To solve the problem, we will follow these steps:
### Step 1: Calculate the depression in freezing point (ΔTf)
The depression in freezing point is given by the formula:
\[
\Delta T_f = T_f^0 - T_f
\]
Where:
- \(T_f^0\) is the freezing point of pure solvent (water), which is \(0^\circ C\).
- \(T_f\) is the freezing point of the solution, which is \(-0.24^\circ C\).
Calculating ΔTf:
\[
\Delta T_f = 0 - (-0.24) = 0.24^\circ C
\]
### Step 2: Use the freezing point depression formula
The freezing point depression can also be expressed as:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
Where:
- \(i\) is the van 't Hoff factor.
- \(K_f\) is the cryoscopic constant for water, which is \(1.86 \, K \, kg \, mol^{-1}\).
- \(m\) is the molality of the solution.
### Step 3: Calculate the molality (m)
Given a \(0.5\%\) solution of KCl, we have:
- Mass of KCl = \(0.5 \, g\)
- Mass of water = \(100 \, g - 0.5 \, g = 99.5 \, g = 0.0995 \, kg\)
To find molality (m):
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
\]
First, calculate the number of moles of KCl:
\[
\text{Molar mass of KCl} = 39.1 \, (K) + 35.5 \, (Cl) = 74.6 \, g/mol
\]
\[
\text{Moles of KCl} = \frac{0.5 \, g}{74.6 \, g/mol} \approx 0.0067 \, mol
\]
Now, calculate molality:
\[
m = \frac{0.0067 \, mol}{0.0995 \, kg} \approx 0.0672 \, mol/kg
\]
### Step 4: Substitute values into the freezing point depression formula
Now we can substitute the values into the freezing point depression formula:
\[
0.24 = i \cdot 1.86 \cdot 0.0672
\]
Rearranging to find \(i\):
\[
i = \frac{0.24}{1.86 \cdot 0.0672} \approx 1.92
\]
### Step 5: Calculate the degree of dissociation (α)
For KCl, it dissociates into \(K^+\) and \(Cl^-\):
\[
KCl \rightarrow K^+ + Cl^-
\]
If we start with 1 mole of KCl, at equilibrium:
- Initial: 1 mole of KCl
- Change: -α (degree of dissociation)
- Final: \(1 - α\) moles of KCl, and α moles of \(K^+\) and α moles of \(Cl^-\).
The total number of particles at equilibrium is:
\[
1 - α + α + α = 1 + α
\]
Setting this equal to the van 't Hoff factor:
\[
1 + α = 1.92
\]
Solving for α:
\[
α = 1.92 - 1 = 0.92
\]
### Step 6: Convert α to percentage
To express the degree of dissociation as a percentage:
\[
\text{Degree of dissociation} = 0.92 \times 100\% = 92\%
\]
### Final Answers
- Van 't Hoff factor (i) = 1.92
- Degree of dissociation (α) = 92%
