A `1.24 M` aqueous solution of `KI` has density of `1.15 g cm^(-3)`.
Answer the following questions about this solution:
The experimental freezing point of the solution is `-4.46^(@)C`.
What percentage of `KI` is dissociated?

Options

A. `82%`
B. `90%`
C. `83%`
D. None

Answer

To determine the percentage of dissociation of potassium iodide (KI) in a 1.24 M aqueous solution with a density of 1.15 g/cm³ and an experimental freezing point of -4.46°C, we will follow these steps:

### Step 1: Calculate the mass of the solution
Using the density and volume of the solution, we can find the mass of the solution.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} \]
Assuming a volume of 1 L (1000 mL):
\[ \text{Mass of solution} = 1.15 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1150 \, \text{g} \]

### Step 2: Calculate the moles of solute (KI)
Using the molarity (M) and the volume of the solution, we can calculate the number of moles of KI.

\[ \text{Moles of KI} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Moles of KI} = 1.24 \, \text{mol/L} \times 1 \, \text{L} = 1.24 \, \text{mol} \]

### Step 3: Calculate the mass of the solute (KI)
Using the molar mass of KI (approximately 166 g/mol), we can find the mass of KI.

\[ \text{Mass of KI} = \text{Moles of KI} \times \text{Molar mass of KI} \]
\[ \text{Mass of KI} = 1.24 \, \text{mol} \times 166 \, \text{g/mol} = 205.84 \, \text{g} \]

### Step 4: Calculate the mass of the solvent (water)
The mass of the solvent can be calculated by subtracting the mass of the solute from the mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of KI} \]
\[ \text{Mass of solvent} = 1150 \, \text{g} - 205.84 \, \text{g} = 944.16 \, \text{g} \]

### Step 5: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

\[ m = \frac{\text{Moles of KI}}{\text{Mass of solvent (in kg)}} \]
\[ m = \frac{1.24 \, \text{mol}}{0.94416 \, \text{kg}} \approx 1.31 \, \text{mol/kg} \]

### Step 6: Calculate the freezing point depression (ΔTf)
Using the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]
Where \( K_f \) for water is approximately 1.86 °C kg/mol. We can rearrange this to find \( i \):

\[ \Delta T_f = 4.46 \, \text{°C} \quad \text{(experimental value)} \]
\[ 4.46 = i \cdot 1.86 \cdot 1.31 \]
Calculating \( i \):
\[ i = \frac{4.46}{1.86 \cdot 1.31} \approx 1.828 \]

### Step 7: Calculate the degree of dissociation (α)
For KI, which dissociates into K⁺ and I⁻, the van 't Hoff factor \( i \) can be expressed as:

\[ i = 1 + \alpha \]
Where \( \alpha \) is the degree of dissociation. Thus:

\[ \alpha = i - 1 = 1.828 - 1 = 0.828 \]

### Step 8: Calculate the percentage of dissociation
To find the percentage of dissociation:

\[ \text{Percentage of dissociation} = \alpha \times 100\% \]
\[ \text{Percentage of dissociation} = 0.828 \times 100\% \approx 82.8\% \]

Rounding this gives us approximately **83%**.

### Final Answer:
The percentage of KI that is dissociated in the solution is **83%**.