`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?

Options

A. `0.93^(@)C`
B. `-0.93^(@)C`
C. `1.86^(@)C`
D. `-1.86^(@)C`

Answer

To solve the problem of finding the freezing point of a 0.50 molal solution of ethylene glycol in water, we will 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:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) is the van 't Hoff factor (for ethylene glycol, \( i = 1 \) since it does not dissociate),
- \( K_f \) is the freezing point constant of the solvent (water in this case),
- \( m \) is the molality of the solution.

### Step 2: Identify the values
From the question, we have:
- \( K_f = 1.86 \, ^\circ C \, \text{mol}^{-1} \, \text{kg} \)
- \( m = 0.50 \, \text{molal} \)

### Step 3: Calculate the freezing point depression
Substituting the values into the formula:
\[ \Delta T_f = 1 \cdot 1.86 \, ^\circ C \, \text{mol}^{-1} \, \text{kg} \cdot 0.50 \, \text{molal} \]
\[ \Delta T_f = 1.86 \cdot 0.50 = 0.93 \, ^\circ C \]

### Step 4: Determine the freezing point of the solution
The freezing point of pure water is \( 0 \, ^\circ C \). The freezing point of the solution will be lower than that of pure water by the amount of the depression calculated:
\[ \text{Freezing point of solution} = 0 \, ^\circ C - \Delta T_f \]
\[ \text{Freezing point of solution} = 0 \, ^\circ C - 0.93 \, ^\circ C = -0.93 \, ^\circ C \]

### Final Answer
The mixture will freeze at \( -0.93 \, ^\circ C \).

---