The molal depression constant for water is `1.86^(@)C`. The freezing point of a `0.05-molal` solution of a non-electrolyte in water is

Options

A. `-1.86^(@)C`
B. `-0.93^(@)C`
C. `0.093^(@)C`
D. `0.93^(@)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 freezing point depression:

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

Where:
- \(\Delta T_f\) = change in freezing point
- \(i\) = van 't Hoff factor (which is 1 for non-electrolytes)
- \(K_f\) = molal freezing point depression constant (for water, \(K_f = 1.86 \, ^\circ C \, \text{kg/mol}\))
- \(m\) = molality of the solution

### Step-by-Step Solution:

1. **Identify the values:**
- \(i = 1\) (since it is a non-electrolyte)
- \(K_f = 1.86 \, ^\circ C \, \text{kg/mol}\)
- \(m = 0.05 \, \text{mol/kg}\)

2. **Calculate the change in freezing point (\(\Delta T_f\)):**
\[ \Delta T_f = i \cdot K_f \cdot m \]
\[ \Delta T_f = 1 \cdot 1.86 \, ^\circ C \cdot 0.05 \, \text{mol/kg} \]
\[ \Delta T_f = 1.86 \cdot 0.05 = 0.093 \, ^\circ C \]

3. **Determine the new freezing point:**
- The freezing point of pure water is \(0 \, ^\circ C\).
- Since the freezing point decreases, we subtract \(\Delta T_f\) from the freezing point of pure water:
\[ T_f = 0 \, ^\circ C - 0.093 \, ^\circ C = -0.093 \, ^\circ C \]

### Final Answer:
The freezing point of a 0.05-molal solution of a non-electrolyte in water is \(-0.093 \, ^\circ C\).

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