To compare the osmotic pressures of the two solutions, we will use the concept of effective molality, which takes into account the dissociation (for electrolytes) or lack of dissociation (for non-electrolytes) of solutes in solution.
### Step-by-Step Solution:
1. Understand the Types of Solutes:
- Calcium Chloride (CaCl₂) - An electrolyte that dissociates in water.
- Sucrose - A non-electrolyte that does not dissociate in water.
2. Determine the Van't Hoff Factor (i):
- CaCl₂: It dissociates completely into 3 ions (1 Ca²⁺ and 2 Cl⁻), hence [tex]\(i = 3\)[/tex].
- Sucrose: It does not dissociate, hence [tex]\(i = 1\)[/tex].
3. Calculate the Effective Molality:
- Effective Molality for [tex]\( CaCl_2: \)[/tex]
[tex]\[
\text{Effective Molality} = \text{molality} \times i = 1.0 \, m \times 3 = 3.0 \, m
\][/tex]
- Effective Molality for Sucrose:
[tex]\[
\text{Effective Molality} = \text{molality} \times i = 3.0 \, m \times 1 = 3.0 \, m
\][/tex]
4. Compare the Effective Molalities:
- Effective Molality of [tex]\( CaCl_2 \)[/tex] is [tex]\( 3.0 \, m \)[/tex].
- Effective Molality of Sucrose is [tex]\( 3.0 \, m \)[/tex].
Both solutions have the same effective molality (3.0 m).
5. Conclusion:
Since both solutions have the same effective molality, their osmotic pressures will be equal.
Therefore, the correct answer is:
```
They have the same osmotic pressure.
```
