What will be the freezing point of 0.2 molal aqueous solution of `MgBr_(2)` ? If salt dissociates 40% in solution and `K_(f)` for water is 1.86 KKg `mol^(-1)`

Options

A. `-3.35^(@)`C
B. `-0.67^(@)`C
C. `-0.6^(@)`C
D. `-0.45^(@)`C

Explanation

To determine the freezing point of a 0.2 molal aqueous solution of MgBr₂, where the salt dissociates 40% in solution, we can follow these steps:

### Step 1: Identify the given values
- Molality (M) = 0.2 mol/kg
- Degree of dissociation (α) = 40% = 0.4
- Freezing point depression constant (Kf) for water = 1.86 K·kg/mol

### Step 2: Determine the van 't Hoff factor (i)
MgBr₂ dissociates in solution as follows:
\[ \text{MgBr}_2 \rightarrow \text{Mg}^{2+} + 2\text{Br}^- \]

For each formula unit of MgBr₂ that dissociates, it produces 1 Mg²⁺ ion and 2 Br⁻ ions, leading to a total of 3 particles. However, since the dissociation is only 40%, we need to account for this in calculating the effective number of particles.

The effective number of particles (i) can be calculated as:
\[ i = 1 + \alpha \cdot (n - 1) \]
where n is the number of particles produced from one formula unit of the solute. Here, n = 3 (1 Mg²⁺ + 2 Br⁻).

Substituting the values:
\[ i = 1 + 0.4 \cdot (3 - 1) \]
\[ i = 1 + 0.4 \cdot 2 \]
\[ i = 1 + 0.8 = 1.8 \]

### Step 3: Calculate the freezing point depression (ΔTf)
The freezing point depression is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Substituting the known values:
\[ \Delta T_f = 1.8 \cdot 1.86 \cdot 0.2 \]

Calculating this:
\[ \Delta T_f = 1.8 \cdot 1.86 \cdot 0.2 = 0.6696 \text{ K} \approx 0.67 \text{ K} \]

### Step 4: Determine the freezing point (Tf)
The normal freezing point of water (Tf₀) is 0 °C. The freezing point of the solution can be calculated using:
\[ T_f = T_f^0 - \Delta T_f \]
Substituting the values:
\[ T_f = 0 - 0.67 = -0.67 \text{ °C} \]

### Final Answer
The freezing point of the 0.2 molal aqueous solution of MgBr₂ is approximately **-0.67 °C**.

---