The freezing point of a `0.05` molal solution of a non-electrolyte in water is:
(`K_(f) = 1.86 "molality"^(-1)`)

Options

A. `-1.86^(@)C`
B. `-0.93^(@)C`
C. `-0.093^(@)C`
D. `0.093^(@)C`

Explanation

To find the freezing point of a 0.05 molal solution of a non-electrolyte in water, we can use the formula for depression in freezing point:

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

Where:
- \(\Delta T_f\) = depression in freezing point
- \(I\) = Van't Hoff factor (for non-electrolytes, \(I = 1\))
- \(K_f\) = freezing point depression constant (given as \(1.86 \, \text{molal}^{-1}\))
- \(m\) = molality of the solution (given as \(0.05 \, \text{molal}\))

### Step 1: Identify the values
- \(I = 1\) (for non-electrolyte)
- \(K_f = 1.86 \, \text{molal}^{-1}\)
- \(m = 0.05 \, \text{molal}\)

### Step 2: Calculate \(\Delta T_f\)
Substituting the values into the formula:

\[ \Delta T_f = 1 \cdot 1.86 \cdot 0.05 \]

Calculating this gives:

\[ \Delta T_f = 1.86 \cdot 0.05 = 0.093 \]

### Step 3: Determine the freezing point of the solution
The freezing point of pure water (\(T_f^0\)) is \(0^\circ C\). The freezing point of the solution (\(T_f\)) can be calculated as:

\[ T_f = T_f^0 - \Delta T_f \]

Substituting the values:

\[ T_f = 0 - 0.093 = -0.093^\circ C \]

### Final Answer
Thus, the freezing point of the 0.05 molal solution of a non-electrolyte in water is:

\[ \boxed{-0.093^\circ C} \]