Vapour pressures of para xylene and dimethylbenezene at `90^@C` are respectivley 150 mm and 400 mm. Calculate the mole fraction of dimethyl benzene in the mixture that boils at `90^@C` , when the pressure is 0.5 atm.

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the pressure from atm to mm Hg.
Given:
- Pressure = 0.5 atm
- Conversion factor: 1 atm = 760 mm Hg

Calculation:
\[ \text{Pressure in mm Hg} = 0.5 \, \text{atm} \times 760 \, \text{mm Hg/atm} = 380 \, \text{mm Hg} \]

### Step 2: Define the mole fractions.
Let:
- Mole fraction of dimethylbenzene = \( x \)
- Mole fraction of paraxylene = \( 1 - x \)

### Step 3: Apply Raoult's Law.
According to Raoult's Law, the vapor pressure of the solution (\( P \)) is given by:
\[ P = P_{A}^0 \cdot (1 - x) + P_{B}^0 \cdot x \]
Where:
- \( P_{A}^0 \) = vapor pressure of paraxylene = 150 mm Hg
- \( P_{B}^0 \) = vapor pressure of dimethylbenzene = 400 mm Hg

Substituting the known values:
\[ 380 = 150 \cdot (1 - x) + 400 \cdot x \]

### Step 4: Simplify the equation.
Expanding the equation:
\[ 380 = 150 - 150x + 400x \]
Combining like terms:
\[ 380 = 150 + 250x \]

### Step 5: Solve for \( x \).
Rearranging the equation:
\[ 250x = 380 - 150 \]
\[ 250x = 230 \]
\[ x = \frac{230}{250} = 0.92 \]

### Step 6: Conclusion.
The mole fraction of dimethylbenzene in the mixture that boils at 90°C when the pressure is 0.5 atm is \( x = 0.92 \).

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