Ethylene glycol, `CH_(2)OH*CH_(2)OH` the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed with 6 gallons of water to protect it from freezing at `-24^(@)C`. (1 gallon `=3.785` lit, `K_(f)=1.86`) ?

Explanation

To solve the problem of determining the minimum weight of ethylene glycol that must be mixed with 6 gallons of water to prevent freezing at -24°C, we can follow these steps:

### Step 1: Calculate the total volume of water in liters
Given that 1 gallon = 3.785 liters, we can convert 6 gallons to liters:
\[ \text{Volume of water} = 6 \, \text{gallons} \times 3.785 \, \text{liters/gallon} = 22.71 \, \text{liters} \]

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