Question

Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $ x_A $ and $ x_B $ are the mole fraction of A and B in solution while $ y_A $ and $ y_B $ are the mole fraction of A and B in vapour phase then :

• A. \( \frac{x_A}{x_B}<\frac{y_A}{y_B} \)
• B. \( \frac{x_A}{x_B} = \frac{y_A}{y_B} \)
• C. \( \frac{x_A}{x_B}>\frac{y_A}{y_B} \)
• D. \( (x_A - y_A)<(x_B - y_B) \)

Hint:

In an ideal solution, the vapour phase is richer in the more volatile component (the one with the higher vapour pressure). Here, B has a higher vapour pressure than A, so the vapour phase will have a higher mole fraction of B compared to the solution. This implies that the ratio of mole fractions of A to B will be smaller in the vapour phase than in the solution.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the behavior of ideal solutions and the concepts of Raoult's law regarding vapour pressure in a solution. 

1. Understanding Raoult's Law:
   • For an ideal solution, the partial vapour pressure of each component is given by Raoult's Law:
   • \(P_A = x_A \cdot P_{A}^{\circ}\) and \(P_B = x_B \cdot P_{B}^{\circ}\), where \(P_{A}^{\circ}\) and \(P_{B}^{\circ}\) are the vapour pressures of pure components A and B, respectively.
2. Given Data:
   • For pure A, \(P_{A}^{\circ} = 350 \, \text{mm Hg}\)
   • For pure B, \(P_{B}^{\circ} = 750 \, \text{mm Hg}\)
3. Total Vapour Pressure:
   • The total vapour pressure \(P_{\text{total}}\) is the sum of the partial pressures:
   • \(P_{\text{total}} = P_A + P_B = x_A \cdot 350 + x_B \cdot 750\)
4. Mole Fractions in Vapour Phase:
   • The mole fraction of A in the vapour phase, \(y_A\) is given by:
   • \(y_A = \frac{P_A}{P_{\text{total}}} = \frac{x_A \cdot 350}{x_A \cdot 350 + x_B \cdot 750}\)
   • Similarly, \(y_B = \frac{P_B}{P_{\text{total}}} = \frac{x_B \cdot 750}{x_A \cdot 350 + x_B \cdot 750}\)
5. Understanding the Concentration Ratio:
   • We need to compare \(\frac{x_A}{x_B}\) with \(\frac{y_A}{y_B}\).
   • Expressing \(\frac{y_A}{y_B}\) using the above equations:
   • \(\frac{y_A}{y_B} = \frac{x_A \cdot 350 / (x_A \cdot 350 + x_B \cdot 750)}{x_B \cdot 750 / (x_A \cdot 350 + x_B \cdot 750)}\)
   • Simplifying gives:
   • \(\frac{y_A}{y_B} = \frac{x_A \cdot 350}{x_B \cdot 750}\)
   • Thus, \(\frac{y_A}{y_B} = \frac{x_A}{x_B} \cdot \frac{350}{750} = \frac{x_A}{x_B} \cdot \frac{7}{15}\)
6. Conclusion:
   • Since \(\frac{7}{15} < 1\), it follows that \(\frac{x_A}{x_B} > \frac{y_A}{y_B}\).
   • Thus, the correct answer is \(\frac{x_A}{x_B} > \frac{y_A}{y_B}\).

Answer 2

To determine the relationship between the mole fractions of components A and B in both the liquid and vapor phases, we can use Raoult's Law for ideal solutions:

Raoult's Law: For a two-component system, the partial vapor pressure of each component is given by: 
\( P_{A} = P_{A}^0 \times x_A \) 
\( P_{B} = P_{B}^0 \times x_B \) 
where \( P_{A}^0 \) and \( P_{B}^0 \) are the vapor pressures of the pure components A and B, respectively.

Total Vapor Pressure: The total vapor pressure of the solution is: 
\( P_{total} = P_{A} + P_{B} = P_{A}^0 \times x_A + P_{B}^0 \times x_B \)

Mole Fraction in Vapor Phase: The mole fractions in the vapor phase can be calculated using Dalton's Law: 
\( y_A = \frac{P_{A}}{P_{total}} = \frac{P_{A}^0 \times x_A}{P_{total}} \) 
\( y_B = \frac{P_{B}}{P_{total}} = \frac{P_{B}^0 \times x_B}{P_{total}} \)

Comparison: We need to compare \( \frac{x_A}{x_B} \) with \( \frac{y_A}{y_B} \).
Given that \( P_{A}^0 = 350 \) mm Hg and \( P_{B}^0 = 750 \) mm Hg, and knowing that \( P_{B}^0 > P_{A}^0 \), it follows that component B is more volatile than component A. Therefore, in the vapor phase, the mole fraction of B will be greater as compared to A, leading to: 
\( \frac{x_A}{x_B} > \frac{y_A}{y_B} \)

Thus, the correct answer is: 
\( \frac{x_A}{x_B}>\frac{y_A}{y_B} \)