The vapour pressure of pure benzene `C_(6)H_(6)` at `50^(@)C` is `268 "torr"`. How many moles of non-volatile solute per mole of benzene is required to prepare a solution of benzene 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 to 167 torr from its pure vapor pressure of 268 torr. We will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure benzene, \( P_0 = 268 \, \text{torr} \)
- Vapor pressure of the solution, \( P = 167 \, \text{torr} \)

2. **Use Raoult's Law:**
According to Raoult's Law:
\[ P = P_0 \cdot x_{\text{solvent}} \]
where \( x_{\text{solvent}} \) is the mole fraction of the solvent (benzene).

3. **Calculate Mole Fraction of Solvent:**
Rearranging the equation to find \( x_{\text{solvent}} \):
\[ x_{\text{solvent}} = \frac{P}{P_0} = \frac{167}{268} \]

4. **Calculate the Value of \( x_{\text{solvent}} \):**
\[ x_{\text{solvent}} \approx 0.622 \]

5. **Determine Mole Fraction of Solute:**
The mole fraction of the solute \( x_{\text{solute}} \) can be calculated as:
\[ x_{\text{solute}} = 1 - x_{\text{solvent}} = 1 - 0.622 = 0.378 \]

6. **Relate Mole Fraction to Moles:**
Let \( n \) be the number of moles of solute per mole of benzene. In this case, the total moles in the solution are \( 1 + n \) (1 mole of benzene and \( n \) moles of solute).

The mole fraction of the solute can be expressed as:
\[ x_{\text{solute}} = \frac{n}{1 + n} \]

7. **Set Up the Equation:**
Now we can set up the equation:
\[ 0.378 = \frac{n}{1 + n} \]

8. **Solve for \( n \):**
Cross-multiplying gives:
\[ 0.378(1 + n) = n \]
Expanding this:
\[ 0.378 + 0.378n = n \]
Rearranging gives:
\[ n - 0.378n = 0.378 \]
\[ 0.622n = 0.378 \]
\[ n = \frac{0.378}{0.622} \approx 0.607 \]

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