A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below:

The following inferences are made:
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inferences is/are:

The vapor pressure of a liquid is determined by the kinetic energy of its molecules and the strength of the attractive forces holding them together in the liquid phase:

• Strong Intermolecular Forces: Molecules are held together tightly, making it harder for them to escape into the gas phase. Consequently, the liquid will have a lower vapor pressure at any given temperature and a higher boiling point.
• Weak Intermolecular Forces: Molecules escape easily into the gas phase, resulting in a higher vapor pressure at a given temperature.

Therefore, at any fixed temperature point, the relationship is inversely proportional:

$$\text{Vapor Pressure} \propto \frac{1}{\text{Strength of Intermolecular Forces}}$$

Step-by-Step Analysis of the Graph:

If we draw a vertical line at a constant temperature (for example, at $$313\text{ K}$$ or $$333\text{ K}$$), we can compare the vapor pressures of the three liquids directly:

$$\text{Vapor Pressure Order at constant } T: \quad P_X > P_Y > P_Z$$

Applying our inverse relationship rule, we can deduce the true relative strength of their intermolecular interactions:

$$\text{Intermolecular Interactions Strength Order:} \quad \mathbf{Z > Y > X}$$

Evaluating the Inferences:

• (A) X has higher intermolecular interactions compared to Y: Incorrect (X has a higher vapor pressure, so its forces are weaker than Y).
• (B) X has lower intermolecular interactions compared to Y: Correct (Matches our derived order where $$Y > X$$).
• (C) Z has lower intermolecular interactions compared to Y: Incorrect (Z has the lowest vapor pressure, meaning it possesses the strongest forces).

Conclusion:

Only statement (B) accurately describes the trend shown in the vapor pressure curves.

Answer: Option C — (B)