The vapour pressure of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is :

Options

A. 0.073
B. 0.027
C. 0.27
D. 0.73

Explanation

To find the mole fraction of benzene in the vapor phase in contact with an equimolar solution of benzene and toluene, we can follow these steps:

### Step 1: Identify the given data
- Vapour pressure of pure benzene (P0A) = 160 mm Hg
- Vapour pressure of pure toluene (P0B) = 60 mm Hg
- The solution is equimolar, meaning we have 1 mole of benzene and 1 mole of toluene.

### Step 2: Calculate the mole fractions of benzene and toluene in the liquid phase
Since the solution is equimolar:
- Mole fraction of benzene (XA) = moles of benzene / (moles of benzene + moles of toluene) = 1 / (1 + 1) = 1/2
- Mole fraction of toluene (XB) = moles of toluene / (moles of benzene + moles of toluene) = 1 / (1 + 1) = 1/2

### Step 3: Calculate the partial pressures of benzene and toluene in the vapor phase
Using Raoult's Law:
- Partial pressure of benzene (PA) = P0A * XA = 160 mm Hg * (1/2) = 80 mm Hg
- Partial pressure of toluene (PB) = P0B * XB = 60 mm Hg * (1/2) = 30 mm Hg

### Step 4: Calculate the total pressure (Ptotal) in the vapor phase
Total pressure (Ptotal) = PA + PB = 80 mm Hg + 30 mm Hg = 110 mm Hg

### Step 5: Calculate the mole fraction of benzene in the vapor phase (YA)
Using the formula:
\[ YA = \frac{PA}{Ptotal} \]
Substituting the values:
\[ YA = \frac{80 \text{ mm Hg}}{110 \text{ mm Hg}} = \frac{8}{11} \approx 0.727 \]

### Conclusion
The mole fraction of benzene in the vapor phase is approximately 0.727.

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