A motor vehicle raditor was filled with `8 L`of water to which `2 L`of methyl alcohol `(density 0.8 g mL^(-1))` was added. What is the lowest temperature at which the vehicle can be parked outdoors without the danger that the water in the raditor will freeze? Given that `K_(f)` for water is `1.86 K kg mol^(-1)`

Explanation

To solve the problem, we need to determine the lowest temperature at which the water in the radiator will not freeze after adding methyl alcohol. We will use the formula for freezing point depression, which is given by:

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

Where:
- \(\Delta T_f\) is the freezing point depression,
- \(K_f\) is the cryoscopic constant for water (given as \(1.86 \, \text{K kg mol}^{-1}\)),
- \(m\) is the molality of the solution.

### Step 1: Calculate the mass of water and methyl alcohol

1. **Mass of water**:
- Volume of water = \(8 \, \text{L} = 8000 \, \text{mL}\)
- Density of water = \(1 \, \text{g/mL}\)
- Mass of water = Volume × Density = \(8000 \, \text{mL} \times 1 \, \text{g/mL} = 8000 \, \text{g} = 8 \, \text{kg}\)

2. **Mass of methyl alcohol**:
- Volume of methyl alcohol = \(2 \, \text{L} = 2000 \, \text{mL}\)
- Density of methyl alcohol = \(0.8 \, \text{g/mL}\)
- Mass of methyl alcohol = Volume × Density = \(2000 \, \text{mL} \times 0.8 \, \text{g/mL} = 1600 \, \text{g} = 1.6 \, \text{kg}\)

### Step 2: Calculate the number of moles of methyl alcohol

- Molar mass of methyl alcohol (CH₃OH) = \(12 + 1 \times 3 + 16 = 32 \, \text{g/mol}\)
- Moles of methyl alcohol = \(\frac{\text{mass}}{\text{molar mass}} = \frac{1600 \, \text{g}}{32 \, \text{g/mol}} = 50 \, \text{mol}\)

### Step 3: Calculate the molality of the solution

- Molality (m) is defined as moles of solute per kg of solvent.
- Mass of solvent (water) = \(8 \, \text{kg}\)
- Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{50 \, \text{mol}}{8 \, \text{kg}} = 6.25 \, \text{mol/kg}\)

### Step 4: Calculate the freezing point depression

Using the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m = 1.86 \, \text{K kg mol}^{-1} \cdot 6.25 \, \text{mol/kg} = 11.625 \, \text{K} \]

### Step 5: Determine the lowest temperature

The normal freezing point of water is \(0 \, \text{°C}\) or \(273 \, \text{K}\). The new freezing point will be:

\[ T_f = 273 \, \text{K} - \Delta T_f = 273 \, \text{K} - 11.625 \, \text{K} = 261.375 \, \text{K} \]

### Conclusion

The lowest temperature at which the vehicle can be parked outdoors without the danger of the water in the radiator freezing is approximately \(261.375 \, \text{K}\).

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