To solve the problem, we need to find the vapor pressure of water above a solution at 338 K, given that the boiling point of the solution is 375.3 K. We will use the concepts of boiling point elevation and vapor pressure lowering.
### Step-by-Step Solution:
1. **Identify the boiling point elevation (ΔTb)**:
- The boiling point of pure water (solvent) at standard pressure is approximately 373 K.
- The boiling point of the solution is given as 375.3 K.
- Therefore, the boiling point elevation (ΔTb) can be calculated as:
\[
\Delta T_b = T_{b,\text{solution}} - T_{b,\text{pure}} = 375.3 \, \text{K} - 373 \, \text{K} = 2.3 \, \text{K}
\]
2. **Use the boiling point elevation formula**:
- The formula for boiling point elevation is given by:
\[
\Delta T_b = K_b \cdot m
\]
- Where \( K_b \) is the ebullioscopic constant and \( m \) is the molality of the solution.
- Given \( K_b \) for water is 0.52 K kg/mol, we can rearrange the formula to find molality (m):
\[
m = \frac{\Delta T_b}{K_b} = \frac{2.3}{0.52} \approx 4.423 \, \text{mol/kg}
\]
3. **Calculate the relative lowering of vapor pressure**:
- The relative lowering of vapor pressure is given by:
\[
\frac{P^0 - P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}
\]
- We can express the number of moles of solute in terms of molality:
\[
n_{\text{solute}} = m \cdot \text{mass of solvent (kg)}
\]
- For simplicity, we can assume we have 1 kg of solvent (water), thus:
\[
n_{\text{solute}} = 4.423 \, \text{mol}
\]
4. **Calculate the number of moles of solvent**:
- The molecular weight of water is approximately 18 g/mol, so for 1 kg (1000 g) of water:
\[
n_{\text{solvent}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol}
\]
5. **Substitute values into the vapor pressure formula**:
- Now substituting \( n_{\text{solute}} \) and \( n_{\text{solvent}} \) into the relative lowering of vapor pressure equation:
\[
\frac{P^0 - P}{P^0} = \frac{4.423}{55.56}
\]
- Rearranging gives:
\[
P^0 - P = P^0 \cdot \frac{4.423}{55.56}
\]
6. **Calculate the vapor pressure of pure water (P^0)**:
- Given \( P^0 \) (vapor pressure of pure water at 338 K) = 0.2467 atm.
- Substitute \( P^0 \) into the equation:
\[
P^0 - P = 0.2467 \cdot \frac{4.423}{55.56} \approx 0.01964
\]
7. **Calculate the vapor pressure of the solution (P)**:
- Now, solve for \( P \):
\[
P = P^0 - (0.2467 - 0.01964) \approx 0.2467 - 0.01964 \approx 0.22706 \, \text{atm}
\]
- Rounding off gives:
\[
P \approx 0.23 \, \text{atm}
\]
### Final Answer:
The vapor pressure of water above the solution at 338 K is approximately **0.23 atm**.
