Question:

Consider a vapour-liquid mixture of components \( A \) and \( B \) that obeys Raoult’s law. The vapour pressure of \( A \) is half that of \( B \).The vapour phase concentrations of \( A \) and \( B \) are \( 3 \, \text{mol} \, \text{m}^{-3} \) and \( 6 \, \text{mol} \, \text{m}^{-3} \), respectively. At equilibrium, the ratio of the liquid phase concentration of \( A \) to that of \( B \) is:

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For equilibrium problems involving Raoult’s law, relate vapour phase concentrations to liquid phase mole fractions using the proportionality of partial pressures and given vapour pressures.
Updated On: Jan 24, 2025
  • 1.0
  • 0.5
  • 2.0
  • 1.5
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The Correct Option is A

Solution and Explanation

Step 1: Understand Raoult’s law and equilibrium condition. According to Raoult’s law, the partial pressure of a component \( i \) in the vapour phase is given by: \[ P_i = x_i P_i^*, \] where: - \( P_i \) is the partial pressure of component \( i \), - \( x_i \) is the mole fraction of \( i \) in the liquid phase, - \( P_i^* \) is the vapour pressure of \( i \) in the pure state. At equilibrium, the partial pressure \( P_i \) is proportional to the vapour phase concentration \( C_i \): \[ P_i \propto C_i. \] Step 2: Relate the concentrations in the vapour phase. The given vapour phase concentrations are: \[ C_A = 3 \, \text{mol} \, \text{m}^{-3}, \quad C_B = 6 \, \text{mol} \, \text{m}^{-3}. \] Thus, the ratio of partial pressures is: \[ \frac{P_A}{P_B} = \frac{C_A}{C_B} = \frac{3}{6} = 0.5. \] Step 3: Relate the liquid phase mole fractions. Using Raoult’s law: \[ \frac{P_A}{P_B} = \frac{x_A P_A^*}{x_B P_B^*}. \] The problem states that \( P_A^* \) is half of \( P_B^* \): \[ P_A^* = \frac{1}{2} P_B^*. \] Substitute this into the equation: \[ \frac{P_A}{P_B} = \frac{x_A (\frac{1}{2} P_B^*)}{x_B P_B^*}. \] Simplify: \[ \frac{P_A}{P_B} = \frac{x_A}{2x_B}. \] Step 4: Solve for the ratio \( \frac{x_A}{x_B} \). Substitute \( \frac{P_A}{P_B} = 0.5 \): \[ 0.5 = \frac{x_A}{2x_B}. \] Solve for \( \frac{x_A}{x_B} \): \[ \frac{x_A}{x_B} = 1.0. \] Step 5: Conclusion. The ratio of the liquid phase concentration of \( A \) to that of \( B \) is \( 1.0 \).
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