Question

Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.
Given: The gas constant \(R = 0.08 \,\text{L atm K}^{-1}\text{ mol}^{-1}\)
Molar mass of A is \(50 \,\text{g mol}^{-1}\)
Molar mass of B is \(57 \,\text{g mol}^{-1}\)
Density of liquid B at \(300\,\text{K}\) is \(0.5 \,\text{g mL}^{-1}\)
\(1\,\text{atm} = 760\,\text{mm Hg}\)
The mole fraction of B in vapour phase which is in equilibrium with this solution is _______.

Hint:

For ideal solutions, first calculate liquid phase mole fraction, then use Raoult’s law to determine vapour pressures and vapour phase composition.

Answer 1

Correct Answer: 0.16

Step 1: Understanding the Question:
We need to calculate the mole fraction of \(B\) in the vapour phase in equilibrium with the solution.
Given: \[ P_{total}=100\ mmHg \] \[ P_A^\circ=105\ mmHg \] The solution is \(5\ molal\) in \(B\).

Step 2: Key Formula or Approach:
Raoult’s law: \[ P_A=x_AP_A^\circ \] Total pressure: \[ P_{total}=P_A+P_B \] Mole fraction in vapour phase: \[ y_B=\frac{P_B}{P_{total}} \]

Step 3: Detailed Explanation:
A \(5\ molal\) solution means: \[ 5\ mol\ B \] in \[ 1000\ g\ A \] Molar mass of \(A\): \[ 50\ g\ mol^{-1} \] Moles of \(A\): \[ \frac{1000}{50}=20 \] Hence: \[ x_A=\frac{20}{20+5} \] \[ =\frac{20}{25} \] \[ =0.8 \] Now: \[ P_A=x_AP_A^\circ \] \[ =0.8\times105 \] \[ =84\ mmHg \] Thus: \[ P_B=100-84 \] \[ =16\ mmHg \] Therefore: \[ y_B=\frac{16}{100} \] \[ =0.16 \]

Step 4: Final Answer:
The mole fraction of \(B\) in vapour phase is: \[ \boxed{0.16} \]