To find the freezing point of a 1.24 M aqueous solution of KI with a density of 1.15 g/cm³, we will follow these steps:
### Step 1: Calculate the mass of the solution
Using the density and volume of the solution, we can calculate the mass of the solution.
\[
\text{Density} = 1.15 \, \text{g/cm}^3
\]
\[
\text{Volume} = 1000 \, \text{cm}^3 = 1 \, \text{L}
\]
\[
\text{Mass of solution} = \text{Density} \times \text{Volume} = 1.15 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1150 \, \text{g}
\]
### Step 2: Calculate the mass of the solute (KI)
To find the mass of KI in the solution, we need to calculate the number of moles of KI using its molarity.
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}}
\]
\[
\text{Moles of KI} = 1.24 \, \text{mol/L} \times 1 \, \text{L} = 1.24 \, \text{mol}
\]
Now, we calculate the mass of KI using its molar mass (molar mass of KI = 166 g/mol).
\[
\text{Mass of KI} = \text{Moles of KI} \times \text{Molar mass of KI} = 1.24 \, \text{mol} \times 166 \, \text{g/mol} = 205.84 \, \text{g}
\]
### Step 3: 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.
\[
\text{Mass of solvent (water)} = \text{Mass of solution} - \text{Mass of solute}
\]
\[
\text{Mass of solvent} = 1150 \, \text{g} - 205.84 \, \text{g} = 944.16 \, \text{g}
\]
### Step 4: Convert the mass of the solvent to kg
To calculate molality, we need the mass of the solvent in kg.
\[
\text{Mass of solvent in kg} = \frac{944.16 \, \text{g}}{1000} = 0.94416 \, \text{kg}
\]
### Step 5: Calculate the molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent.
\[
\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.24 \, \text{mol}}{0.94416 \, \text{kg}} \approx 1.31 \, \text{mol/kg}
\]
### Step 6: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[
\Delta T_f = K_f \times m \times i
\]
Where:
- \( K_f \) for water = 1.86 °C kg/mol
- \( i \) (van 't Hoff factor for KI) = 2 (since KI dissociates into K⁺ and I⁻)
\[
\Delta T_f = 1.86 \, \text{°C kg/mol} \times 1.31 \, \text{mol/kg} \times 2 \approx 4.88 \, \text{°C}
\]
### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution is:
\[
\text{Freezing point} = 0 \, \text{°C} - \Delta T_f = 0 \, \text{°C} - 4.88 \, \text{°C} \approx -4.88 \, \text{°C}
\]
### Final Answer
The freezing point of the solution is approximately **-4.88 °C**.
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