A solution of a non-volatile solute in water has a boiling point of 375.3 K. The vapour pressure of water above this solution at 338 K is: [Given `p^(0)` (water) = 0.2467 atm at 338 K and `K_(b)` for wate = 0.52 K kg `mol^(-1)`]

Options

A. 0.18 atm
B. 0.23 atm
C. 0.34 atm
D. 0.42 atm

Answer

To solve the problem step by step, we will follow the process of calculating the vapor pressure of water above the solution using the given data.

### Step 1: Determine the elevation in boiling point (ΔTb)
The elevation in boiling point is calculated using the formula:
\[ \Delta T_b = T_b (\text{solution}) - T_b (\text{solvent}) \]
Where:
- \(T_b (\text{solution}) = 375.3 \, \text{K}\)
- \(T_b (\text{solvent}) = 373 \, \text{K}\) (boiling point of pure water)

Calculating ΔTb:
\[ \Delta T_b = 375.3 \, \text{K} - 373 \, \text{K} = 2.3 \, \text{K} \]

### Step 2: Use the elevation in boiling point to find molality (m)
The elevation in boiling point is also given by:
\[ \Delta T_b = K_b \cdot m \]
Where:
- \(K_b = 0.52 \, \text{K kg mol}^{-1}\)

Rearranging the formula to find molality (m):
\[ m = \frac{\Delta T_b}{K_b} = \frac{2.3}{0.52} \approx 4.42 \, \text{mol/kg} \]

### Step 3: Calculate the number of moles of water
To find the number of moles of water, we use the formula:
\[ \text{Number of moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} \]
Given:
- Mass of water = 1000 g = 1 kg
- Molar mass of water = 18 g/mol

Calculating the number of moles of water:
\[ \text{Number of moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.55 \, \text{mol} \]

### Step 4: Calculate the number of moles of solute
From the molality calculated in Step 2, we know:
- Molality (m) = 4.42 mol/kg, which means there are 4.42 moles of solute in 1 kg of solvent.

### Step 5: Calculate the total number of moles in the solution
Total moles = moles of water + moles of solute:
\[ \text{Total moles} = 55.55 + 4.42 = 59.97 \, \text{mol} \]

### Step 6: Calculate the mole fraction of water
The mole fraction of water (X_water) is given by:
\[ X_{\text{water}} = \frac{\text{moles of water}}{\text{total moles}} = \frac{55.55}{59.97} \approx 0.926 \]

### Step 7: Calculate the vapor pressure of water above the solution
The vapor pressure of water above the solution (P_solution) can be calculated using Raoult's Law:
\[ P_{\text{solution}} = X_{\text{water}} \cdot P^{0}_{\text{water}} \]
Where:
- \(P^{0}_{\text{water}} = 0.2467 \, \text{atm}\)

Calculating the vapor pressure:
\[ P_{\text{solution}} = 0.926 \cdot 0.2467 \approx 0.228 \, \text{atm} \]

### Step 8: Round off the answer
Rounding off the vapor pressure gives:
\[ P_{\text{solution}} \approx 0.23 \, \text{atm} \]

### Final Answer
The vapor pressure of water above the solution at 338 K is approximately **0.23 atm**.

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