Compound `PdCl_(4).6H_(2)O` is a hydrated complex, `1 m` aqueous solution of it has freezing point `269.28 K`. Assuming `100%` ionization of complex, calculate the number of ions furnished by complex in the solution.

Options

A. 1
B. 2
C. 4
D. 0

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the change in freezing point (ΔTf)
The freezing point of pure water (solvent) is 273 K. The freezing point of the solution is given as 269.28 K. Therefore, we can calculate ΔTf as follows:

\[ \Delta T_f = T_f^0 - T_f = 273 \, \text{K} - 269.28 \, \text{K} = 3.72 \, \text{K} \]

### Step 2: Use the freezing point depression formula
The freezing point depression can be expressed using the formula:

\[ \Delta T_f = K_f \cdot m \cdot i \]

Where:
- \( K_f \) (freezing point depression constant for water) = 1.86 K kg/mol
- \( m \) (molality of the solution) = 1 m
- \( i \) (Van't Hoff factor, number of particles in solution)

Substituting the known values into the equation:

\[ 3.72 = 1.86 \cdot 1 \cdot i \]

### Step 3: Solve for the Van't Hoff factor (i)
Rearranging the equation to find \( i \):

\[ i = \frac{3.72}{1.86} = 2 \]

### Step 4: Relate the Van't Hoff factor to the number of ions
The relationship between the degree of dissociation (α), the Van't Hoff factor (i), and the number of ions (n) is given by the formula:

\[ \alpha = \frac{i - 1}{n - 1} \]

Since the problem states that there is 100% ionization, we have:

\[ \alpha = 1 \]

Substituting the values we have:

\[ 1 = \frac{2 - 1}{n - 1} \]

### Step 5: Solve for the number of ions (n)
Rearranging the equation gives:

\[ 1(n - 1) = 1 \quad \Rightarrow \quad n - 1 = 1 \quad \Rightarrow \quad n = 2 \]

### Conclusion
The number of ions furnished by the complex \( PdCl_4 \cdot 6H_2O \) in the solution is **2**.

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