How much ethyl alcohol must be added to `1.00 L` of water so that the solution will not freeze at `-4^(@)F` ? `K_f` of water = `1.86 ^(@)C/m`

Options

A. 20 g
B. 10.75 g
C. 494.5 g
D. 492 g

Answer

To solve the problem of how much ethyl alcohol must be added to 1.00 L of water so that the solution will not freeze at -4°F, we will follow these steps:

### Step 1: Convert the freezing point from Fahrenheit to Celsius
The formula to convert Fahrenheit to Celsius is:
\[ °C = \frac{5}{9}(°F - 32) \]
Substituting -4°F into the formula:
\[ °C = \frac{5}{9}(-4 - 32) = \frac{5}{9}(-36) = -20°C \]

### Step 2: Calculate the change in freezing point (ΔTf)
The change in freezing point (ΔTf) is calculated as:
\[ ΔTf = Tf^0 - Tf \]
Where \(Tf^0\) is the freezing point of pure water (0°C) and \(Tf\) is the freezing point of the solution (-20°C):
\[ ΔTf = 0 - (-20) = 20°C \]

### Step 3: Use the freezing point depression formula
The freezing point depression formula is:
\[ ΔTf = K_f \cdot i \cdot m \]
Where:
- \(K_f\) is the freezing point depression constant (1.86 °C/m for water),
- \(i\) is the van 't Hoff factor (1 for non-electrolytes like ethyl alcohol),
- \(m\) is the molality of the solution.

Since \(i = 1\), we can rearrange the formula to find molality \(m\):
\[ m = \frac{ΔTf}{K_f} = \frac{20}{1.86} \approx 10.75 \, \text{mol/kg} \]

### Step 4: Calculate the weight of water
The weight of 1.00 L of water can be calculated using its density (1 g/mL):
\[ \text{Weight of water} = \text{Density} \times \text{Volume} = 1 \, \text{g/mL} \times 1000 \, \text{mL} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 5: Calculate the weight of ethyl alcohol needed
Using the formula for molality:
\[ m = \frac{W_2}{M \cdot W_1} \]
Where:
- \(W_2\) is the weight of ethyl alcohol,
- \(M\) is the molar mass of ethyl alcohol (46 g/mol),
- \(W_1\) is the weight of water (1 kg).

Rearranging the formula to find \(W_2\):
\[ W_2 = m \cdot M \cdot W_1 \]
Substituting the values:
\[ W_2 = 10.75 \, \text{mol/kg} \times 46 \, \text{g/mol} \times 1 \, \text{kg} = 10.75 \times 46 \approx 495 \, \text{g} \]

### Final Answer
Approximately 495 grams of ethyl alcohol must be added to 1.00 L of water to prevent the solution from freezing at -4°F.

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