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

Options

A. 4.75atm
B. 3.44atm
C. 9.43 atm
D. 2.68atm

Explanation

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

### Step 1: Calculate the total number of moles of solute in each solution.

1. **For Urea:**
- Volume (V1) = 400 mL = 0.4 L
- Molarity (C1) = 0.1 M
- Moles of urea (n1) = C1 × V1 = 0.1 mol/L × 0.4 L = 0.04 moles

2. **For Glucose:**
- Volume (V2) = 300 mL = 0.3 L
- Molarity (C2) = 0.2 M
- Moles of glucose (n2) = C2 × V2 = 0.2 mol/L × 0.3 L = 0.06 moles

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

- Total volume (V_total) = V1 + V2 = 400 mL + 300 mL = 700 mL = 0.7 L

### Step 3: Calculate the concentration of the mixed solution.

- Total moles of solute (n_total) = n1 + n2 = 0.04 moles + 0.06 moles = 0.10 moles
- Concentration (C_net) = n_total / V_total = 0.10 moles / 0.7 L = 0.142857 M (approximately 0.14 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 = concentration of the solution (C_net)
- R = ideal gas constant = 0.0821 L·atm/(K·mol)
- T = temperature in Kelvin = 300 K

Substituting the values:
\[ \pi = 0.14 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K} \]

### Step 5: Perform the calculation.

Calculating the osmotic pressure:
\[ \pi = 0.14 \times 0.0821 \times 300 \]
\[ \pi = 3.43 \, \text{atm} \] (approximately 3.44 atm)

### Final Answer:
The osmotic pressure of the mixed solution is approximately **3.44 atm**.

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