`K_(f)` for waer 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. `27g`
B. `72g`
C. `93g`
D. `39g`

Explanation

To solve the problem, we will use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

where:
- \(\Delta T_f\) is the change in freezing point,
- \(K_f\) is the freezing point depression constant,
- \(m\) is the molality of the solution.

### Step 1: Calculate the change in freezing point (\(\Delta T_f\))

The freezing point of pure water is \(0^\circ C\). The freezing point of the solution is given as \(-2.8^\circ C\). Therefore:

\[ \Delta T_f = 0 - (-2.8) = 2.8^\circ C \]

### Step 2: Substitute values into the freezing point depression formula

We know that \(K_f\) for water is \(1.86 \, K \, kg \, mol^{-1}\). Now, substituting the values into the formula:

\[ 2.8 = 1.86 \cdot m \]

### Step 3: Solve for molality (m)

Rearranging the equation to find \(m\):

\[ m = \frac{2.8}{1.86} \approx 1.5054 \, mol/kg \]

### Step 4: Calculate the number of moles of solute

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. Since we have \(1.0 \, kg\) of water, the number of moles of ethylene glycol (\(C_2H_6O_2\)) can be calculated as:

\[ \text{moles of solute} = m \cdot \text{weight of solvent} = 1.5054 \, mol/kg \cdot 1.0 \, kg = 1.5054 \, mol \]

### Step 5: Calculate the mass of ethylene glycol

To find the mass of ethylene glycol, we need its molar mass. The molar mass of ethylene glycol (\(C_2H_6O_2\)) can be calculated as follows:

- Carbon (C): \(2 \times 12.01 \, g/mol = 24.02 \, g/mol\)
- Hydrogen (H): \(6 \times 1.008 \, g/mol = 6.048 \, g/mol\)
- Oxygen (O): \(2 \times 16.00 \, g/mol = 32.00 \, g/mol\)

Adding these together gives:

\[ \text{Molar mass of } C_2H_6O_2 = 24.02 + 6.048 + 32.00 \approx 62.068 \, g/mol \]

Now, we can calculate the mass of ethylene glycol:

\[ \text{mass} = \text{moles} \times \text{molar mass} = 1.5054 \, mol \times 62.068 \, g/mol \approx 93.38 \, g \]

### Final Answer

Therefore, the mass of ethylene glycol that must be added is approximately **93.38 grams**.

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