The osmotic pressure of equimolar solutions of `BaCl_(2)`,`NaCl`,and glucose follow the order

Options

A. `BaCl2 gt NaCl gt Glucose`
B. `Glucose gt NaCl gt BaCl_(2)`
C. `NaCl gt BaCl_(2) gt Glucose`
D. `NaCl gt Glucose gt BaCl_(2)`

Answer

To determine the order of osmotic pressure for equimolar solutions of BaCl₂, NaCl, and glucose, we need to consider the dissociation of these compounds in solution and how that affects the number of particles present, which in turn influences osmotic pressure.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have three compounds: Barium Chloride (BaCl₂), Sodium Chloride (NaCl), and Glucose.

2. **Understand Dissociation**:
- **Barium Chloride (BaCl₂)**: This compound dissociates in water to form 3 ions:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
Thus, for BaCl₂, the van 't Hoff factor (i) = 3.

- **Sodium Chloride (NaCl)**: This compound dissociates in water to form 2 ions:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Thus, for NaCl, the van 't Hoff factor (i) = 2.

- **Glucose**: Glucose does not dissociate in solution; it remains as one molecule:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 \]
Thus, for glucose, the van 't Hoff factor (i) = 1.

3. **Determine Osmotic Pressure**: The osmotic pressure (π) is directly proportional to the number of particles in solution:
\[ \pi \propto i \cdot C \]
where C is the molarity of the solution. Since we are considering equimolar solutions, we can focus on the van 't Hoff factor (i) alone.

4. **Rank the Compounds**:
- For BaCl₂: i = 3
- For NaCl: i = 2
- For Glucose: i = 1

5. **Order of Osmotic Pressure**: Based on the values of i, the order of osmotic pressure for equimolar solutions is:
\[ \text{BaCl}_2 > \text{NaCl} > \text{Glucose} \]

### Final Answer:
The osmotic pressure of equimolar solutions of BaCl₂, NaCl, and glucose follows the order:
\[ \text{BaCl}_2 > \text{NaCl} > \text{Glucose} \]