A solution of sucrose (molar mass =342) is prepared by dissolving 68.2 g in 1000 g of water. Calculate
The osmotic pressure at 273 K.

Options

A. `2`
B. `3`
C. `4`
D. `4.805`

Explanation

To calculate the osmotic pressure of a sucrose solution at 273 K, we can follow these steps:

### Step 1: Calculate the number of moles of sucrose
To find the number of moles of sucrose, we use the formula:

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

Given:
- Mass of sucrose = 68.2 g
- Molar mass of sucrose = 342 g/mol

Calculating the number of moles:

\[ \text{Number of moles} = \frac{68.2 \, \text{g}}{342 \, \text{g/mol}} \approx 0.1994 \, \text{mol} \]

### Step 2: Calculate the volume of the solution
Since we are dissolving 1000 g of water and the density of water is 1 g/mL, the volume of the solution is:

\[ \text{Volume} = 1000 \, \text{g} = 1000 \, \text{mL} = 1 \, \text{L} \]

### Step 3: Calculate the concentration of the solution
Concentration (C) in mol/L is given by:

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

Substituting the values:

\[ C = \frac{0.1994 \, \text{mol}}{1 \, \text{L}} = 0.1994 \, \text{mol/L} \]

### Step 4: Calculate the osmotic pressure using the formula
The osmotic pressure (\(\pi\)) can be calculated using the formula:

\[ \pi = C R T \]

Where:
- \(C\) = concentration in mol/L = 0.1994 mol/L
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin = 273 K

Substituting the values:

\[ \pi = 0.1994 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 273 \, \text{K} \]

Calculating:

\[ \pi \approx 0.1994 \times 0.0821 \times 273 \approx 4.46 \, \text{atm} \]

### Final Answer
The osmotic pressure of the sucrose solution at 273 K is approximately **4.46 atm**.

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