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

Options

A. 3.44atm
B. 2.48atm
C. 1.48atm
D. None of the above

Explanation

To calculate the osmotic pressure of the solution formed by mixing urea and glucose, we can follow these steps:

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

We need to find the concentration of the mixed solution using the formula:

\[ C_{net} = \frac{C_1V_1 + C_2V_2}{V_1 + V_2} \]

Where:
- \(C_1 = 0.1 \, M\) (concentration of urea)
- \(V_1 = 300 \, mL = 0.3 \, L\) (volume of urea solution)
- \(C_2 = 0.2 \, M\) (concentration of glucose)
- \(V_2 = 200 \, mL = 0.2 \, L\) (volume of glucose solution)

Substituting the values into the equation:

\[ C_{net} = \frac{(0.1 \, M \times 0.3 \, L) + (0.2 \, M \times 0.2 \, L)}{0.3 \, L + 0.2 \, L} \]

Calculating the numerator:

\[ C_{net} = \frac{(0.03 \, M \cdot L) + (0.04 \, M \cdot L)}{0.5 \, L} = \frac{0.07 \, M \cdot L}{0.5 \, L} \]

Calculating \(C_{net}\):

\[ C_{net} = 0.14 \, M \]

### Step 2: Calculate the osmotic pressure using the formula

The osmotic pressure (\(\pi\)) can be calculated using the formula:

\[ \pi = C_{net}RT \]

Where:
- \(R = 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1}\) (universal gas constant)
- \(T = 300 \, K\) (temperature)

Substituting the values:

\[ \pi = 0.14 \, M \times 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \times 300 \, K \]

Calculating:

\[ \pi = 0.14 \times 0.0821 \times 300 \]

\[ \pi \approx 3.44 \, \text{atm} \]

### Final Answer

The osmotic pressure of the mixed solution is approximately **3.44 atm**.

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