Van't Hoff factor is 1.92 for `MgI_(2)` solution with concentraition 0.2M then the degree of dissociation of salt at this concentration is :-

Options

A. `46%`
B. `96%`
C. `30.67%`
D. `64%`

Explanation

To solve the problem, we need to find the degree of dissociation (α) of magnesium iodide (MgI₂) in a solution where the Van't Hoff factor (i) is given as 1.92 and the concentration is 0.2 M.

### Step-by-Step Solution:

1. **Understanding the Dissociation of MgI₂:**
Magnesium iodide dissociates in solution as follows:
\[ \text{MgI}_2 \rightarrow \text{Mg}^{2+} + 2 \text{I}^- \]
From this dissociation, we can see that one formula unit of MgI₂ produces one magnesium ion (Mg²⁺) and two iodide ions (I⁻), resulting in a total of three particles.

2. **Identifying the Van't Hoff Factor (i):**
The Van't Hoff factor (i) is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved. For MgI₂, the expected value of i (if fully dissociated) would be 3 (1 Mg²⁺ + 2 I⁻).

3. **Calculating the Degree of Dissociation (α):**
The degree of dissociation (α) can be calculated using the formula:
\[ i = 1 + \frac{n - 1}{n} \cdot \alpha \]
Where:
- \(i\) = Van't Hoff factor (1.92)
- \(n\) = number of particles expected from one formula unit of MgI₂ (which is 3)

Rearranging the formula gives us:
\[ \alpha = \frac{i - 1}{n - 1} \]

4. **Substituting the Values:**
Now substituting the known values into the equation:
\[ \alpha = \frac{1.92 - 1}{3 - 1} = \frac{0.92}{2} = 0.46 \]

5. **Converting to Percentage:**
To express the degree of dissociation as a percentage, we multiply by 100:
\[ \text{Degree of dissociation} = 0.46 \times 100 = 46\% \]

### Final Answer:
The degree of dissociation of MgI₂ at a concentration of 0.2 M is **46%**.