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 osmotic pressure of the solution at 298 K will be

Options

A. 4.29 atm
B. 4.49 atm
C. 4.69 atm
D. 4.89 atm

Explanation

To calculate the osmotic pressure of the sucrose solution, we will follow these steps:

### Step 1: Calculate the number of moles of sucrose
The number of moles of sucrose can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{mass of sucrose (g)}}{\text{molar mass of sucrose (g/mol)}} \]

Given:
- Mass of sucrose = 68.4 g
- Molar mass of sucrose = 342 g/mol

\[ \text{Number of moles} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 2: Calculate the concentration of the solution
Concentration (C) is defined as the number of moles of solute per liter of solution. Since we dissolved sucrose in 1 kg of water, we can assume the volume of the solution is approximately 1 L (since the density of water is about 1 kg/L).

\[ C = \frac{\text{Number of moles of sucrose}}{\text{Volume of solution (L)}} \]

\[ C = \frac{0.2 \, \text{mol}}{1 \, \text{L}} = 0.2 \, \text{mol/L} \]

### Step 3: Use the osmotic pressure formula
The osmotic pressure (\(\pi\)) can be calculated using the formula:

\[ \pi = CRT \]

Where:
- \(C\) = concentration of the solution (mol/L)
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin = 298 K

Substituting the values:

\[ \pi = (0.2 \, \text{mol/L}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (298 \, \text{K}) \]

\[ \pi = 0.2 \times 0.0821 \times 298 \]

\[ \pi = 4.887 \, \text{atm} \]

### Step 4: Round the result
Rounding the result to two decimal places gives:

\[ \pi \approx 4.89 \, \text{atm} \]

### Final Answer
The osmotic pressure of the sucrose solution at 298 K is approximately **4.89 atm**.

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