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

Options

A. 3.44atm
B. 4.33atm
C. 2.88atm
D. 1.77atm

Explanation

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

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

We have two solutions mixed together:
- 100 mL of 0.1 M urea
- 300 mL of 0.2 M glucose

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

**For urea:**
- Volume (V1) = 100 mL = 0.1 L
- Concentration (C1) = 0.1 M

Number of moles of urea (n1) = C1 × V1 = 0.1 mol/L × 0.1 L = 0.01 moles

**For glucose:**
- Volume (V2) = 300 mL = 0.3 L
- Concentration (C2) = 0.2 M

Number of moles of glucose (n2) = C2 × V2 = 0.2 mol/L × 0.3 L = 0.06 moles

Now, we can find the total number of moles in the solution:
Total moles (n_total) = n1 + n2 = 0.01 + 0.06 = 0.07 moles

Next, we calculate the total volume of the mixed solution:
Total volume (V_total) = V1 + V2 = 100 mL + 300 mL = 400 mL = 0.4 L

Now we can calculate the total concentration (C_net):
\[ C_{net} = \frac{n_{total}}{V_{total}} = \frac{0.07 \text{ moles}}{0.4 \text{ L}} = 0.175 \text{ M} \]

### Step 2: Calculate the osmotic pressure

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

Substituting the values:
\[ \pi = 0.175 \text{ M} \cdot 0.0821 \text{ L·atm/(K·mol)} \cdot 300 \text{ K} \]

Calculating this gives:
\[ \pi = 0.175 \cdot 0.0821 \cdot 300 \]
\[ \pi = 4.31 \text{ atm} \]

### Final Answer:
The osmotic pressure of the mixed solution is approximately **4.31 atm**.

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