The vapour pressure of pure benzene at `25^@C` is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution of

Options

A. 0.2 m
B. 0.4 m
C. 0.5 m
D. 0.1 m

Answer

To solve the problem, we need to calculate the molality of the solution using the given vapor pressures of pure benzene and the solution containing solute A.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Vapor pressure of pure benzene (P₀) = 640 mm Hg
- Vapor pressure of solution (Pₛ) = 630 mm Hg

2. **Calculate the Relative Lowering of Vapor Pressure**:
\[ \Delta P = P₀ - Pₛ = 640 \, \text{mm Hg} - 630 \, \text{mm Hg} = 10 \, \text{mm Hg} \]

3. **Calculate the Relative Lowering of Vapor Pressure**:
\[ \text{Relative Lowering} = \frac{\Delta P}{P₀} = \frac{10 \, \text{mm Hg}}{640 \, \text{mm Hg}} = \frac{1}{64} \]

4. **Relate Relative Lowering to Mole Fraction of Solute**:
The relative lowering of vapor pressure is equal to the mole fraction of the solute (xₗ):
\[ xₗ = \frac{n_B}{n_A + n_B} \approx \frac{1}{64} \]
where \( n_A \) is the number of moles of solvent (benzene) and \( n_B \) is the number of moles of solute A.

5. **Calculate the Mole Fraction of Solvent**:
Since \( x_A + x_B = 1 \):
\[ x_A = 1 - x_B = 1 - \frac{1}{64} = \frac{63}{64} \]

6. **Set Up the Ratio of Moles**:
From the mole fractions, we can write:
\[ \frac{n_B}{n_A} = \frac{x_B}{x_A} = \frac{\frac{1}{64}}{\frac{63}{64}} = \frac{1}{63} \]

7. **Express Moles in Terms of Mass**:
The molality (m) is defined as:
\[ m = \frac{n_B}{\text{mass of solvent in kg}} = \frac{n_B}{\frac{W_A}{1000}} \]
where \( W_A \) is the mass of the solvent (benzene) in grams.

8. **Relate Moles to Mass**:
We know:
\[ n_B = \frac{W_B}{M_B} \quad \text{and} \quad n_A = \frac{W_A}{M_A} \]
Thus, we can express:
\[ \frac{W_B}{M_B} = \frac{1}{63} \cdot \frac{W_A}{M_A} \]

9. **Substituting Values**:
Rearranging gives:
\[ m = \frac{W_B}{M_B} \cdot \frac{1000}{W_A} \]
Substituting \( n_B = \frac{1}{63} n_A \):
\[ m = \frac{1}{63} \cdot \frac{W_A}{M_A} \cdot \frac{1000}{W_A} = \frac{1000}{63 M_A} \]

10. **Calculate the Molar Mass of Benzene**:
The molar mass of benzene (C₆H₆) is:
\[ M_A = 78 \, \text{g/mol} \]

11. **Final Calculation of Molality**:
\[ m = \frac{1000}{63 \times 78} \approx 0.203 \, \text{mol/kg} \]

12. **Conclusion**:
The molality of the solution is approximately 0.203 mol/kg, which can be rounded to 0.2 mol/kg.