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
Where:
PA and
PB are the partial pressures of A and B in the vapor phase,
xA and
xB are the mole fractions of A and B in the liquid phase,
PA0 and
PB0 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=21
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=21×500=250mm Hg
PB=xBPB0=21×400=200mm 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
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=PtotalPA=450250=0.555
yB=PtotalPB=450200=0.444
Thus, the mole fractions of A and B in the vapor phase are approximately 0.555 and 0.444, respectively.