If 200mL of 0.1M urea solution is mixed with 400mL of 0.2M glucose solution at 300K. Calculate osmotic pressure?

Explanation

To calculate the osmotic pressure of the mixed solution of urea and glucose, we will follow these steps:

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

1. **Identify the volumes and concentrations of the solutions:**
- Volume of urea solution (V1) = 200 mL = 0.200 L
- Concentration of urea solution (C1) = 0.1 M
- Volume of glucose solution (V2) = 400 mL = 0.400 L
- Concentration of glucose solution (C2) = 0.2 M

2. **Calculate the number of moles of each solute:**
- Moles of urea = C1 × V1 = 0.1 mol/L × 0.200 L = 0.020 moles
- Moles of glucose = C2 × V2 = 0.2 mol/L × 0.400 L = 0.080 moles

3. **Calculate the total volume of the solution:**
- Total volume (V_total) = V1 + V2 = 200 mL + 400 mL = 600 mL = 0.600 L

4. **Calculate the total number of moles of solute:**
- Total moles of solute = Moles of urea + Moles of glucose = 0.020 + 0.080 = 0.100 moles

5. **Calculate the concentration of the mixed solution (C_net):**
- C_net = Total moles of solute / Total volume = 0.100 moles / 0.600 L = 0.1667 M (approximately 0.16 M)

### Step 2: Calculate the osmotic pressure

1. **Use the formula for osmotic pressure (π):**
\[ \pi = C_{\text{net}} \cdot R \cdot T \]
where:
- \(C_{\text{net}} = 0.16 \, \text{M}\)
- \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\) (the ideal gas constant)
- \(T = 300 \, \text{K}\)

2. **Substitute the values into the formula:**
\[ \pi = 0.16 \, \text{M} \cdot 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 300 \, \text{K} \]

3. **Calculate the osmotic pressure:**
\[ \pi = 0.16 \cdot 0.0821 \cdot 300 = 3.93768 \, \text{atm} \]
Rounding this to three significant figures, we get:
\[ \pi \approx 3.94 \, \text{atm} \]

### Final Answer:
The osmotic pressure of the mixed solution is approximately **3.94 atm**.

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