`10.0 g` of glucose `(pi_(1))`, `10.0 g` of (urea`(pi_(2))`, and `10.0 g` of sucrose `(pi_(3))` are dissolved in `250.0 mL` of water at `273 K (pi= "osmotic pressure of a solution")`. The relationship between the osmotic pressure pressure of the solutions is

Options

A. `pi_(1) gt pi_(2) gt pi_(3)`
B. `pi_(3) gt pi_(1) gt pi_(2)`
C. `pi_(2) gt pi_(1) gt pi_(3)`
D. `pi_(2) gt pi_(3) gt pi_(1)`

Explanation

To solve the problem, we need to determine the relationship between the osmotic pressures of three solutions: glucose, urea, and sucrose, each dissolved in 250 mL of water at 273 K. The osmotic pressure (\( \pi \)) is directly proportional to the number of moles of solute present in the solution.

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- Glucose (C₆H₁₂O₆): Molar mass = 180 g/mol
- Urea (NH₂CONH₂): Molar mass = 60 g/mol
- Sucrose (C₁₂H₂₂O₁₁): Molar mass = 342 g/mol

2. **Calculate the Number of Moles for Each Solute**:
- For glucose:
\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{180 \, \text{g/mol}} = 0.0556 \, \text{mol} \]
- For urea:
\[ n_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{60 \, \text{g/mol}} = 0.1667 \, \text{mol} \]
- For sucrose:
\[ n_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{342 \, \text{g/mol}} = 0.0292 \, \text{mol} \]

3. **Compare the Number of Moles**:
- From the calculations:
- \( n_1 \) (glucose) = 0.0556 mol
- \( n_2 \) (urea) = 0.1667 mol
- \( n_3 \) (sucrose) = 0.0292 mol
- Order of moles: \( n_2 > n_1 > n_3 \)

4. **Establish the Relationship of Osmotic Pressures**:
- Since osmotic pressure (\( \pi \)) is directly proportional to the number of moles of solute:
\[ \pi_2 > \pi_1 > \pi_3 \]
- Therefore, the relationship between the osmotic pressures is:
\[ \pi_2 : \pi_1 : \pi_3 \]

5. **Conclusion**:
- The correct relationship is \( \pi_2 > \pi_1 > \pi_3 \), which corresponds to option C.