One molal solution of a complex of cobalt chloride with `NH_(3)` in water showed an elevation in boiling point equal to `2.08^(@)`. Assuming that the complex is completely ionized in the solution, the complex is (`K_(b)` for water = `"0.52 K kg mol"^(-1)`)

Options

A. `[Co(NH_(3))_(6)]Cl_(3)`
B. `[Co(NH_(3))_(5)Cl]Cl_(2)`
C. `[Co(NH_(3))_(4)Cl_(2)]Cl`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- We have a one molal solution of a complex of cobalt chloride with ammonia.
- The elevation in boiling point (ΔT_b) is given as 2.08°C.
- The ebullioscopic constant (K_b) for water is 0.52 K kg mol⁻¹.

### Step 2: Use the formula for boiling point elevation
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 breaks into)
- \( K_b \) = ebullioscopic constant
- \( m \) = molality of the solution

### Step 3: Substitute the known values into the formula
We know:
- \( \Delta T_b = 2.08 \, °C \)
- \( K_b = 0.52 \, K \, kg \, mol^{-1} \)
- \( m = 1 \, mol/kg \)

Substituting these values into the formula:
\[ 2.08 = i \cdot 0.52 \cdot 1 \]

### Step 4: Solve for \( i \)
Rearranging the equation to solve for \( i \):
\[ i = \frac{2.08}{0.52} \]
Calculating this gives:
\[ i = 4 \]

### Step 5: Interpret the value of \( i \)
The value of \( i = 4 \) indicates that the complex dissociates into 4 particles in solution.

### Step 6: Determine the possible complex
Now we need to find a cobalt chloride complex that dissociates into 4 ions. The possible dissociation can be represented as:
\[ CoCl_3 \cdot 6NH_3 \rightarrow Co^{3+} + 3Cl^{-} + 6NH_3 \]
This gives us a total of 4 ions (1 \( Co^{3+} \) and 3 \( Cl^{-} \)).

### Conclusion
The complex that corresponds to the dissociation into 4 ions is \( CoCl_3 \cdot 6NH_3 \).