An aqueous solution of a salt `MX_(2)` at certain temperature has a van'f Hoff factor of 2. The degree of dissociation for this solution of the salt is:

Options

A. `0.50`
B. `0.80`
C. `0.67`
D. `0.33`

Answer

To find the degree of dissociation (α) for the aqueous solution of the salt \( MX_2 \) with a van 't Hoff factor (i) of 2, we can follow these steps:

### Step 1: Understand the dissociation of the salt
The salt \( MX_2 \) dissociates in water to give:
- 1 ion of \( M^{2+} \)
- 2 ions of \( X^{-} \)

Thus, the total number of ions produced (n) when one formula unit of \( MX_2 \) dissociates is:
\[ n = 1 + 2 = 3 \]

### Step 2: Use the van 't Hoff factor
The van 't Hoff factor (i) is related to the degree of dissociation (α) and the number of particles produced (n) by the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]

### Step 3: Substitute known values into the equation
We know:
- \( i = 2 \)
- \( n = 3 \)

Substituting these values into the equation:
\[ 2 = 1 + (3 - 1) \cdot \alpha \]

### Step 4: Simplify the equation
Now simplify the equation:
\[ 2 = 1 + 2\alpha \]
Subtract 1 from both sides:
\[ 2 - 1 = 2\alpha \]
\[ 1 = 2\alpha \]

### Step 5: Solve for α
Now, divide both sides by 2:
\[ \alpha = \frac{1}{2} \]
Thus, the degree of dissociation (α) is:
\[ \alpha = 0.5 \]

### Conclusion
The degree of dissociation for the solution of the salt \( MX_2 \) is 0.5.

### Final Answer
The correct choice for this question is option A: 0.50.

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