KBr is 80% dissociated in aqueous solution of 0.5M concentration (Given `K_(f)` for water `=1.86 K kg mol^(-1)`). The solution freezes at

Options

A. 271.326 K
B. 272 K
C. 270.5 K
D. 268.5 K

Answer

To solve the problem of finding the freezing point of an aqueous solution of KBr that is 80% dissociated, we will follow these steps:

### Step 1: Determine the degree of dissociation and van 't Hoff factor (i)
Given that KBr is 80% dissociated, we can express this as:
- Degree of dissociation (α) = 0.8

KBr dissociates into K⁺ and Br⁻ ions:
\[ \text{KBr} \rightarrow \text{K}^+ + \text{Br}^- \]

The number of particles (n) produced from the dissociation of KBr is 2 (1 K⁺ and 1 Br⁻). The van 't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
Substituting the values:
\[ i = 1 + 0.8(2 - 1) = 1 + 0.8 = 1.8 \]

### Step 2: Calculate the molality (m)
Given the molarity (M) of the solution is 0.5 M, we can assume that for dilute solutions, molality (m) is approximately equal to molarity (M):
\[ m \approx 0.5 \, \text{mol/kg} \]

### Step 3: Calculate the depression in freezing point (ΔTf)
Using the formula for depression in freezing point:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) for water = 1.86 K kg mol⁻¹
- \( i = 1.8 \)
- \( m = 0.5 \)

Substituting the values:
\[ \Delta T_f = 1.8 \cdot 1.86 \cdot 0.5 \]
\[ \Delta T_f = 1.674 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution (Ts)
The freezing point of pure water (T₀f) is 0°C. The freezing point of the solution (Ts) can be calculated as:
\[ T_s = T_{0f} - \Delta T_f \]
Substituting the values:
\[ T_s = 0 - 1.674 = -1.674 \, \text{°C} \]

### Final Answer
The solution freezes at approximately **-1.674 °C**.

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