KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is :
`K_f (H_2O) = 1.86 (K kg)/("mole")`

Options

A. 274.339 K
B. `-1.339K`
C. `257.3 K `
D. `-1.339^@C`

Explanation

To solve the problem of finding the freezing point of a 0.4 molal solution of KBr that is 80% ionized, we will follow these steps:

### Step 1: Understand the dissociation of KBr
KBr dissociates in water into K⁺ and Br⁻ ions. The dissociation can be represented as:
\[ \text{KBr} \rightarrow \text{K}^+ + \text{Br}^- \]

### Step 2: Calculate the van 't Hoff factor (i)
Since KBr is 80% ionized, we can calculate the van 't Hoff factor (i) using the formula:
\[ i = 1 + \alpha \]
Where:
- \( \alpha \) is the degree of ionization.

Given that KBr is 80% ionized:
\[ \alpha = 0.80 \]
Thus,
\[ i = 1 + 0.80 = 1.80 \]

### Step 3: Use the freezing point depression formula
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) is the freezing point depression constant for water, which is given as \( 1.86 \, \text{K kg/mol} \).
- \( m \) is the molality of the solution, which is given as \( 0.4 \, \text{mol/kg} \).

### Step 4: Substitute values into the formula
Now we can substitute the known values into the equation:
\[ \Delta T_f = 1.80 \cdot 1.86 \cdot 0.4 \]

### Step 5: Perform the calculation
Calculating \( \Delta T_f \):
\[ \Delta T_f = 1.80 \cdot 1.86 \cdot 0.4 = 1.3392 \, \text{K} \]

### Step 6: Calculate the new freezing point
The freezing point of pure water is \( 0^\circ C \) (or 273 K). The freezing point of the solution (\( T_f \)) can be calculated as:
\[ T_f = T_{f, \text{pure}} - \Delta T_f \]
\[ T_f = 0 - 1.3392 = -1.3392^\circ C \]

### Final Answer
Thus, the freezing point of the 0.4 molal solution of KBr is approximately:
\[ T_f \approx -1.34^\circ C \]

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