`K_(f)` for water is `1.86 K kg mol^(-1)`. If your automobile radiator holds `1.0 kg` of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?

Options

A. 72g
B. 93g
C. 39g
D. 27g

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Data
- **Freezing point depression constant (Kf)** for water = 1.86 K kg mol⁻¹
- **Mass of water (solvent)** = 1.0 kg = 1000 g
- **Desired freezing point** = -2.8 °C
- **Molecular mass of ethylene glycol (C₂H₆O₂)** = 62 g/mol

### Step 2: Calculate the Depression in Freezing Point (ΔTf)
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Where:
- \(T_f^{\text{pure}}\) (freezing point of pure water) = 0 °C
- \(T_f^{\text{solution}}\) = -2.8 °C

Calculating ΔTf:
\[ \Delta T_f = 0 - (-2.8) = 2.8 \, \text{°C} \]

### Step 3: Use the Freezing Point Depression Formula
The formula relating freezing point depression to molality (m) is:
\[ \Delta T_f = K_f \times m \]
Where:
- \(m\) is the molality of the solution.

Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} = \frac{2.8}{1.86} \]

Calculating molality:
\[ m \approx 1.505 \, \text{mol/kg} \]

### Step 4: Relate Molality to Moles of Solute
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Therefore:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Given that the mass of the solvent is 1 kg:
\[ \text{moles of solute} = m \times \text{mass of solvent (kg)} = 1.505 \times 1 = 1.505 \, \text{mol} \]

### Step 5: Calculate the Mass of Ethylene Glycol
To find the mass of ethylene glycol (C₂H₆O₂), we use the formula:
\[ \text{mass of solute} = \text{moles of solute} \times \text{molar mass of solute} \]
Substituting the values:
\[ \text{mass of ethylene glycol} = 1.505 \, \text{mol} \times 62 \, \text{g/mol} \]

Calculating the mass:
\[ \text{mass of ethylene glycol} \approx 93.31 \, \text{g} \]

### Final Answer
Approximately **93 grams** of ethylene glycol must be added to lower the freezing point of the solution to -2.8 °C.

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