The vapour pressure of pure benzene is `639.7 mm Hg` and the vapour pressure of solution of a solute in benzene at the temperature is `631.9 mm Hg`. Calculate the molality of the solution.

Explanation

To solve the problem, we need to calculate the molality of the solution using the given vapor pressures of pure benzene and the solution. Here’s the step-by-step solution:

### Step 1: Identify the Given Values
- Vapor pressure of pure benzene (P₀) = 639.7 mm Hg
- Vapor pressure of the solution (P) = 631.9 mm Hg

### Step 2: Calculate the Change in Vapor Pressure
The change in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P₀ - P \]
Substituting the values:
\[ \Delta P = 639.7 \, \text{mm Hg} - 631.9 \, \text{mm Hg} = 7.8 \, \text{mm Hg} \]

### Step 3: Calculate the Mole Fraction of the Solute
Using Raoult's Law, the mole fraction of the solute (x_solute) can be calculated using the formula:
\[ \frac{\Delta P}{P₀} = x_{\text{solute}} \]
Substituting the values:
\[ x_{\text{solute}} = \frac{7.8 \, \text{mm Hg}}{639.7 \, \text{mm Hg}} \approx 0.0122 \]

### Step 4: Calculate the Mole Fraction of the Solvent
The mole fraction of the solvent (x_solvent) can be calculated as:
\[ x_{\text{solvent}} = 1 - x_{\text{solute}} \]
Substituting the value of x_solute:
\[ x_{\text{solvent}} = 1 - 0.0122 \approx 0.9878 \]

### Step 5: Relate Mole Fraction to Molality
The relationship between mole fraction and molality (m) can be expressed as:
\[ m = \frac{x_{\text{solute}}}{x_{\text{solvent}}} \times \frac{1000}{M_{\text{solvent}}} \]
Where M_solvent is the molar mass of the solvent (benzene, C₆H₆), which is 78 g/mol.

### Step 6: Calculate the Molality
Substituting the values:
\[ m = \frac{0.0122}{0.9878} \times \frac{1000}{78} \]
Calculating the value:
\[ m \approx 0.0123 \times 12.82 \approx 0.158 \]

### Final Answer
The molality of the solution is approximately **0.158 mol/kg**.

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