The freezing poing of an aqueous solution of a non-electrolyte is `-0.14^(@)C`. The molality of this solution is `[K_(f) (H_(2)O) = 1.86 kg mol^(-1)]`:

Options

A. `1.86m`
B. `1.00m`
C. `0.15m`
D. `0.075m`

Explanation

To find the molality of the aqueous solution given its freezing point, we can follow these steps:

### Step 1: Understand the Freezing Point Depression Formula
The freezing point depression can be expressed using the formula:
\[ \Delta T_f = K_f \times m \]
where:
- \(\Delta T_f\) is the change in freezing point,
- \(K_f\) is the freezing point depression constant,
- \(m\) is the molality of the solution.

### Step 2: Identify the Values
From the question, we have:
- The freezing point of pure water, \(T_f^0 = 0^\circ C\),
- The freezing point of the solution, \(T_f = -0.14^\circ C\),
- The value of \(K_f\) for water, \(K_f = 1.86 \, \text{kg mol}^{-1}\).

### Step 3: Calculate \(\Delta T_f\)
Calculate the change in freezing point:
\[ \Delta T_f = T_f^0 - T_f = 0^\circ C - (-0.14^\circ C) = 0.14^\circ C \]

### Step 4: Substitute Values into the Formula
Now, substitute \(\Delta T_f\) and \(K_f\) into the freezing point depression formula:
\[ 0.14 = 1.86 \times m \]

### Step 5: Solve for Molality \(m\)
Rearranging the equation to solve for \(m\):
\[ m = \frac{0.14}{1.86} \]

### Step 6: Perform the Calculation
Now calculate the value:
\[ m = \frac{0.14}{1.86} \approx 0.0753 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately \(0.0753 \, \text{mol/kg}\).

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