The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at `14^@F` is `(K_f ` of water = 1.86 K kg `mol^(-1)` )

Options

A. 263.11 g
B. 247.31 g
C. 236.11 g
D. 281.01 g

Answer

To solve the problem of determining the weight of ethyl alcohol that must be added to 1.0 L of water so that the solution will freeze at 14°F, we will follow these steps:

### Step 1: Convert the freezing point from Fahrenheit to Celsius
The formula to convert Fahrenheit (°F) to Celsius (°C) is:
\[ °C = \frac{5}{9} (°F - 32) \]
Substituting the given freezing point:
\[ °C = \frac{5}{9} (14 - 32) = \frac{5}{9} (-18) = -10°C \]

### Step 2: Calculate the change in freezing point (ΔTf)
The change in freezing point (ΔTf) is calculated as:
\[ ΔTf = Tf (pure solvent) - Tf (solution) \]
Since the freezing point of pure water is 0°C:
\[ ΔTf = 0 - (-10) = 10°C \]

### Step 3: Use the freezing point depression formula
The formula for freezing point depression is:
\[ ΔTf = K_f \cdot m \]
Where:
- \(K_f\) for water = 1.86 K kg/mol
- \(m\) is the molality of the solution.

### Step 4: Calculate the molality (m)
We can rearrange the formula to find molality:
\[ m = \frac{ΔTf}{K_f} = \frac{10}{1.86} \approx 5.376 \, \text{mol/kg} \]

### Step 5: Calculate the weight of the solvent (water)
Since we have 1.0 L of water and the density of water is approximately 1 g/mL:
\[ \text{Weight of water} = 1.0 \, \text{L} \times 1000 \, \text{g/L} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 6: Relate molality to moles of solute
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
Thus:
\[ \text{moles of ethyl alcohol} = m \cdot \text{kg of solvent} = 5.376 \, \text{mol/kg} \times 1 \, \text{kg} = 5.376 \, \text{mol} \]

### Step 7: Calculate the weight of ethyl alcohol
The molecular weight of ethyl alcohol (C2H5OH) is approximately 46 g/mol. Therefore, the weight of ethyl alcohol can be calculated as:
\[ \text{Weight of ethyl alcohol} = \text{moles} \times \text{molecular weight} = 5.376 \, \text{mol} \times 46 \, \text{g/mol} \approx 247.36 \, \text{g} \]

### Final Answer
The weight of ethyl alcohol that must be added is approximately **247.31 g**.

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