The vapour pressure of benzene at`80^(@)C` is lowered by `10mm` by dissoving `2g`of a non-volatile substance in`78g`of benzene .The vapour pressure of pure benzene at `80^(@)C`is `750mm`.The molecular weight of the substance will be:

Options

A. 15
B. 150
C. 1500
D. 148

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure benzene (P₀) = 750 mm
- Decrease in vapor pressure (ΔP) = 10 mm
- Mass of non-volatile solute (W_solute) = 2 g
- Mass of benzene (W_solvent) = 78 g

### Step 2: Calculate the vapor pressure of the solution
The vapor pressure of the solution (P_solution) can be calculated using the formula:
\[ P_{\text{solution}} = P_0 - \Delta P \]
Substituting the values:
\[ P_{\text{solution}} = 750 \, \text{mm} - 10 \, \text{mm} = 740 \, \text{mm} \]

### Step 3: Use the relative lowering of vapor pressure
The relative lowering of vapor pressure is given by:
\[ \frac{P_0 - P_{\text{solution}}}{P_0} = \text{mole fraction of solute} \]
Substituting the values:
\[ \frac{750 \, \text{mm} - 740 \, \text{mm}}{750 \, \text{mm}} = \frac{10 \, \text{mm}}{750 \, \text{mm}} = \frac{1}{75} \]

### Step 4: Relate mole fraction to moles of solute and solvent
Let \( n_{\text{solute}} \) be the number of moles of the solute and \( n_{\text{solvent}} \) be the number of moles of the solvent (benzene). The mole fraction of solute can be expressed as:
\[ \text{mole fraction of solute} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
Since the mass of the solute is much smaller than that of the solvent, we can approximate:
\[ \text{mole fraction of solute} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]

### Step 5: Calculate moles of solvent
The number of moles of benzene (solvent) can be calculated using its molar mass (78 g/mol):
\[ n_{\text{solvent}} = \frac{W_{\text{solvent}}}{M_{\text{solvent}}} = \frac{78 \, \text{g}}{78 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 6: Express moles of solute
The number of moles of the solute can be expressed as:
\[ n_{\text{solute}} = \frac{W_{\text{solute}}}{M_{\text{solute}}} = \frac{2 \, \text{g}}{M_{\text{solute}}} \]

### Step 7: Set up the equation
From the mole fraction approximation:
\[ \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{1}{75} \]
Substituting the values:
\[ \frac{\frac{2}{M_{\text{solute}}}}{1} = \frac{1}{75} \]

### Step 8: Solve for molecular weight of the solute
Cross-multiplying gives:
\[ 2 \cdot 75 = M_{\text{solute}} \]
\[ M_{\text{solute}} = 150 \]

### Step 9: Conclusion
The molecular weight of the non-volatile substance is 150 g/mol.