Benzene `(C_(6)H_(6), 78 g//mol)` and toluene `(C_(7)H_(8),92g//mol)` form an ideal solution. At `60^(@)C` the vapour pressure of pure benzene and pure toluene are `0.507` atm and `0.184` atm, respectively. The mole fraction of benezen in a solution of these two chemicals that has a vapour preesure of `0.350` atm at `60^(@)C`, will be :

Options

A. `0.514`
B. `0.690`
C. `0.486`
D. `0.190`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Vapor pressure of pure benzene, \( P^0_B = 0.507 \, \text{atm} \)
- Vapor pressure of pure toluene, \( P^0_T = 0.184 \, \text{atm} \)
- Total vapor pressure of the solution, \( P_T = 0.350 \, \text{atm} \)

### Step 2: Define the mole fractions
Let:
- \( x_B \) = mole fraction of benzene
- \( x_T \) = mole fraction of toluene

Since the sum of the mole fractions in a solution is equal to 1, we have:
\[ x_B + x_T = 1 \]

### Step 3: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_T = P^0_B \cdot x_B + P^0_T \cdot x_T \]

Substituting the known values:
\[ 0.350 = 0.507 \cdot x_B + 0.184 \cdot x_T \]

### Step 4: Substitute \( x_T \) in terms of \( x_B \)
From the mole fraction equation, we can express \( x_T \) as:
\[ x_T = 1 - x_B \]

Substituting this into the vapor pressure equation gives:
\[ 0.350 = 0.507 \cdot x_B + 0.184 \cdot (1 - x_B) \]

### Step 5: Simplify the equation
Now, expand and simplify the equation:
\[ 0.350 = 0.507 \cdot x_B + 0.184 - 0.184 \cdot x_B \]
\[ 0.350 = (0.507 - 0.184) \cdot x_B + 0.184 \]
\[ 0.350 - 0.184 = 0.323 \cdot x_B \]
\[ 0.166 = 0.323 \cdot x_B \]

### Step 6: Solve for \( x_B \)
Now, solving for \( x_B \):
\[ x_B = \frac{0.166}{0.323} \]
\[ x_B \approx 0.514 \]

### Step 7: Conclusion
Thus, the mole fraction of benzene in the solution is approximately \( 0.514 \).