`25 mL` of an aqueous solution of `KCl` was found to requires `20 mL` of `1M AgNO_(3)` solution when titrated using a `K_(2)CrO_(4)` as indicator. Depression in freezing point of `KCl` solution with `100%` ionisation will be :
`(K_(f) = 2.0 mol^(-1) kg "and molarity = molality")`

Options

A. `5.0^@`
B. `3.2^(@)`
C. `1.6^(@)`
D. `0.8^(@)`

Explanation

To solve the problem, we need to find the depression in freezing point of the KCl solution using the given data. Here’s the step-by-step solution:

### Step 1: Determine the moles of AgNO₃ used
Given that 20 mL of 1 M AgNO₃ is used, we can calculate the moles of AgNO₃:
\[ \text{Moles of AgNO}_3 = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.020 \, \text{L} \times 1 \, \text{mol/L} = 0.020 \, \text{mol} \]

### Step 2: Relate moles of KCl to moles of AgNO₃
Since KCl reacts with AgNO₃ in a 1:1 ratio to form AgCl (a precipitate), the moles of KCl will also be 0.020 mol.

### Step 3: Calculate the molarity of KCl solution
We know that the volume of KCl solution is 25 mL. To find the molarity of KCl:
\[ \text{Molarity of KCl} = \frac{\text{Moles of KCl}}{\text{Volume of solution (L)}} = \frac{0.020 \, \text{mol}}{0.025 \, \text{L}} = 0.8 \, \text{M} \]

### Step 4: Calculate the molality of KCl solution
Since the problem states that molarity and molality are equal, we have:
\[ \text{Molality of KCl} = 0.8 \, \text{mol/kg} \]

### Step 5: Determine the van 't Hoff factor (i)
For KCl, which dissociates completely in solution:
\[ \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \]
The van 't Hoff factor \(i\) for KCl is 2 (since it dissociates into two ions).

### Step 6: Use the freezing point depression formula
The depression in freezing point (\(\Delta T_f\)) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(i = 2\)
- \(K_f = 2.0 \, \text{mol}^{-1} \text{kg}\)
- \(m = 0.8 \, \text{mol/kg}\)

Substituting the values:
\[ \Delta T_f = 2 \cdot 2.0 \cdot 0.8 = 3.2 \, \text{°C} \]

### Final Answer
The depression in freezing point of the KCl solution is \(3.2 \, \text{°C}\).

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