To find the freezing point of the solution given its boiling point, we can follow these steps:
### Step 1: Identify the given data
- Boiling point of the solution (T_b) = 100.18 °C
- K_b (boiling point elevation constant) = 0.52 K kg mol⁻¹
- K_f (freezing point depression constant) = 1.86 K kg mol⁻¹
### Step 2: Convert the boiling point to Kelvin
To convert Celsius to Kelvin, we add 273.15:
\[ T_b = 100.18 + 273.15 = 373.33 \, \text{K} \]
### Step 3: Determine the boiling point of pure water
The boiling point of pure water (T_b°) is:
\[ T_b° = 373.15 \, \text{K} \]
### Step 4: Calculate the change in boiling point (ΔT_b)
Using the formula for boiling point elevation:
\[ \Delta T_b = T_b - T_b° \]
\[ \Delta T_b = 373.33 \, \text{K} - 373.15 \, \text{K} = 0.18 \, \text{K} \]
### Step 5: Calculate the molality (m) of the solution
Using the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot m \]
Rearranging gives:
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the values:
\[ m = \frac{0.18 \, \text{K}}{0.52 \, \text{K kg mol}^{-1}} \]
\[ m = 0.3462 \, \text{mol kg}^{-1} \]
### Step 6: Calculate the change in freezing point (ΔT_f)
Using the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \]
Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg mol}^{-1} \cdot 0.3462 \, \text{mol kg}^{-1} \]
\[ \Delta T_f = 0.644 \, \text{K} \]
### Step 7: Determine the freezing point of the solution (T_f)
The freezing point of pure water (T_f°) is:
\[ T_f° = 273.15 \, \text{K} \]
Using the formula:
\[ T_f = T_f° - \Delta T_f \]
Substituting the values:
\[ T_f = 273.15 \, \text{K} - 0.644 \, \text{K} \]
\[ T_f = 272.506 \, \text{K} \]
### Step 8: Convert the freezing point back to Celsius
To convert from Kelvin to Celsius:
\[ T_f = 272.506 - 273.15 \]
\[ T_f = -0.644 \, \text{°C} \]
### Final Answer
The freezing point of the solution is approximately:
\[ T_f \approx -0.64 \, \text{°C} \]
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