For `[CrCl_(3).xNH_(3)]` elevation in boiling point of one molal solution is double of one molal urea solution, hence x is (Given that the complex is 100% ionised)

Options

A. 4
B. 5
C. 6
D. none of these

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 given about the elevation in boiling point of the solution.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the elevation in boiling point of a one molal solution of the complex is double that of a one molal urea solution. Since urea is a non-electrolyte, it does not dissociate in solution.

2. **Boiling Point Elevation Formula**:
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_b \) = ebullioscopic constant of the solvent
- \( m \) = molality of the solution

3. **Calculating for Urea**:
For urea (which does not dissociate):
- \( i = 1 \)
- \( m = 1 \) (one molal)
- Therefore, the elevation in boiling point for urea is:
\[ \Delta T_b^{urea} = 1 \cdot K_b \cdot 1 = K_b \]

4. **Calculating for the Complex**:
For the complex \([CrCl_3 \cdot xNH_3]\), we know:
- The elevation in boiling point is double that of urea:
\[ \Delta T_b^{complex} = 2 \cdot K_b \]
- Let \( i \) be the van 't Hoff factor for the complex. Then:
\[ \Delta T_b^{complex} = i \cdot K_b \cdot 1 = i \cdot K_b \]

5. **Setting Up the Equation**:
From the above, we can set up the equation:
\[ i \cdot K_b = 2 \cdot K_b \]
Dividing both sides by \( K_b \) (assuming \( K_b \neq 0 \)):
\[ i = 2 \]

6. **Understanding the Ionization**:
The problem states that the complex is 100% ionized. The van 't Hoff factor \( i \) is related to the number of particles produced upon dissociation. Since \( i = 2 \), this means that the complex dissociates into 2 particles.

7. **Structure of the Complex**:
The complex can be represented as:
\[ [CrCl_3 \cdot xNH_3] \]
Given that \( Cr \) has a coordination number of 6, and it is combined with 3 chloride ions, the remaining coordination sites must be filled with ammonia ligands.

8. **Finding \( x \)**:
Since the total coordination number is 6 and we have 3 chloride ions, the number of ammonia ligands \( x \) can be calculated as:
\[ x = 6 - 3 = 3 \]

9. **Final Answer**:
Therefore, the value of \( x \) is \( 3 \).

### Summary:
The value of \( x \) in the complex \([CrCl_3 \cdot xNH_3]\) is \( 3 \).