At `25^(@)C,` the vapour pressure of pure water is 23.76 mm of Hg and that of an aqueous dilute solution of urea is 22.98 mm of Hg. Calculate the molality of this solution?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Vapor pressure of pure water (P₀) = 23.76 mmHg
- Vapor pressure of the aqueous dilute solution of urea (P) = 22.98 mmHg

### Step 2: Calculate the change in vapor pressure
Using Raoult's Law, we can find the change in vapor pressure (ΔP):
\[ \Delta P = P₀ - P = 23.76 \, \text{mmHg} - 22.98 \, \text{mmHg} = 0.78 \, \text{mmHg} \]

### Step 3: Calculate the mole fraction of the solute (urea)
According to Raoult's Law:
\[ \frac{\Delta P}{P₀} = x_b \]
Where \(x_b\) is the mole fraction of the solute (urea). Substituting the values:
\[ x_b = \frac{0.78}{23.76} \approx 0.0328 \]

### Step 4: Calculate the mole fraction of the solvent (water)
The mole fraction of the solvent (water) can be calculated as:
\[ x_a = 1 - x_b = 1 - 0.0328 \approx 0.9672 \]

### Step 5: Relate mole fraction to molality
The mole fraction of the solute can be expressed in terms of molality (m) and the mass of the solvent. The formula for molality (m) is:
\[ m = \frac{x_b}{x_a} \cdot \frac{1000}{M_a} \]
Where \(M_a\) is the molar mass of the solvent (water), which is approximately 18 g/mol or 0.018 kg/mol.

### Step 6: Substitute the values to find molality
Substituting the values we have:
\[ m = \frac{0.0328}{0.9672} \cdot \frac{1000}{0.018} \]
Calculating this gives:
\[ m \approx \frac{0.0328}{0.9672} \cdot 55555.56 \approx 1.89 \, \text{mol/kg} \]

### Final Result
The molality of the aqueous dilute solution of urea is approximately **1.89 mol/kg**.

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