For `[CrCl_(3).xNH_(3)]`, elevation in boiling point of the one molal solution is triple of one molal aqueous solution of urea. Assuming 100% ionisation of complex molecule, calculate the value of x.

Answer

To solve the problem, we need to determine the value of \( x \) in the complex \( [CrCl_3 \cdot xNH_3] \) based on the information provided about the elevation in boiling point. Here’s a step-by-step solution:

### Step 1: Understand the relationship between boiling point elevation and Van't Hoff factor
The elevation in boiling point (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( i \) = Van't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = molal elevation constant of the solvent
- \( m \) = molality of the solution

### Step 2: Set up the equation for the complex and urea
According to the problem, the elevation in boiling point of the one molal solution of the complex is triple that of a one molal aqueous solution of urea. Therefore, we can write:
\[ \Delta T_b \text{ (complex)} = 3 \cdot \Delta T_b \text{ (urea)} \]
Substituting the formula for both solutions, we have:
\[ i_{complex} \cdot K_b \cdot 1 = 3 \cdot (i_{urea} \cdot K_b \cdot 1) \]

### Step 3: Simplify the equation
Since \( K_b \) is the same for both solutions, it cancels out:
\[ i_{complex} = 3 \cdot i_{urea} \]

### Step 4: Determine the Van't Hoff factor for urea
Urea does not dissociate in solution, so:
\[ i_{urea} = 1 \]
Thus:
\[ i_{complex} = 3 \cdot 1 = 3 \]

### Step 5: Analyze the dissociation of the complex
The complex \( [CrCl_3 \cdot xNH_3] \) can dissociate into ions. Given that \( i_{complex} = 3 \), we can express the dissociation as:
\[ [CrCl_3 \cdot xNH_3] \rightarrow CrCl_xNH_3^{2+} + 2Cl^- \]
This means that the complex must dissociate into a total of 3 particles.

### Step 6: Determine the coordination number of chromium
Chromium typically has a coordination number of 6. In the complex \( [CrCl_3 \cdot xNH_3] \), there are 3 chloride ions. To satisfy the coordination number of 6, the remaining ligands must be ammonia (NH₃) molecules.

### Step 7: Calculate the value of \( x \)
Since there are 3 chloride ions, the number of ammonia ligands must be:
\[ x + 3 = 6 \]
Solving for \( x \):
\[ x = 6 - 3 = 3 \]

### Conclusion
The value of \( x \) is 3.