The osmotic pressure of 0.2 molar solution of urea at `300 K(R = 0.082)` litre atm `mol^(-1)K^(-1)` is

Options

A. 4.92 atm
B. 1 atm
C. 0.25 atm
D. 27 atm

Explanation

To find the osmotic pressure of a 0.2 molar solution of urea at 300 K, we can use the formula derived from van't Hoff's law of osmotic pressure:

\[ \pi = CRT \]

where:
- \(\pi\) = osmotic pressure
- \(C\) = concentration of the solution (in mol/L)
- \(R\) = ideal gas constant (in L·atm/(mol·K))
- \(T\) = temperature (in K)

### Step-by-Step Solution:

1. **Identify the given values:**
- Concentration \(C = 0.2 \, \text{mol/L}\)
- Temperature \(T = 300 \, \text{K}\)
- Ideal gas constant \(R = 0.082 \, \text{L·atm/(mol·K)}\)

2. **Substitute the values into the osmotic pressure formula:**
\[ \pi = (0.2 \, \text{mol/L}) \times (0.082 \, \text{L·atm/(mol·K)}) \times (300 \, \text{K}) \]

3. **Calculate the product:**
- First, calculate \(0.2 \times 0.082\):
\[ 0.2 \times 0.082 = 0.0164 \]
- Next, multiply this result by 300:
\[ 0.0164 \times 300 = 4.92 \]

4. **Determine the units:**
- The units will be:
\[ \text{mol/L} \times \text{L·atm/(mol·K)} \times \text{K} = \text{atm} \]
- Therefore, the osmotic pressure \(\pi\) is:
\[ \pi = 4.92 \, \text{atm} \]

### Final Answer:
The osmotic pressure of the 0.2 molar solution of urea at 300 K is **4.92 atm**.