Animated Solution for Chemistry - Solutions: For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here $x_L$ and $x_M$ represent mole fractions of L and M, respectively, in the solution. the correct statement(s) applicable to this system is(are) –

When dealing with binary liquid solutions, graphs can often be tricky if you don't read the axes carefully. Let's break down this classic JEE problem step by step.

First, take a close look at the x-axis. It represents the mole fraction of component M, denoted as $x_M$. Notice that it decreases from $1$ on the left to $0$ on the right. Because this is a binary mixture of liquids L and M, the sum of their mole fractions must always be one ($x_L+x_M=1$). This means that as we move from left to right, the mole fraction of L ($x_L$) is actually increasing from $0$ to $1$.

Now, let's focus on point Z. This point lies on the far right of the graph, exactly where $x_M=0$. If there is no M in the solution, it means the liquid is 100% pure L. Therefore, the y-coordinate at point Z represents the vapor pressure of pure liquid L, which we denote as $p_L^0$.

Raoult's Law

To understand what the curve is telling us, we need a benchmark. What if this solution behaved perfectly ideally? According to Raoult's Law, the partial vapor pressure of a component in an ideal solution is directly proportional to its mole fraction:

If we were to plot this ideal behavior on our graph, it would be a perfectly straight line starting from the origin (where $x_L=0$ and $p_L=0$) and ending exactly at point Z (where $x_L=1$ and $p_L=p_L^0$).

Now, compare our ideal straight line with the actual curve given in the problem. The actual curve bulges upwards, lying entirely above the ideal straight line. Mathematically, this means:

In thermodynamics, we call this a positive deviation from Raoult's Law. But what does this mean physically?

If the vapor pressure is higher than expected, it means the molecules are escaping from the liquid phase into the gas phase much more easily than they would in an ideal scenario. This happens when the new intermolecular forces between the different molecules (L-M interactions) are weaker than the original forces holding the pure liquids together (L-L and M-M interactions). Because the molecules don't hold onto each other as tightly, they vaporize readily. This perfectly validates Statement (A).

Finally, let's look at the shape of the curve as it approaches point Z. As $x_L\to 1$, the solution becomes almost entirely pure liquid L, with just a tiny trace of M. In this highly dilute regime, the few molecules of M are completely surrounded by L molecules and do not significantly disrupt the bulk L-L interactions.

Because of this, the solvent (liquid L) begins to behave ideally. Graphically, you can see the real curve smoothly merging and becoming tangent to the ideal straight line as it reaches Z. This confirms that Raoult's Law is obeyed by the solvent as its mole fraction approaches $1$. This makes Statement (C) absolutely correct.

By carefully reading the axes and understanding the physical meaning behind the curve's shape, we can confidently conclude that the correct statements are (A) and (C).