The osmotic pressures of equimolar solutions of urea , `BaCl_(2) and AlCl_(3)`will be in the order :

Options

A. (a) `AlCl_(3) gt BaCl_(2)`` gt urea`
B. (b) `BaCl_(2) gt AlCl_(3)` `gturea`
C. (c) `urea gt``BaCl_(2) gt AlCl_(3)`
D. (d) `BaCl_(2) gt "urea" gtAlCl_(3)`

Answer

To determine the order of osmotic pressures of equimolar solutions of urea, BaCl₂, and AlCl₃, we will follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) is given by the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
Where:
- \( i \) = Van't Hoff factor (number of particles the solute dissociates into)
- \( C \) = concentration of the solution (in molarity)
- \( R \) = universal gas constant (constant)
- \( T \) = temperature (constant)

Since the solutions are equimolar (same concentration) and at the same temperature, the osmotic pressure will depend primarily on the Van't Hoff factor \( i \).

### Step 2: Calculate the Van't Hoff factor for each solute
1. **Urea**: Urea is a non-electrolyte, meaning it does not dissociate in solution. Therefore, the Van't Hoff factor \( i \) for urea is:
\[ i_{\text{urea}} = 1 \]

2. **BaCl₂**: Barium chloride dissociates completely in solution:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
For every 1 mole of BaCl₂, it produces 3 moles of ions (1 Ba²⁺ and 2 Cl⁻). Thus, the Van't Hoff factor \( i \) for BaCl₂ is:
\[ i_{\text{BaCl}_2} = 1 + 2 = 3 \]

3. **AlCl₃**: Aluminum chloride also dissociates completely:
\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]
For every 1 mole of AlCl₃, it produces 4 moles of ions (1 Al³⁺ and 3 Cl⁻). Thus, the Van't Hoff factor \( i \) for AlCl₃ is:
\[ i_{\text{AlCl}_3} = 1 + 3 = 4 \]

### Step 3: Compare the Van't Hoff factors
Now we have the Van't Hoff factors for each solute:
- \( i_{\text{urea}} = 1 \)
- \( i_{\text{BaCl}_2} = 3 \)
- \( i_{\text{AlCl}_3} = 4 \)

### Step 4: Determine the order of osmotic pressures
Since osmotic pressure is directly proportional to the Van't Hoff factor, we can arrange the osmotic pressures in the following order:
- Highest osmotic pressure: AlCl₃ (i = 4)
- Middle osmotic pressure: BaCl₂ (i = 3)
- Lowest osmotic pressure: Urea (i = 1)

Thus, the order of osmotic pressures of equimolar solutions of urea, BaCl₂, and AlCl₃ is:
\[ \text{AlCl}_3 > \text{BaCl}_2 > \text{Urea} \]

### Final Answer
The osmotic pressures of equimolar solutions of urea, BaCl₂, and AlCl₃ will be in the order:
\[ \text{AlCl}_3 > \text{BaCl}_2 > \text{Urea} \]