Question:

At 300 K, the vapour pressures of A and B liquids are 500 and 400 mm Hg respectively. Equal moles of A and B are mixed to form an ideal solution. The mole fraction of A and B in vapor state is respectively.

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Mole fraction in ideal solutions can be found by dividing the vapor pressure of the component by the total vapor pressure.
Updated On: Mar 13, 2025
  • 0.5,0.5 0.5, 0.5
  • 0.666,0.333 0.666, 0.333
  • 0.444,0.555 0.444, 0.555
  • 0.555,0.444 0.555, 0.444
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The Correct Option is D

Solution and Explanation

We are given the following information:
Vapour pressures of A and B at 300 K are 500 mm Hg and 400 mm Hg, respectively.
Equal moles of A and B are mixed to form an ideal solution. We are asked to find the mole fraction of A and B in the vapor phase. Step 1: Use Raoult's Law Raoult's Law states that the partial pressure of each component in the vapor phase is directly proportional to its mole fraction in the liquid phase: PA=xAPA0andPB=xBPB0 P_A = x_A P_A^0 \quad \text{and} \quad P_B = x_B P_B^0 Where:
PA P_A and PB P_B are the partial pressures of A and B in the vapor phase,
xA x_A and xB x_B are the mole fractions of A and B in the liquid phase,
PA0 P_A^0 and PB0 P_B^0 are the vapor pressures of pure A and B at 300 K. Since equal moles of A and B are mixed, the mole fraction of A and B in the liquid phase is: xA=xB=12 x_A = x_B = \frac{1}{2}
Step 2: Calculate the partial pressures of A and B in the vapor phase Using Raoult's Law, the partial pressures of A and B in the vapor phase are: PA=xAPA0=12×500=250mm Hg P_A = x_A P_A^0 = \frac{1}{2} \times 500 = 250 \, \text{mm Hg} PB=xBPB0=12×400=200mm Hg P_B = x_B P_B^0 = \frac{1}{2} \times 400 = 200 \, \text{mm Hg}
Step 3: Calculate the total pressure The total pressure of the system is the sum of the partial pressures of A and B: Ptotal=PA+PB=250+200=450mm Hg P_{\text{total}} = P_A + P_B = 250 + 200 = 450 \, \text{mm Hg}
Step 4: Calculate the mole fraction of A and B in the vapor phase The mole fraction of A and B in the vapor phase is given by: yA=PAPtotal=250450=0.555 y_A = \frac{P_A}{P_{\text{total}}} = \frac{250}{450} = 0.555 yB=PBPtotal=200450=0.444 y_B = \frac{P_B}{P_{\text{total}}} = \frac{200}{450} = 0.444 Thus, the mole fractions of A and B in the vapor phase are approximately 0.555 and 0.444, respectively.
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