The osmotic pressure of 1 M solution of sucrose is 22.4 atm. What will be the osmotic potential of 0.1 M sucrose solution?

Options

A. 22.4 atm
B. 2.2 atm
C. `-2.24` bars
D. 0

Answer

To solve the problem of finding the osmotic potential of a 0.1 M sucrose solution given that the osmotic pressure of a 1 M sucrose solution is 22.4 atm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Osmotic Pressure**:
The osmotic pressure (π) can be calculated using the formula:
\[ \pi = MRT \]
where:
- \( \pi \) = osmotic pressure
- \( M \) = molar concentration of the solution
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

2. **Identify Given Values**:
From the problem, we know:
- For the 1 M sucrose solution:
- \( M_1 = 1 \, \text{M} \)
- \( \pi_1 = 22.4 \, \text{atm} \)
- For the 0.1 M sucrose solution:
- \( M_2 = 0.1 \, \text{M} \)
- \( \pi_2 = ? \) (this is what we need to find)

3. **Set Up the Relationship**:
Since \( R \) and \( T \) are constants, we can set up the following relationship between the two solutions:
\[ \frac{\pi_1}{\pi_2} = \frac{M_1}{M_2} \]

4. **Substitute the Known Values**:
Plugging in the known values into the equation:
\[ \frac{22.4 \, \text{atm}}{\pi_2} = \frac{1 \, \text{M}}{0.1 \, \text{M}} \]

5. **Simplify the Equation**:
This simplifies to:
\[ \frac{22.4}{\pi_2} = 10 \]

6. **Solve for \( \pi_2 \)**:
Rearranging the equation to solve for \( \pi_2 \):
\[ \pi_2 = \frac{22.4}{10} \]
\[ \pi_2 = 2.24 \, \text{atm} \]

7. **Final Answer**:
The osmotic potential of a 0.1 M sucrose solution is approximately:
\[ \pi_2 \approx 2.2 \, \text{atm} \]