Water is used in car radiators. In winter season, ethylene glycol is added to water so that water may not freeze. Assuming ethylene glycol to be non-volatile, calculate minimum amount of ethylene glycol that must be added to 6.0 kg of water to prevent it from freezing at `-0.3^(@)C`. The molal depression constant of water is `1.86^(@)`.

Explanation

To solve the problem of calculating the minimum amount of ethylene glycol that must be added to 6.0 kg of water to prevent it from freezing at -0.3°C, we can follow these steps:

### Step 1: Identify the given data
- Freezing point depression constant of water (Kf) = 1.86 °C kg/mol
- Mass of water (solvent) = 6.0 kg = 6000 g
- Freezing point of pure water (Tf°) = 0 °C
- Freezing point of the solution (Tf) = -0.3 °C

### Step 2: Calculate the change in freezing point (ΔTf)
\[ \Delta Tf = Tf° - Tf = 0 °C - (-0.3 °C) = 0.3 °C \]

### Step 3: Use the freezing point depression formula
The formula for freezing point depression is given by:
\[ \Delta Tf = Kf \cdot m \]
where \( m \) is the molality of the solution. The molality can be expressed as:
\[ m = \frac{W_2}{M_2 \cdot W_1} \]
where:
- \( W_2 \) = mass of the solute (ethylene glycol)
- \( M_2 \) = molar mass of the solute (ethylene glycol)
- \( W_1 \) = mass of the solvent (water)

### Step 4: Calculate the molar mass of ethylene glycol (C2H6O2)
The molar mass can be calculated as follows:
- Carbon (C) = 12 g/mol, and there are 2 carbon atoms: \( 12 \times 2 = 24 \) g/mol
- Hydrogen (H) = 1 g/mol, and there are 6 hydrogen atoms: \( 1 \times 6 = 6 \) g/mol
- Oxygen (O) = 16 g/mol, and there are 2 oxygen atoms: \( 16 \times 2 = 32 \) g/mol

Adding these together:
\[ M_2 = 24 + 6 + 32 = 62 \text{ g/mol} \]

### Step 5: Substitute values into the freezing point depression formula
Rearranging the formula for molality:
\[ \Delta Tf = Kf \cdot \frac{W_2}{M_2 \cdot W_1} \]
Substituting the known values:
\[ 0.3 = 1.86 \cdot \frac{W_2}{62 \cdot 6.0} \]

### Step 6: Solve for \( W_2 \)
Rearranging gives:
\[ W_2 = \frac{0.3 \cdot 62 \cdot 6.0}{1.86} \]
Calculating:
\[ W_2 = \frac{11.16}{1.86} \approx 6.0 \text{ g} \]

### Step 7: Final Calculation
Now, calculating the mass of ethylene glycol:
\[ W_2 \approx 60 \text{ g} \]

### Conclusion
The minimum amount of ethylene glycol that must be added to 6.0 kg of water to prevent it from freezing at -0.3°C is approximately **60 grams**.

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