The osmotic pressure of a solution containing 0.6 mol of solute per litre at 300 K is

Explanation

To find the osmotic pressure of a solution, we can use the formula:

\[ \pi = \frac{n}{V} \cdot R \cdot T \]

where:
- \(\pi\) = osmotic pressure
- \(n\) = number of moles of solute
- \(V\) = volume of the solution in liters
- \(R\) = universal gas constant (0.0821 atm·L/(K·mol))
- \(T\) = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of moles of solute (\(n\)) = 0.6 mol
- Volume of solution (\(V\)) = 1 L (since it is given as per liter)
- Temperature (\(T\)) = 300 K
- Gas constant (\(R\)) = 0.0821 atm·L/(K·mol)

2. **Substitute the values into the formula:**
\[ \pi = \frac{0.6 \, \text{mol}}{1 \, \text{L}} \cdot 0.0821 \, \text{atm·L/(K·mol)} \cdot 300 \, \text{K} \]

3. **Calculate the osmotic pressure:**
- First, calculate the product of \(R\) and \(T\):
\[ R \cdot T = 0.0821 \cdot 300 = 24.63 \, \text{atm·L/mol} \]
- Now, substitute this back into the equation for \(\pi\):
\[ \pi = 0.6 \cdot 24.63 = 14.778 \, \text{atm} \]

4. **Round to appropriate significant figures:**
- The final answer can be rounded to 14.78 atm.

### Final Answer:
The osmotic pressure of the solution is approximately **14.78 atm**.