The vapour pressure of mixture of toluene and xylene at `90^(@)`C is 0.5 atm. If at this temperature 400 mm and 150 mm respectively
then what will be the mole fraction of toluene in mixture ?

Options

A. 0.08
B. 0.92
C. 0.88
D. 0.46

Explanation

To find the mole fraction of toluene in the mixture of toluene and xylene at 90°C, we will use Raoult's Law. Here are the steps to solve the problem:

### Step 1: Understand Raoult's Law
Raoult's Law states that the vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

### Step 2: Write the equation using Raoult's Law
For the mixture of toluene (T) and xylene (X), the total vapor pressure (P_total) can be expressed as:
\[ P_{\text{total}} = P^0_T \cdot X_T + P^0_X \cdot X_X \]
Where:
- \( P^0_T \) = vapor pressure of pure toluene = 400 mm Hg
- \( P^0_X \) = vapor pressure of pure xylene = 150 mm Hg
- \( X_T \) = mole fraction of toluene
- \( X_X \) = mole fraction of xylene

### Step 3: Express the mole fraction of xylene
Since the sum of the mole fractions is equal to 1:
\[ X_X = 1 - X_T \]

### Step 4: Substitute the mole fraction of xylene into the equation
Substituting \( X_X \) into the Raoult's Law equation gives:
\[ P_{\text{total}} = P^0_T \cdot X_T + P^0_X \cdot (1 - X_T) \]

### Step 5: Substitute known values
We know that \( P_{\text{total}} = 0.5 \) atm. We need to convert this to mm Hg:
\[ 0.5 \, \text{atm} = 0.5 \times 760 \, \text{mm Hg} = 380 \, \text{mm Hg} \]

Now substituting the values:
\[ 380 = 400 \cdot X_T + 150 \cdot (1 - X_T) \]

### Step 6: Simplify the equation
Expanding the equation:
\[ 380 = 400 \cdot X_T + 150 - 150 \cdot X_T \]
Combine like terms:
\[ 380 = (400 - 150) \cdot X_T + 150 \]
\[ 380 = 250 \cdot X_T + 150 \]

### Step 7: Solve for \( X_T \)
Subtract 150 from both sides:
\[ 380 - 150 = 250 \cdot X_T \]
\[ 230 = 250 \cdot X_T \]
Now, divide both sides by 250:
\[ X_T = \frac{230}{250} \]
\[ X_T = 0.92 \]

### Conclusion
The mole fraction of toluene in the mixture is \( 0.92 \).

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