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

Options

A. 3atm
B. 2atm
C. 4atm
D. 1atm

Explanation

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

### Step 1: Calculate the total concentration (C_total) of the solution.
We have two solutions being mixed:
- Urea solution: 300 mL of 0.1 M
- Glucose solution: 100 mL of 0.2 M

First, we need to find the number of moles of each solute.

**For Urea:**
- Volume (V1) = 300 mL = 0.300 L
- Molarity (C1) = 0.1 M

Number of moles of urea (n1) = C1 × V1 = 0.1 mol/L × 0.300 L = 0.03 moles

**For Glucose:**
- Volume (V2) = 100 mL = 0.100 L
- Molarity (C2) = 0.2 M

Number of moles of glucose (n2) = C2 × V2 = 0.2 mol/L × 0.100 L = 0.02 moles

### Step 2: Calculate the total volume of the mixed solution.
Total volume (V_total) = V1 + V2 = 300 mL + 100 mL = 400 mL = 0.400 L

### Step 3: Calculate the total concentration (C_total).
C_total = (n1 + n2) / V_total
C_total = (0.03 moles + 0.02 moles) / 0.400 L
C_total = 0.05 moles / 0.400 L = 0.125 M

### Step 4: Calculate the osmotic pressure (π).
The formula for osmotic pressure is given by:
\[ \pi = C \cdot R \cdot T \]
Where:
- C = total concentration (C_total) = 0.125 M
- R = gas constant = 0.0821 atm·L/(K·mol)
- T = temperature in Kelvin = 300 K

Substituting the values:
\[ \pi = 0.125 \, \text{mol/L} \times 0.0821 \, \text{atm·L/(K·mol)} \times 300 \, \text{K} \]

Calculating:
\[ \pi = 0.125 \times 0.0821 \times 300 \]
\[ \pi = 3.07 \, \text{atm} \]

### Final Answer:
The osmotic pressure of the mixed solution is approximately **3.07 atm**.

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