If 300mL of 0.1M urea solution is mixed with 300mL of 0.2M glucose solution at 300K. Calculate osmotic pressure?

Explanation

To calculate the osmotic pressure of the solution formed by mixing 300 mL of 0.1 M urea solution with 300 mL of 0.2 M glucose solution at 300 K, we will follow these steps:

### Step 1: Calculate the total concentration of the solution

We have two solutions being mixed:
- Urea solution: 0.1 M, volume = 300 mL
- Glucose solution: 0.2 M, volume = 300 mL

The total concentration \( C_{\text{net}} \) can be calculated using the formula:

\[ C_{\text{net}} = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2} \]

Where:
- \( C_1 = 0.1 \, \text{M} \) (concentration of urea)
- \( V_1 = 300 \, \text{mL} = 0.3 \, \text{L} \)
- \( C_2 = 0.2 \, \text{M} \) (concentration of glucose)
- \( V_2 = 300 \, \text{mL} = 0.3 \, \text{L} \)

Substituting the values:

\[ C_{\text{net}} = \frac{(0.1 \, \text{M} \times 0.3 \, \text{L}) + (0.2 \, \text{M} \times 0.3 \, \text{L})}{0.3 \, \text{L} + 0.3 \, \text{L}} \]

Calculating the numerator:

\[ = \frac{(0.03 \, \text{mol}) + (0.06 \, \text{mol})}{0.6 \, \text{L}} = \frac{0.09 \, \text{mol}}{0.6 \, \text{L}} = 0.15 \, \text{M} \]

### Step 2: Use the osmotic pressure formula

The osmotic pressure \( \pi \) is given by the formula:

\[ \pi = C_{\text{net}} R T \]

Where:
- \( C_{\text{net}} = 0.15 \, \text{M} \)
- \( R = 0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \) (gas constant)
- \( T = 300 \, \text{K} \)

Substituting the values into the formula:

\[ \pi = 0.15 \, \text{M} \times 0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \times 300 \, \text{K} \]

Calculating:

\[ \pi = 0.15 \times 0.0821 \times 300 \]

\[ \pi = 3.69 \, \text{atm} \]

### Final Answer
The osmotic pressure of the solution is **3.69 atm**.