A solution of sucross (molar mass = 342 g `"mol"^(-1)`) has been prepared by dissolving 68.4 g of sucrose in one kg of water. `K_(f)` for water is `"1.86 K kg mol"^(-1)` and vapour pressure of water at 298 K is 0.024 atm.
The vapour pressure of the solution at 298 K will be

Options

A. 0.230 atm
B. 0.233 atm
C. 0.236 atm
D. 0.0239 atm

Explanation

To find the vapor pressure of the sucrose solution at 298 K, we will use Raoult's Law, which states that the vapor pressure of a solution (P_s) is related to the vapor pressure of the pure solvent (P_0) and the mole fraction of the solvent in the solution.

### Step 1: Write down the given information
- Mass of sucrose (solute), W2 = 68.4 g
- Molar mass of sucrose, M2 = 342 g/mol
- Mass of water (solvent), W1 = 1000 g
- Molar mass of water, M1 = 18 g/mol
- Vapor pressure of pure water at 298 K, P0 = 0.024 atm

### Step 2: Calculate the number of moles of sucrose (N2)
Using the formula for the number of moles:
\[ N_2 = \frac{W_2}{M_2} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} \approx 0.200 \, \text{mol} \]

### Step 3: Calculate the number of moles of water (N1)
Using the formula for the number of moles:
\[ N_1 = \frac{W_1}{M_1} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 4: Calculate the mole fraction of the solvent (water)
The mole fraction of the solvent (X1) is given by:
\[ X_1 = \frac{N_1}{N_1 + N_2} = \frac{55.56}{55.56 + 0.200} \approx \frac{55.56}{55.76} \approx 0.995 \]

### Step 5: Apply Raoult's Law
According to Raoult's Law:
\[ P_s = P_0 \cdot X_1 \]
Substituting the values:
\[ P_s = 0.024 \, \text{atm} \cdot 0.995 \approx 0.02388 \, \text{atm} \]

### Step 6: Final Answer
The vapor pressure of the solution at 298 K is approximately:
\[ P_s \approx 0.0239 \, \text{atm} \]

### Conclusion
Thus, the vapor pressure of the sucrose solution is approximately 0.0239 atm.

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