Vapour pressure of benzene at `30^(@)C` is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute is (mol. Weight of solvent = 78)

Options

A. `356.2`
B. `456.8`
C. `530.1`
D. `656.7`

Explanation

To find the molecular weight of the non-volatile solute, we can use Raoult's Law, which states that the vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution. The formula for the lowering of vapor pressure can be expressed as:

\[ \Delta P = P^0 - P_s \]

Where:
- \( P^0 \) = vapor pressure of the pure solvent (benzene) = 121.8 mm
- \( P_s \) = vapor pressure of the solution = 120.2 mm
- \( \Delta P \) = lowering of vapor pressure = \( P^0 - P_s \)

### Step 1: Calculate the lowering of vapor pressure (\( \Delta P \))
\[ \Delta P = 121.8 \, \text{mm} - 120.2 \, \text{mm} = 1.6 \, \text{mm} \]

### Step 2: Calculate the mole fraction of the solvent (benzene)
Using Raoult's Law, we can express the lowering of vapor pressure in terms of mole fractions:
\[ \frac{\Delta P}{P^0} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]

Where:
- \( n_{solute} \) = number of moles of the solute
- \( n_{solvent} \) = number of moles of the solvent (benzene)

### Step 3: Calculate the moles of benzene
Given that the mass of benzene is 250 g and its molar mass is 78 g/mol:
\[ n_{solvent} = \frac{250 \, \text{g}}{78 \, \text{g/mol}} \approx 3.21 \, \text{mol} \]

### Step 4: Substitute values into the equation
Substituting the known values into the equation:
\[ \frac{1.6}{121.8} = \frac{n_{solute}}{n_{solute} + 3.21} \]

### Step 5: Solve for \( n_{solute} \)
Let \( n_{solute} = \frac{15 \, \text{g}}{M} \) where \( M \) is the molecular weight of the solute. Thus, we rewrite the equation:
\[ \frac{1.6}{121.8} = \frac{\frac{15}{M}}{\frac{15}{M} + 3.21} \]

Cross-multiplying gives:
\[ 1.6 \left( \frac{15}{M} + 3.21 \right) = 121.8 \cdot \frac{15}{M} \]

### Step 6: Simplify and solve for \( M \)
Expanding and rearranging:
\[ 1.6 \cdot 3.21 + \frac{24}{M} = 121.8 \cdot \frac{15}{M} \]
\[ 5.136 + \frac{24}{M} = \frac{1827}{M} \]

Multiplying through by \( M \) to eliminate the fraction:
\[ 5.136M + 24 = 1827 \]
\[ 5.136M = 1827 - 24 \]
\[ 5.136M = 1803 \]
\[ M = \frac{1803}{5.136} \approx 351.5 \, \text{g/mol} \]

### Final Answer:
The molecular weight of the non-volatile solute is approximately **351.5 g/mol**.