The osmotic pressure of a 5% (wt/vol) solution of cane sugar at `150^@C` is

Options

A. 2.45 atm
B. 5.078 atm
C. 3.4 atm
D. 4 atm

Answer

To calculate the osmotic pressure of a 5% (wt/vol) solution of cane sugar at 150°C, we can follow these steps:

### Step 1: Understand the given data
- **Weight of cane sugar (solute)**: 5 grams
- **Volume of solution**: 100 mL = 0.1 L
- **Temperature**: 150°C = 150 + 273 = 423 K
- **Molar mass of cane sugar (C₁₂H₂₂O₁₁)**: 342 g/mol
- **Ideal gas constant (R)**: 0.0821 L·atm/(K·mol)

### Step 2: Calculate the number of moles of cane sugar
To find the number of moles (n) of cane sugar, we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{5 \text{ g}}{342 \text{ g/mol}} \approx 0.0146 \text{ mol} \]

### Step 3: Calculate the molarity (C) of the solution
Molarity (C) is calculated as:
\[ C = \frac{n}{V} \]
Where \( V \) is the volume in liters. Thus:
\[ C = \frac{0.0146 \text{ mol}}{0.1 \text{ L}} = 0.146 \text{ mol/L} \]

### Step 4: Use the osmotic pressure formula
The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \]
Substituting the values:
\[ \pi = 0.146 \text{ mol/L} \cdot 0.0821 \text{ L·atm/(K·mol)} \cdot 423 \text{ K} \]

### Step 5: Calculate the osmotic pressure
Now, calculating the osmotic pressure:
\[ \pi = 0.146 \cdot 0.0821 \cdot 423 \approx 5.078 \text{ atm} \]

### Conclusion
The osmotic pressure of the 5% (wt/vol) solution of cane sugar at 150°C is approximately **5.078 atm**.