If at certain temperature, the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molality of the solution is

Options

A. `0.02`
B. `1.2`
C. `1.11`
D. `0.08`

Explanation

To solve the problem, we need to find the molality of a very dilute aqueous urea solution given the vapor pressures of pure water and the solution.

### Step-by-step Solution:

1. **Identify Given Values**:
- Vapor pressure of pure water (P) = 25 mm Hg
- Vapor pressure of the urea solution (P_s) = 24.5 mm Hg

2. **Calculate the Change in Vapor Pressure**:
\[ \Delta P = P - P_s = 25 \, \text{mm Hg} - 24.5 \, \text{mm Hg} = 0.5 \, \text{mm Hg} \]

3. **Calculate the Relative Lowering of Vapor Pressure (RLVP)**:
\[ \text{RLVP} = \frac{\Delta P}{P} = \frac{0.5 \, \text{mm Hg}}{25 \, \text{mm Hg}} = 0.02 \]

4. **Use Raoult's Law**:
According to Raoult's Law for a non-volatile solute:
\[ \text{RLVP} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
Since the solution is very dilute, we can approximate:
\[ n_{\text{solute}} \ll n_{\text{solvent}} \implies n_{\text{solute}} + n_{\text{solvent}} \approx n_{\text{solvent}} \]
Therefore:
\[ \text{RLVP} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]

5. **Relate Moles to Molality**:
The molality (m) is defined as:
\[ m = \frac{n_{\text{solute}}}{\text{mass of solvent (kg)}} \]
We can express the moles of solvent (water) as:
\[ n_{\text{solvent}} = \frac{m_{\text{solvent}}}{M_{\text{water}}} \]
where \(M_{\text{water}} = 18 \, \text{g/mol}\).

6. **Rearranging the Equation**:
From the RLVP equation:
\[ \text{RLVP} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \implies n_{\text{solute}} = \text{RLVP} \cdot n_{\text{solvent}} \]
Substituting for \(n_{\text{solvent}}\):
\[ n_{\text{solute}} = \text{RLVP} \cdot \frac{m_{\text{solvent}}}{M_{\text{water}}} \]

7. **Substituting Values**:
We know RLVP = 0.02, and let’s assume we have 1 kg of water (1000 g):
\[ n_{\text{solvent}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]
Now substituting back:
\[ n_{\text{solute}} = 0.02 \cdot 55.56 \approx 1.11 \, \text{mol} \]

8. **Calculating Molality**:
Using the definition of molality:
\[ m = \frac{n_{\text{solute}}}{\text{mass of solvent (kg)}} = \frac{1.11 \, \text{mol}}{1 \, \text{kg}} = 1.11 \, \text{mol/kg} \]

### Final Answer:
The molality of the urea solution is approximately **1.11 mol/kg**.

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