To find the depression in freezing point of the KCl solution, we can follow these steps:
### Step 1: Calculate the moles of AgNO₃ used in the titration.
Given that the concentration of AgNO₃ is 1 M and the volume used is 20 mL, we can calculate the moles of AgNO₃:
\[
\text{Moles of AgNO}_3 = \text{Concentration} \times \text{Volume} = 1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.020 \, \text{mol}
\]
### Step 2: Determine the moles of KCl in the solution.
From the titration reaction, we know that KCl reacts with AgNO₃ in a 1:1 ratio. Therefore, the moles of KCl will also be equal to the moles of AgNO₃ used:
\[
\text{Moles of KCl} = 0.020 \, \text{mol}
\]
### Step 3: Calculate the molality of the KCl solution.
To find the molality, we need to know the mass of the solvent (water) in kilograms. First, we need to find the mass of the solution. Assuming the density of the solution is approximately that of water (1 g/mL), the mass of 25 mL of solution is:
\[
\text{Mass of solution} = 25 \, \text{g}
\]
Now, we can calculate the mass of KCl in grams. The molar mass of KCl is approximately 74.55 g/mol. Thus, the mass of KCl is:
\[
\text{Mass of KCl} = \text{Moles} \times \text{Molar mass} = 0.020 \, \text{mol} \times 74.55 \, \text{g/mol} = 1.491 \, \text{g}
\]
Next, we find the mass of the solvent (water):
\[
\text{Mass of water} = \text{Mass of solution} - \text{Mass of KCl} = 25 \, \text{g} - 1.491 \, \text{g} \approx 23.509 \, \text{g} \approx 0.023509 \, \text{kg}
\]
Now we can calculate the molality (m):
\[
\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.020 \, \text{mol}}{0.023509 \, \text{kg}} \approx 0.851 \, \text{mol/kg}
\]
### Step 4: Calculate the depression in freezing point.
The depression in freezing point (\(\Delta T_f\)) can be calculated using the formula:
\[
\Delta T_f = K_f \times m \times i
\]
Where:
- \(K_f = \frac{10}{9} \, \text{K kg/mol}\)
- \(i\) is the van 't Hoff factor, which is 2 for KCl (since it dissociates into K⁺ and Cl⁻ ions).
Substituting the values:
\[
\Delta T_f = \left(\frac{10}{9}\right) \times 0.851 \, \text{mol/kg} \times 2
\]
Calculating this gives:
\[
\Delta T_f \approx \frac{10}{9} \times 0.851 \times 2 \approx \frac{17.02}{9} \approx 1.89 \, \text{K}
\]
### Final Answer:
The depression in freezing point of the KCl solution is approximately **1.89 K**.
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