`A 0.50` molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water is `1.86^(@)` per molal, at which temperature will the mixture freeze?

Explanation

To solve the problem of determining the freezing point of a 0.50 molal solution of ethylene glycol in water, we can follow these steps:

### Step 1: Understand the Formula for Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
where:
- \( \Delta T_f \) = freezing point depression
- \( K_f \) = freezing point constant of the solvent (water in this case)
- \( m \) = molality of the solution

### Step 2: Identify the Given Values
From the problem, we know:
- \( K_f \) for water = 1.86 °C/m
- \( m \) (molality of the solution) = 0.50 molal

### Step 3: Calculate the Freezing Point Depression
Substituting the values into the formula:
\[ \Delta T_f = 1.86 \, \text{°C/m} \times 0.50 \, \text{m} = 0.93 \, \text{°C} \]

### Step 4: Determine the New Freezing Point
The normal freezing point of pure water (\( T_{0f} \)) is 0 °C. The new freezing point (\( T_f \)) of the solution can be calculated using the formula:
\[ T_f = T_{0f} - \Delta T_f \]
Substituting the known values:
\[ T_f = 0 \, \text{°C} - 0.93 \, \text{°C} = -0.93 \, \text{°C} \]

### Final Answer
The freezing point of the 0.50 molal solution of ethylene glycol in water is:
\[ T_f = -0.93 \, \text{°C} \]

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