Vapour Pressure Deviation and Raoults Law Behavior for Binary Liquid Mixture

To determine the correct statements regarding the given liquid mixture of $\mathbf{L}$ and $\mathbf{M}$, we analyze the vapour pressure curve using thermodynamic principles and Raoult's law.

1. Analysis of Point $\mathbf{Z}$ and Raoult's Law Behavior

The given graph plots the partial vapour pressure of liquid $\mathbf{L}$ ($p_{\mathbf{L}}$) against the mole fraction of liquid $\mathbf{M}$ ($x_{\mathbf{M}}$).

For a binary liquid mixture, the mole fractions of the two components are related by: $x_{\mathbf{L}} + x_{\mathbf{M}} = 1$

At point $\mathbf{Z}$, the graph shows $x_{\mathbf{M}} = 0$. Substituting this into the relation above gives: $x_{\mathbf{L}} = 1 - 0 = 1$

• Since $x_{\mathbf{L}} = 1$, point $\mathbf{Z}$ represents pure liquid $\mathbf{L}$, and the vapour pressure at $\mathbf{Z}$ corresponds to the vapour pressure of pure liquid $\mathbf{L}$, denoted as $p_{\mathbf{L}}^\circ$.
• According to Raoult's law, the partial vapour pressure of a component is given by: $p_{\mathbf{L}} = x_{\mathbf{L}} p_{\mathbf{L}}^\circ$
• For any real solution, as the concentration of a component approaches purity ($x_{\mathbf{L}} \rightarrow 1$), its behavior approaches that of an ideal solution and obeys Raoult's law tangentially.

Therefore, Point $\mathbf{Z}$ signifies the vapour pressure of pure liquid $\mathbf{L}$, and Raoult's law is obeyed as $x_{\mathbf{L}} \rightarrow 1$. This makes Statement B correct and Statements A and C incorrect.

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2. Intermolecular Forces and Deviation from Raoult's Law

For an ideal solution, the plot of $p_{\mathbf{L}}$ versus $x_{\mathbf{M}}$ would be a linear relationship given by: $p_{\mathbf{L}} = (1 - x_{\mathbf{M}}) p_{\mathbf{L}}^\circ$

In the provided curve:

• The graph lies above the ideal straight line connecting $x_{\mathbf{M}} = 1$ ($p_{\mathbf{L}} = 0$) and $x_{\mathbf{M}} = 0$ ($p_{\mathbf{L}} = p_{\mathbf{L}}^\circ$).
• This upward curvature signifies a positive deviation from Raoult's law ($p_{\mathbf{L}} > x_{\mathbf{L}} p_{\mathbf{L}}^\circ$).

A positive deviation occurs when molecules have a higher tendency to escape into the vapour phase than in an ideal solution. Physically, this means that the solute-solvent interactions ($\mathbf{L}-\mathbf{M}$) are weaker than the solute-solute ($\mathbf{M}-\mathbf{M}$) and solvent-solvent ($\mathbf{L}-\mathbf{L}$) interactions in the pure components: $\text{Interaction Strength: } (\mathbf{L}-\mathbf{L}) \text{ and } (\mathbf{M}-\mathbf{M}) > (\mathbf{L}-\mathbf{M})$

Therefore, the intermolecular attractive forces between $\mathbf{L}-\mathbf{L}$ and $\mathbf{M}-\mathbf{M}$ in pure liquids are stronger than $\mathbf{L}-\mathbf{M}$ interactions in the solution. This makes Statement D correct.

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Final Conclusion

The correct statements are B and D.