concentrated in the vapour phase!