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 will use the concept of vapor pressure lowering and the formula for calculating molality. Here’s a 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 (Pₛ) = 631.9 mm Hg

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

### Step 3: Use the formula for relative lowering of vapor pressure
The relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P₀} = \frac{n_b}{n_a} \]
Where:
- \( n_b \) = number of moles of solute
- \( n_a \) = number of moles of solvent (benzene)

### Step 4: Relate moles to mass and molality
The number of moles can be expressed in terms of mass and molar mass:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass of the substance (in grams)
- \( M \) = molar mass (in g/mol)

For benzene, the molar mass (Mₐ) = 78 g/mol.

The molality (m) is defined as:
\[ m = \frac{n_b}{\text{mass of solvent (kg)}} \]

### Step 5: Rearranging the formula
We can rearrange the relative lowering of vapor pressure formula:
\[ \frac{\Delta P}{P₀} = \frac{n_b}{\frac{w_a}{M_a}} = \frac{n_b \cdot M_a}{w_a} \]
Where \( w_a \) is the mass of the solvent in grams.

### Step 6: Substitute values into the formula
Now, substituting the values we have:
\[ \frac{7.8}{639.7} = \frac{n_b \cdot 78}{1000} \]
Where \( w_a \) is taken as 1000 g (1 kg of solvent).

### Step 7: Solve for \( n_b \)
Rearranging gives:
\[ n_b = \frac{7.8 \cdot 1000}{639.7 \cdot 78} \]
Calculating this:
\[ n_b = \frac{7800}{49983.6} \approx 0.156 \, \text{mol} \]

### Step 8: Calculate molality
Since we have 1 kg of solvent:
\[ \text{molality} = \frac{n_b}{1 \, \text{kg}} = 0.156 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately **0.156 mol/kg**.

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