For `[CrCl_(3).xNH_(3)]`, elevation in boiling point of one molal solution is double of one molal solution of glucose , hence `x` is if complex is 100% ionised :

Options

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

Answer

To solve the problem, we need to analyze the boiling point elevation of the solution and the dissociation of the complex compound `[CrCl_(3).xNH_(3)]`. Here’s a step-by-step solution:

### Step 1: Understand the Colligative Property
The elevation in boiling point is a colligative property, which means it depends on the number of solute particles in a solution rather than their identity. The formula for the elevation in boiling point 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 breaks into)
- \(K_b\) = ebullioscopic constant of the solvent
- \(m\) = molality of the solution

### Step 2: Analyze the Given Information
We are given that the elevation in boiling point of a one molal solution of glucose is half that of a one molal solution of `[CrCl_(3).xNH_(3)]`.

For glucose:
- Since glucose does not ionize, \(i = 1\).
- Therefore, the elevation in boiling point for glucose (\(\Delta T_b\)) can be expressed as:

\[ \Delta T_{b, \text{glucose}} = 1 \cdot K_b \cdot 1 = K_b \]

For `[CrCl_(3).xNH_(3)]`:
- Let’s denote the van 't Hoff factor for this complex as \(i\).
- The elevation in boiling point for `[CrCl_(3).xNH_(3)]` can be expressed as:

\[ \Delta T_{b, \text{complex}} = i \cdot K_b \cdot 1 = i \cdot K_b \]

### Step 3: Set Up the Relationship
According to the problem, the elevation in boiling point of the complex is double that of glucose:

\[ \Delta T_{b, \text{complex}} = 2 \cdot \Delta T_{b, \text{glucose}} \]

Substituting the expressions we derived:

\[ i \cdot K_b = 2 \cdot K_b \]

### Step 4: Solve for \(i\)
Dividing both sides by \(K_b\) (assuming \(K_b \neq 0\)):

\[ i = 2 \]

### Step 5: Determine the Composition of the Complex
The complex `[CrCl_(3).xNH_(3)]` dissociates into ions in solution. Given that \(i = 2\), this means that the complex must dissociate into 2 particles.

The dissociation can be represented as:

\[ [CrCl_3 \cdot xNH_3] \rightarrow Cr^{3+} + 3Cl^{-} + xNH_3 \]

From the dissociation, we can see that if \(x\) ammonia molecules are present, the total number of particles will be:

\[ 1 \text{ (Cr}^{3+\text{)}} + 3 \text{ (Cl}^{-}\text{)} + x \text{ (NH}_3\text{)} = 4 + x \]

Since we found \(i = 2\):

\[ 4 + x = 2 \]

### Step 6: Solve for \(x\)
Rearranging gives:

\[ x = 2 - 4 = -2 \]

This indicates that there is an error in our assumption. Since the complex is 100% ionized, we need to consider the coordination number of chromium, which is 6.

### Step 7: Correct the Calculation
We know that \(Cr^{3+}\) has a coordination number of 6, and it can bind with 4 ammonia molecules and 2 chloride ions:

\[ x + 3 = 6 \implies x = 3 \]

### Final Answer
Thus, the value of \(x\) in the complex `[CrCl_(3).xNH_(3)]` is:

\[ \boxed{3} \]