A `0.001` molal solution of `[Pt(NH_(3))_(4)Cl_(4)]` in water had freezing point depression of `0.0054^(@)C`. If `K_(f)` for water is `1.80`, calculating the number f `Cl^(-)` ions furnished.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have:
- Molality (m) of the solution = 0.001 molal
- Freezing point depression (ΔTf) = 0.0054 °C
- Freezing point depression constant (Kf) for water = 1.80 °C/m

### Step 2: Use the formula for freezing point depression
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) is the freezing point depression,
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into),
- \( K_f \) is the freezing point depression constant,
- \( m \) is the molality of the solution.

### Step 3: Rearrange the formula to find \( i \)
We can rearrange the formula to solve for \( i \):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 4: Substitute the values into the equation
Now, substituting the known values into the equation:
\[ i = \frac{0.0054}{1.80 \cdot 0.001} \]

### Step 5: Calculate \( i \)
Calculating the right-hand side:
\[ i = \frac{0.0054}{0.0018} = 3 \]

### Step 6: Interpret the value of \( i \)
The value of \( i = 3 \) indicates that the solute dissociates into 3 particles in solution.

### Step 7: Write the dissociation equation for \([Pt(NH_3)_4Cl_4]\)
The compound \([Pt(NH_3)_4Cl_4]\) dissociates in water as follows:
\[ [Pt(NH_3)_4Cl_4] \rightarrow [Pt(NH_3)_4]^{2+} + 2Cl^{-} \]

### Step 8: Count the ions produced
From the dissociation equation, we see that:
- 1 ion of \([Pt(NH_3)_4]^{2+}\) and
- 2 ions of \(Cl^{-}\)

Thus, the total number of ions produced is:
\[ 1 + 2 = 3 \]

### Conclusion
The number of \(Cl^{-}\) ions furnished by the dissociation of \([Pt(NH_3)_4Cl_4]\) is **2**.

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