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 find the mole fraction of benzene in a solution of benzene and toluene that has a vapor pressure of 0.350 atm at 60°C, we can use Raoult's Law. Here’s a step-by-step solution:

### Step 1: Write down Raoult's Law
Raoult's Law states that the total vapor pressure of a solution (P_total) is the sum of the partial pressures of each component:
\[ P_{total} = P_{A0} \cdot X_A + P_{B0} \cdot X_B \]
Where:
- \( P_{A0} \) = vapor pressure of pure benzene = 0.507 atm
- \( P_{B0} \) = vapor pressure of pure toluene = 0.184 atm
- \( X_A \) = mole fraction of benzene
- \( X_B \) = mole fraction of toluene

### Step 2: Express \( X_B \) in terms of \( X_A \)
Since the sum of the mole fractions in a solution equals 1:
\[ X_A + X_B = 1 \]
Thus, we can express \( X_B \) as:
\[ X_B = 1 - X_A \]

### Step 3: Substitute \( X_B \) into Raoult's Law
Now, substitute \( X_B \) into the Raoult's Law equation:
\[ P_{total} = P_{A0} \cdot X_A + P_{B0} \cdot (1 - X_A) \]
Given that \( P_{total} = 0.350 \) atm, we can write:
\[ 0.350 = 0.507 \cdot X_A + 0.184 \cdot (1 - X_A) \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 0.350 = 0.507 \cdot X_A + 0.184 - 0.184 \cdot X_A \]
Combine the terms involving \( X_A \):
\[ 0.350 = (0.507 - 0.184) \cdot X_A + 0.184 \]
\[ 0.350 - 0.184 = (0.507 - 0.184) \cdot X_A \]
\[ 0.166 = 0.323 \cdot X_A \]

### Step 5: Solve for \( X_A \)
Now, solve for \( X_A \):
\[ X_A = \frac{0.166}{0.323} \]
Calculating this gives:
\[ X_A \approx 0.514 \]

### Final Answer
The mole fraction of benzene in the solution is approximately \( 0.514 \).

---