A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of solution, what is its osmotic pressure at 273 K?
`(R = 0.081 L atm K^(-1) mol^(-1))`

Options

A. 6.02 atm
B. 4.92 atm
C. 4.04 atm
D. 5.32 atm

Answer

To find the osmotic pressure of the sucrose solution, we can use the formula for osmotic pressure:

\[ \Pi = C \cdot R \cdot T \]

where:
- \(\Pi\) is the osmotic pressure,
- \(C\) is the molarity of the solution,
- \(R\) is the ideal gas constant, and
- \(T\) is the temperature in Kelvin.

### Step 1: Calculate the number of moles of sucrose

First, we need to calculate the number of moles of sucrose in the solution. The number of moles (\(n\)) can be calculated using the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

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

Substituting the values:

\[ n = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 2: Calculate the molarity of the solution

Since the solution is prepared by dissolving the sucrose in 1 litre of solution, the molarity (\(C\)) is given by:

\[ C = \frac{n}{V} \]

where \(V\) is the volume of the solution in litres. Here, \(V = 1 \, \text{L}\):

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

### Step 3: Substitute values into the osmotic pressure formula

Now we can substitute the values into the osmotic pressure formula:

- \(C = 0.2 \, \text{mol/L}\)
- \(R = 0.081 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- \(T = 273 \, \text{K}\)

\[ \Pi = 0.2 \, \text{mol/L} \cdot 0.081 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 273 \, \text{K} \]

Calculating this gives:

\[ \Pi = 0.2 \cdot 0.081 \cdot 273 = 4.48 \, \text{atm} \]

### Final Answer

The osmotic pressure of the sucrose solution at 273 K is approximately:

\[ \Pi \approx 4.48 \, \text{atm} \]