Animated Solution for Chemistry - Solutions: Liquid $M$ and liquid $N$ form an ideal solution. The vapour pressures of pure liquids $M$ and $N$ are $450$ and $700\text{ mmHg}$, respectively, at the same temperature. Then correct statement is $x_M=\text{mole fraction of }M\text{ in solution;}$ $x_N=\text{mole fraction of }N\text{ in solution;}$ $y_M=\text{mole fraction of }M\text{ in vapour phase;}$ $y_N=\text{mole fraction of }N\text{ in vapour phase}$

Welcome, future engineers and doctors! Today, we are going to dive deep into a beautiful problem from the world of physical chemistry. This isn't just a question about plugging numbers into a formula; it's a story about two liquids, M and N, and their journey from the liquid phase into the vapour phase.

Imagine you are standing inside a closed container. At your feet is a perfectly mixed, ideal solution of liquid M and liquid N. Above your head, the space is filled with the vapours of these two liquids, constantly evaporating and condensing in a dynamic, invisible dance.

Our mission? To figure out how the composition of the liquid mixture relates to the composition of the vapour mixture. Let's break it down!

Liquid Meets Vapour

In our liquid pool, the molecules of M and N are mixed together. We describe their presence using mole fractions. Let's call the mole fraction of M in the liquid $x_M$, and the mole fraction of N in the liquid $x_N$.

But these molecules don't just stay in the liquid. Because they are volatile, they escape into the space above, creating a vapour phase. In this vapour phase, they also have mole fractions, which we'll call $y_M$ and $y_N$.

The core question is: How do $x_M$ and $y_M$ talk to each other? How does the liquid phase communicate with the vapour phase? To answer this, we need to call upon two legendary scientists: François-Marie Raoult and John Dalton.

Raoult and Dalton

Let's start with the liquid phase. Raoult's Law is the bridge that connects the liquid mole fraction to the pressure it exerts as a vapour. It states that the partial pressure of a component is equal to its pure vapour pressure multiplied by its mole fraction in the liquid.

For our liquid M, we can write:

And for liquid N:

Here, $p_M^{\circ }$ and $p_N^{\circ }$ are the pure vapour pressures. Think of them as the "eagerness" of the liquid to evaporate. The question tells us that $p_M^{\circ }=450\text{ mmHg}$ and $p_N^{\circ }=700\text{ mmHg}$. Clearly, liquid N is much more eager to become a gas!

Now, let's look at the vapour phase. Dalton's Law of Partial Pressures tells us how to calculate the partial pressure if we only know about the gas mixture. It states that the partial pressure of a gas is the total pressure multiplied by its mole fraction in the vapour phase.

So, for M and N in the vapour phase, we write:

Now, we have two different ways to express the exact same physical quantity—the partial pressure! This is a classic physics and chemistry trick. When you have two equations for the same thing, you equate them.

Let's equate Raoult's Law and Dalton's Law for component M:

And for component N:

We want to compare the ratios of the mole fractions. The most elegant way to do this is to divide the two equations. Watch what happens to the total pressure, $P_{\text{total}}$.

The $P_{\text{total}}$ beautifully cancels out from the left side! We don't even need to know what the total pressure inside the container is.

We are almost there. The question asks us to compare the ratio $\frac{x_M}{x_N}$ with the ratio $\frac{y_M}{y_N}$. Let's rearrange our equation to isolate $\frac{x_M}{x_N}$ on one side.

This is our master relation. Now, we bring in the numbers given in the problem. We know that $p_N^{\circ }=700\text{ mmHg}$ and $p_M^{\circ }=450\text{ mmHg}$. Let's substitute these values into our equation.

Look closely at the fraction $\frac{700}{450}$. Because the numerator is larger than the denominator, this fraction is strictly greater than 1.

If you multiply a number by something greater than 1, the result is larger than the original number. Therefore, it mathematically guarantees that:

And there we have it! The correct option is (a).

Physical Intuition: Let's take a step back and appreciate what this means physically. Because liquid N has a higher pure vapour pressure ($700>450$), it is more volatile. It escapes into the vapour phase much more easily than M. Therefore, the vapour phase will always be "richer" in N compared to the liquid phase. This means $y_N$ will be relatively larger than $x_N$, which perfectly aligns with our mathematical conclusion!

Before we wrap up, let's talk about a crucial phrase in the question: "ideal solution." Why did the examiner specifically mention this?

In an ideal solution, the intermolecular forces between the different molecules (M-N interactions) are exactly the same as the forces between the identical molecules (M-M and N-N interactions). Because the molecules don't "feel" any difference when they are mixed, they evaporate exactly as Raoult's Law predicts. There are no surprises, no sudden spikes or drops in vapour pressure.

If this were a non-ideal solution showing a positive deviation, the M and N molecules would repel each other slightly, making it even easier for them to escape into the vapour phase. The total pressure would be higher than expected. Conversely, a negative deviation would mean they attract each other strongly, holding each other back in the liquid phase.

But because we are dealing with an ideal solution, our mathematical bridge between Raoult and Dalton stands perfectly solid. The elegance of this problem lies in how it strips away the complexities and leaves us with a pure, logical comparison of volatilities.

Keep visualizing, keep questioning, and never stop loving the beautiful logic of chemistry!