Question

For two pure volatile liquids X and Y, attractive intermolecular interactions of both X-X and Y-Y are weaker than those of X-Y. The total vapour pressure of an equimolar solution of X and Y is p$_{\text{total}$. The vapour pressure of pure X and pure Y are p$^0_{\text{X}}$ and p$^0_{\text{Y}}$, respectively. Which one of the following relations is correct?

• A. p$_{\text{total}} < $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2
• B. p$_{\text{total}} = $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2
• C. p$_{\text{total}} = $ p$^0_{\text{X}}$ + p$^0_{\text{Y}}$
• D. p$_{\text{total}} > $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2

Hint:

Intermolecular forces rules:
- Stronger $\text{X}-\text{Y}$ interactions $\rightarrow$ negative deviation $\rightarrow$ actual $P < P_{\text{ideal}}$ $\rightarrow$ $p_{\text{total}} < \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2}$.
- Weaker $\text{X}-\text{Y}$ interactions $\rightarrow$ positive deviation $\rightarrow$ actual $P > P_{\text{ideal}}$ $\rightarrow$ $p_{\text{total}} > \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question concerns Raoult's Law and non-ideal solutions.
The strength of intermolecular forces determines the deviation from ideality.

Step 2: Detailed Explanation:
$\bullet$ We are told that X-Y interactions are stronger than X-X and Y-Y interactions.
$\bullet$ This means molecules in the mixture are held more tightly together than in their pure states.
$\bullet$ Stronger attraction results in fewer molecules escaping into the vapor phase (less evaporation).
$\bullet$ This leads to a negative deviation from Raoult's Law.
$\bullet$ For an equimolar solution ($x_{X} = x_{Y} = 0.5$), the ideal pressure would be $p_{ideal} = 0.5 p_{X}^{0} + 0.5 p_{Y}^{0} = \frac{p_{X}^{0} + p_{Y}^{0}}{2}$.
$\bullet$ Due to the negative deviation, the actual pressure $p_{total}$ will be less than the ideal pressure.

Step 3: Final Answer:
$p_{total} < \frac{p_{X}^{0} + p_{Y}^{0}}{2}$.
This matches option (A).