`1 xx 10^(-3)` m solution of `Pt(NH_(3))_(4)Cl_(4)` in `H_(2)O` shows depression in freezing point of `0.0054^(@)C`. The formula of the compound will be [Given `K_(f) (H_(2)O) = 1.86^(@)C m^(-1)`]

Options

A. `[Pt(NH_(3))_(4)]Cl_(4)`
B. `[Pt(NH_(3))Cl]Cl_(3)`
C. `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`
D. `[Pt(NH_(3))Cl_(3)]Cl`

Answer

To solve the problem, we need to determine the formula of the compound `Pt(NH3)4Cl4` based on the given data regarding the depression in freezing point. Here’s a step-by-step solution:

### Step 1: Understand the given data
We have a solution of `Pt(NH3)4Cl4` with a molality of `1 x 10^(-3)` m. The depression in freezing point (ΔTf) is `0.0054 °C`, and the cryoscopic constant (Kf) for water is `1.86 °C kg/mol`.

### Step 2: Use the formula for depression in freezing point
The formula for depression in freezing point is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- ΔTf = depression in freezing point
- i = Van't Hoff factor (number of particles in solution after dissociation)
- Kf = cryoscopic constant
- m = molality of the solution

### Step 3: Substitute the known values into the formula
We can rearrange the formula to solve for the Van't Hoff factor (i):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the known values:
\[ i = \frac{0.0054}{1.86 \cdot (1 \times 10^{-3})} \]

### Step 4: Calculate the value of i
Calculating the denominator:
\[ 1.86 \cdot (1 \times 10^{-3}) = 0.00186 \]
Now substituting this back into the equation for i:
\[ i = \frac{0.0054}{0.00186} \approx 2.903 \]
Rounding this gives us:
\[ i \approx 3 \]

### Step 5: Interpret the value of i
The value of i represents the number of particles into which the compound dissociates in solution. Since i is approximately 3, this means that the compound dissociates into 3 ions in solution.

### Step 6: Determine the formula of the compound
Initially, we have one molecule of `Pt(NH3)4Cl4`. If it dissociates into 3 particles, we can assume it dissociates as follows:
- 1 complex ion `Pt(NH3)4Cl2` (which remains intact)
- 2 chloride ions `2 Cl-`

Thus, the dissociation can be represented as:
\[ Pt(NH3)4Cl4 \rightarrow Pt(NH3)4Cl2 + 2 Cl^- \]
This means that the compound can be represented in a way that shows it forms 3 ions in total.

### Conclusion
The final formula of the compound is:
\[ \text{Pt(NH}_3\text{)}_4\text{Cl}_2 + 2 \text{Cl}^- \]