The vapour pressure of pure benzene at `25^(@)C` is 639.7 mm Hg and vapour pressure of a solution of solute in benzene at the same temperature is 631.9 mm Hg. Calculate the molality of the solution.

Explanation

To solve the problem of calculating the molality of a solution using the given vapor pressures of pure benzene and the solution, we will follow these steps:

### 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: Use Raoult's Law
According to Raoult's Law, the decrease in vapor pressure is related to the mole fraction of the solute:
\[ \frac{P₀ - P}{P₀} = \text{mole fraction of solute (X₂)} \]

### Step 3: Calculate the decrease in vapor pressure
\[ P₀ - P = 639.7 \, \text{mm Hg} - 631.9 \, \text{mm Hg} = 7.8 \, \text{mm Hg} \]

### Step 4: Substitute values into Raoult's Law
\[ \frac{7.8 \, \text{mm Hg}}{639.7 \, \text{mm Hg}} = X₂ \]

### Step 5: Calculate the mole fraction of the solute (X₂)
\[ X₂ = \frac{7.8}{639.7} \approx 0.0122 \]

### Step 6: Relate the mole fraction to moles of solute and solvent
Let:
- n₁ = moles of solvent (benzene)
- n₂ = moles of solute

Using the approximation for dilute solutions:
\[ X₂ = \frac{n₂}{n₁ + n₂} \approx \frac{n₂}{n₁} \]
Thus, we have:
\[ n₂ = 0.0122 \cdot n₁ \]

### Step 7: Calculate molality
Molality (m) is defined as:
\[ m = \frac{n₂}{\text{mass of solvent in kg}} \]

Assuming we have 1000 g of benzene (solvent):
\[ \text{mass of solvent in kg} = \frac{1000 \, \text{g}}{1000} = 1 \, \text{kg} \]

### Step 8: Substitute n₂ in terms of n₁
We know that:
\[ n₂ = 0.0122 \cdot n₁ \]
Thus, substituting into the molality equation:
\[ m = \frac{0.0122 \cdot n₁}{1} \]

### Step 9: Calculate n₁ (moles of benzene)
To find n₁, we need the molar mass of benzene (C₆H₆):
- Molar mass of benzene = 78 g/mol

So, for 1000 g of benzene:
\[ n₁ = \frac{1000 \, \text{g}}{78 \, \text{g/mol}} \approx 12.82 \, \text{mol} \]

### Step 10: Substitute n₁ back into the molality equation
\[ m = 0.0122 \cdot 12.82 \approx 0.156 \]

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

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