To solve the problem of determining how much glycerin must be added to 40 dm³ of water to prevent it from freezing at -10°C, we can follow these steps:
### Step 1: Understand the Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[
\Delta T_f = T_f^{\text{solvent}} - T_f^{\text{solution}}
\]
Where:
- \( T_f^{\text{solvent}} \) is the freezing point of pure solvent (water), which is 0°C.
- \( T_f^{\text{solution}} \) is the freezing point of the solution, which is -10°C.
Calculating \( \Delta T_f \):
\[
\Delta T_f = 0 - (-10) = 10°C
\]
### Step 2: Use the Freezing Point Depression Formula
The formula for freezing point depression is:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
Where:
- \( i \) is the van 't Hoff factor (for glycerin, a non-electrolyte, \( i = 1 \)).
- \( K_f \) is the cryoscopic constant for water (which we will calculate).
- \( m \) is the molality of the solution.
### Step 3: Calculate the Cryoscopic Constant \( K_f \)
The formula for \( K_f \) is:
\[
K_f = \frac{R \cdot \Delta H_{fusion} \cdot M_{solvent}}{1000}
\]
Where:
- \( R = 8.314 \, \text{J/mol·K} \)
- \( \Delta H_{fusion} = 6.01 \, \text{kJ/mol} = 6010 \, \text{J/mol} \)
- \( M_{solvent} = 18 \, \text{g/mol} \) (molar mass of water)
Calculating \( K_f \):
\[
K_f = \frac{8.314 \cdot 6010 \cdot 18}{1000} = 1.85 \, \text{kg/mol}
\]
### Step 4: Calculate the Molality \( m \)
Rearranging the freezing point depression formula gives:
\[
m = \frac{\Delta T_f}{i \cdot K_f}
\]
Substituting the known values:
\[
m = \frac{10}{1 \cdot 1.85} \approx 5.41 \, \text{mol/kg}
\]
### Step 5: Calculate the Amount of Glycerin Needed
Molality is defined as:
\[
m = \frac{n}{mass \, of \, solvent \, (kg)}
\]
Where \( n \) is the number of moles of glycerin. Rearranging gives:
\[
n = m \cdot mass \, of \, solvent
\]
The mass of solvent (water) is 40 kg (since 40 dm³ of water has a mass of 40 kg).
Calculating \( n \):
\[
n = 5.41 \cdot 40 \approx 216.4 \, \text{mol}
\]
### Step 6: Calculate the Mass of Glycerin
Using the molar mass of glycerin (\( C_3H_8O_3 \)), which is 92 g/mol:
\[
\text{mass of glycerin} = n \cdot \text{molar mass of glycerin} = 216.4 \cdot 92 \approx 19900.8 \, \text{g} \approx 19.9 \, \text{kg}
\]
### Final Answer
The amount of glycerin that must be added to 40 dm³ of water to prevent it from freezing at -10°C is approximately **19.9 kg**.
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