The osmotic pressure of a 5% solution of cane sugar (molecular mass 342) at `15^(@)C` is :

Options

A. 3.46 atm
B. 3.64 atm
C. 4.0 atm
D. 2.45 atm

Answer

To find the osmotic pressure of a 5% solution of cane sugar (sucrose) at 15°C, we can use the formula for osmotic pressure:

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

Where:
- \(\Pi\) = osmotic pressure
- \(C\) = concentration in molarity (mol/L)
- \(R\) = universal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Convert the percentage concentration to molarity

A 5% solution means there are 5 grams of solute (cane sugar) in 100 mL of solution.

1. **Calculate the number of moles of cane sugar:**
\[ \text{Moles of cane sugar} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ = \frac{5 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0146 \, \text{mol} \]

2. **Convert the volume of the solution to liters:**
\[ 100 \, \text{mL} = 0.1 \, \text{L} \]

3. **Calculate the molarity (C):**
\[ C = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.0146 \, \text{mol}}{0.1 \, \text{L}} = 0.146 \, \text{mol/L} \]

### Step 2: Convert the temperature to Kelvin

The temperature in Celsius is 15°C. To convert to Kelvin:
\[ T(K) = T(°C) + 273.15 = 15 + 273.15 = 288.15 \, K \]

### Step 3: Calculate the osmotic pressure

Now we can substitute the values into the osmotic pressure formula:
\[ \Pi = C \cdot R \cdot T \]
\[ = 0.146 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 288.15 \, K \]

Calculating this gives:
\[ \Pi \approx 3.43 \, \text{atm} \]

### Final Answer
The osmotic pressure of a 5% solution of cane sugar at 15°C is approximately **3.43 atm**.

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