If 100mL of 0.1M urea solution is mixed with 200mL of 0.2M glucose solution at 300K. Calculate osmotic pressure?

Options

A. 3.94atm
B. 4.86atm
C. Both (A) and (B)
D. None of the above

Explanation

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

### Step 1: Calculate the moles of urea and glucose

1. **For urea:**
- Volume of urea solution (V1) = 100 mL = 0.1 L
- Concentration of urea (C1) = 0.1 M
- Moles of urea = C1 × V1 = 0.1 mol/L × 0.1 L = 0.01 moles

2. **For glucose:**
- Volume of glucose solution (V2) = 200 mL = 0.2 L
- Concentration of glucose (C2) = 0.2 M
- Moles of glucose = C2 × V2 = 0.2 mol/L × 0.2 L = 0.04 moles

### Step 2: Calculate the total volume of the mixed solution

- Total volume (V_total) = V1 + V2 = 100 mL + 200 mL = 300 mL = 0.3 L

### Step 3: Calculate the total concentration of the solution

- Total moles = Moles of urea + Moles of glucose = 0.01 moles + 0.04 moles = 0.05 moles
- Total concentration (C_net) = Total moles / Total volume = 0.05 moles / 0.3 L = 0.1667 M (approximately 0.16 M)

### Step 4: Calculate the osmotic pressure using the formula

- The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \]
where:
- C = total concentration (C_net) = 0.16 M
- R = gas constant = 0.0821 atm·L/(K·mol)
- T = temperature = 300 K

- Substituting the values:
\[ \pi = 0.16 \, \text{M} \times 0.0821 \, \text{atm·L/(K·mol)} \times 300 \, \text{K} \]

### Step 5: Calculate the osmotic pressure

- Performing the calculation:
\[ \pi = 0.16 \times 0.0821 \times 300 \approx 3.94 \, \text{atm} \]

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

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