The osmotic pressure of solution containing `34.2 g` of cane sugar (molar mass = 342 g `mol^(-1)` ) in 1 L of solution at `20^(@)C` is (Given `R = 0.082` L atm `K^(-1) mol^(-1)` )

Options

A. 2.40 atm
B. 3.6 atm
C. 24 atm
D. 0.0024 atm

Answer

To solve the problem of finding the osmotic pressure of a solution containing 34.2 g of cane sugar, we can follow these steps:

### Step 1: Calculate the number of moles of cane sugar
To find the number of moles (n) of cane sugar, we use the formula:

\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of cane sugar = 34.2 g
- Molar mass of cane sugar = 342 g/mol

Substituting the values:

\[ n = \frac{34.2 \, \text{g}}{342 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Calculate the molarity (C) of the solution
Molarity (C) is defined as the number of moles of solute per liter of solution. Since the volume of the solution is 1 L:

\[ C = \frac{n}{\text{Volume (L)}} = \frac{0.1 \, \text{mol}}{1 \, \text{L}} = 0.1 \, \text{mol/L} \]

### Step 3: Convert the temperature to Kelvin
To use the ideal gas law, we need the temperature in Kelvin. The given temperature is:

\[ T = 20^\circ C + 273 = 293 \, K \]

### Step 4: Use the osmotic pressure formula
The formula for osmotic pressure (\(\pi\)) is given by:

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

Where:
- \(C\) = concentration (0.1 mol/L)
- \(R\) = ideal gas constant (0.082 L atm K\(^{-1}\) mol\(^{-1}\))
- \(T\) = temperature in Kelvin (293 K)

Substituting the values into the formula:

\[ \pi = 0.1 \, \text{mol/L} \cdot 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 293 \, K \]

### Step 5: Calculate the osmotic pressure
Now, calculating the osmotic pressure:

\[ \pi = 0.1 \cdot 0.082 \cdot 293 \]

Calculating this:

\[ \pi = 0.1 \cdot 24.106 = 2.4106 \, \text{atm} \]

Rounding to three significant figures, we get:

\[ \pi \approx 2.41 \, \text{atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **2.41 atm**.

---