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

Options

A. 3.69atm
B. 4.86atm
C. 2atm
D. None

Explanation

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

### Step 1: Identify the given data
- Volume of urea solution (V1) = 200 mL = 0.200 L
- Molarity of urea solution (C1) = 0.1 M
- Volume of glucose solution (V2) = 200 mL = 0.200 L
- Molarity of glucose solution (C2) = 0.2 M
- Temperature (T) = 300 K
- Gas constant (R) = 0.0821 L·atm/(K·mol)

### Step 2: Calculate the total concentration (C_net) of the mixed solution
The total concentration can be calculated using the formula:
\[ C_{\text{net}} = \frac{C_1 \cdot V_1 + C_2 \cdot V_2}{V_1 + V_2} \]

Substituting the values:
\[ C_{\text{net}} = \frac{(0.1 \, \text{mol/L} \cdot 0.200 \, \text{L}) + (0.2 \, \text{mol/L} \cdot 0.200 \, \text{L})}{0.200 \, \text{L} + 0.200 \, \text{L}} \]

Calculating the numerator:
\[ = (0.1 \cdot 0.200) + (0.2 \cdot 0.200) = 0.020 + 0.040 = 0.060 \, \text{mol} \]

Calculating the denominator:
\[ = 0.200 + 0.200 = 0.400 \, \text{L} \]

Thus,
\[ C_{\text{net}} = \frac{0.060 \, \text{mol}}{0.400 \, \text{L}} = 0.15 \, \text{mol/L} \]

### Step 3: Calculate the osmotic pressure (π)
Using the formula for osmotic pressure:
\[ \pi = C_{\text{net}} \cdot R \cdot T \]

Substituting the values:
\[ \pi = 0.15 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 300 \, \text{K} \]

Calculating:
\[ \pi = 0.15 \cdot 0.0821 \cdot 300 \]
\[ \pi = 0.15 \cdot 24.63 \approx 3.69 \, \text{atm} \]

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

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