What approximate proportions by volume of water `(d=1g//"cc")` and ethylene glycol `C_(2)H_(6)O_(2)(d=1.12g//"cc")` must be mixed to ensure protection of an automobile cooling system to `-10^(@)C` ?

Answer

To solve the problem of determining the approximate proportions by volume of water and ethylene glycol needed to protect an automobile cooling system to -10°C, we can follow these steps:

### Step 1: Understand the Problem
We need to find the volume ratio of water (density = 1 g/cc) and ethylene glycol (density = 1.12 g/cc) that will lower the freezing point of the solution to -10°C.

### Step 2: Define Variables
Let:
- \( V_w \) = volume of water
- \( V_{EG} \) = volume of ethylene glycol
- The ratio \( \frac{V_w}{V_{EG}} = X \)

### Step 3: Calculate Masses
Using the densities, we can express the masses of water and ethylene glycol in terms of their volumes:
- Mass of water, \( m_w = V_w \times 1 \, \text{g/cc} = V_w \)
- Mass of ethylene glycol, \( m_{EG} = V_{EG} \times 1.12 \, \text{g/cc} = 1.12 V_{EG} \)

### Step 4: Relate the Volumes
From the ratio defined earlier, we can express \( V_{EG} \) in terms of \( V_w \):
\[ V_{EG} = \frac{V_w}{X} \]

### Step 5: Substitute and Calculate
Substituting \( V_{EG} \) into the mass equation for ethylene glycol:
\[ m_{EG} = 1.12 \left( \frac{V_w}{X} \right) \]

### Step 6: Calculate Molality
The molality \( m \) of the solution can be calculated using the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

The molar mass of ethylene glycol \( C_2H_6O_2 \) is 62 g/mol. Therefore, the moles of ethylene glycol can be expressed as:
\[ \text{moles of } C_2H_6O_2 = \frac{m_{EG}}{62} = \frac{1.12 \left( \frac{V_w}{X} \right)}{62} \]

The mass of water in kg is:
\[ \text{mass of } H_2O = V_w \, \text{g} = \frac{V_w}{1000} \, \text{kg} \]

Thus, the molality \( m \) becomes:
\[ m = \frac{1.12 \left( \frac{V_w}{X} \right)/62}{V_w/1000} = \frac{1.12 \times 1000}{62X} \]

### Step 7: Use Freezing Point Depression
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \]
Where \( K_f \) for water is approximately 1.86 °C kg/mol. Given that \( \Delta T_f = 10 \) °C (from 0 °C to -10 °C), we can set up the equation:
\[ 10 = 1.86 \cdot \frac{1.12 \times 1000}{62X} \]

### Step 8: Solve for X
Rearranging the equation gives:
\[ X = \frac{1.86 \times 1.12 \times 1000}{10 \times 62} \]
Calculating this will yield:
\[ X \approx 3.36 \]

### Conclusion
The approximate volume ratio of water to ethylene glycol is:
\[ \frac{V_w}{V_{EG}} \approx 3.36 \]