If 100mL of 0.1M urea solution is mixed with 100mL of 0.2M glucose solution at 300K. Calculate osmotic pressure?

Options

A. 4.86atm
B. 3.69atm
C. 2.94atm
D. 2.74atm

Explanation

To calculate the osmotic pressure of the solution formed by mixing urea and glucose, we can follow these steps:

### Step 1: Calculate the total concentration of solute (C_total)

We have two solutions:
- Urea: 100 mL of 0.1 M
- Glucose: 100 mL of 0.2 M

First, we need to find the total concentration of solutes after mixing.

1. **Calculate the moles of urea:**
\[ \text{Moles of urea} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

2. **Calculate the moles of glucose:**
\[ \text{Moles of glucose} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

3. **Total moles of solute:**
\[ \text{Total moles} = \text{Moles of urea} + \text{Moles of glucose} = 0.01 \, \text{mol} + 0.02 \, \text{mol} = 0.03 \, \text{mol} \]

4. **Total volume after mixing:**
\[ \text{Total volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

5. **Calculate the total concentration (C_total):**
\[ C_{\text{total}} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.03 \, \text{mol}}{0.2 \, \text{L}} = 0.15 \, \text{M} \]

### Step 2: Calculate the osmotic pressure (π)

The osmotic pressure can be calculated using the formula:
\[ \pi = C_{\text{total}} \cdot R \cdot T \]
where:
- \(C_{\text{total}} = 0.15 \, \text{M}\)
- \(R = 0.0821 \, \text{atm L mol}^{-1} \text{K}^{-1}\) (ideal gas constant)
- \(T = 300 \, \text{K}\)

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

Calculating:
\[ \pi = 0.15 \cdot 0.0821 \cdot 300 = 3.693 \, \text{atm} \approx 3.69 \, \text{atm} \]

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

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