The vapour pressure of pure benzene, `C_(6)H_(6)` at `50^(@)C` is 268 Torr. How many moles of non-volatile solute per mol of benzene is required to prepare a solution of benzene is required to prepare a solution of benzenen having a vapour pressure of `167` Torr at `50^(@)C` ?

Options

A. `0.377`
B. `0.605`
C. `0.623`
D. `0.395`

Explanation

To solve the problem, we need to determine how many moles of a non-volatile solute are required to lower the vapor pressure of benzene from its pure state to a specified value. We will use Raoult's Law, which relates the vapor pressure of a solvent in a solution to the mole fraction of the solvent.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure benzene, \( P_0 = 268 \) Torr
- Vapor pressure of the solution, \( P_s = 167 \) Torr

2. **Calculate the Vapor Pressure Lowering:**
\[ \Delta P = P_0 - P_s = 268 \, \text{Torr} - 167 \, \text{Torr} = 101 \, \text{Torr} \]

3. **Use Raoult’s Law:**
According to Raoult's Law, the relationship between the vapor pressures is given by:
\[ \frac{P_s}{P_0} = X_B \]
where \( X_B \) is the mole fraction of benzene in the solution.

4. **Calculate the Mole Fraction of Benzene:**
\[ X_B = \frac{P_s}{P_0} = \frac{167}{268} \approx 0.622 \]

5. **Calculate the Mole Fraction of the Solute:**
Since the total mole fraction must equal 1:
\[ X_A = 1 - X_B = 1 - 0.622 = 0.378 \]
where \( X_A \) is the mole fraction of the solute.

6. **Relate Mole Fraction to Moles:**
The mole fraction \( X_A \) can also be expressed in terms of moles:
\[ X_A = \frac{n_A}{n_A + n_B} \]
where \( n_A \) is the number of moles of solute and \( n_B \) is the number of moles of benzene. Assuming we have 1 mole of benzene:
\[ n_B = 1 \]
Thus,
\[ 0.378 = \frac{n_A}{n_A + 1} \]

7. **Solve for \( n_A \):**
Rearranging the equation:
\[ 0.378(n_A + 1) = n_A \]
\[ 0.378n_A + 0.378 = n_A \]
\[ n_A - 0.378n_A = 0.378 \]
\[ 0.622n_A = 0.378 \]
\[ n_A = \frac{0.378}{0.622} \approx 0.607 \]

8. **Conclusion:**
The number of moles of non-volatile solute required per mole of benzene is approximately \( 0.607 \).

### Final Answer:
The number of moles of non-volatile solute required per mole of benzene is approximately \( 0.607 \).