2.25 g of a Non volatile substance dissolved in 250 g of `C_6H_6`.This solution shows depression in F.P. by 0.256 K.Which of the following is/are correct :
Given that : (`K_b` and `K_f` for `C_6H_6` is 2.53 `"Kmolal"^(-1)` and `5.12 "Kmolal"^(-1),BP` of `C_6H_6=353.3K`)

Options

A. Molar mass of substances is =180
B. B.P. of solution is =180
C. Relative lowering in vapour pressure of solvent =0.0038
D. All are not correct

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Calculate the Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Weight of solute (non-volatile substance)** = 2.25 g
2. **Weight of solvent (C6H6)** = 250 g = 0.250 kg

To find the number of moles of solute, we need the molecular weight (M) of the solute, which we will calculate later.

The formula for molality is:
\[ m = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}} = \frac{\frac{\text{Weight of solute}}{M}}{0.250} \]

### Step 2: Use the Freezing Point Depression Formula
The depression in freezing point (\( \Delta T_f \)) is given by:
\[ \Delta T_f = m \cdot K_f \]
Where:
- \( \Delta T_f \) = 0.256 K
- \( K_f \) for C6H6 = 5.12 K kg/mol

Substituting the values:
\[ 0.256 = \left(\frac{\frac{2.25}{M}}{0.250}\right) \cdot 5.12 \]

### Step 3: Rearranging to Find Molecular Weight (M)
Rearranging the equation to solve for M:
\[ 0.256 = \frac{2.25 \cdot 5.12}{0.250 \cdot M} \]
\[ M = \frac{2.25 \cdot 5.12}{0.256 \cdot 0.250} \]

Calculating the right side:
\[ M = \frac{11.52}{0.064} = 180 \text{ g/mol} \]

### Step 4: Calculate the Elevation in Boiling Point
Next, we calculate the elevation in boiling point using:
\[ \Delta T_b = m \cdot K_b \]
Where:
- \( K_b \) for C6H6 = 2.53 K kg/mol

Using the molality calculated earlier:
\[ m = \frac{2.25}{180} \cdot \frac{1000}{250} = \frac{2.25 \cdot 4}{180} = \frac{9}{180} = 0.050 \text{ mol/kg} \]

Now substituting into the boiling point elevation formula:
\[ \Delta T_b = 0.050 \cdot 2.53 = 0.1265 \text{ K} \]

### Step 5: Calculate the New Boiling Point
The boiling point of the solution is:
\[ \text{New boiling point} = \text{Initial boiling point} + \Delta T_b \]
\[ = 353.3 + 0.1265 = 353.4265 \text{ K} \]

### Step 6: Convert to Celsius
Convert Kelvin to Celsius:
\[ 353.4265 - 273.15 = 80.4265 \text{ °C} \]

### Step 7: Calculate the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is given by:
\[ \frac{P^0 - P_s}{P^0} = \text{Mole fraction of solute} \]
Where:
- Number of moles of solute = \(\frac{2.25}{180}\)
- Number of moles of solvent = \(\frac{250}{78}\)

Calculating mole fraction:
\[ \text{Mole fraction of solute} = \frac{\frac{2.25}{180}}{\frac{2.25}{180} + \frac{250}{78}} \]

Calculating the denominator:
\[ \frac{250}{78} \approx 3.205 \]
Thus,
\[ \text{Mole fraction of solute} \approx \frac{0.0125}{0.0125 + 3.205} \approx 0.0038 \]

### Conclusion
1. The molecular weight of the solute is **180 g/mol**.
2. The boiling point of the solution is approximately **80.43 °C**.
3. The relative lowering of vapor pressure is approximately **0.0038**.