The Van't Hoff factor `0.1M Ba(NO_(3))_(2)` solution is found to be `2.74` the percentage dissociation of the salt is:

Options

A. `85%`
B. `58%`
C. `65.8%`
D. `56.8%`

Explanation

To solve the problem of finding the percentage dissociation of the salt \( Ba(NO_3)_2 \) given its Van't Hoff factor \( i = 2.74 \) for a \( 0.1M \) solution, we can follow these steps:

### Step 1: Understand the dissociation of \( Ba(NO_3)_2 \)
The dissociation of barium nitrate in solution can be represented as:
\[ Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^{-} \]
From this equation, we can see that one formula unit of \( Ba(NO_3)_2 \) dissociates into 3 ions (1 \( Ba^{2+} \) ion and 2 \( NO_3^{-} \) ions).

### Step 2: Identify the number of ions (n)
From the dissociation equation, we find:
- \( n = 3 \) (the total number of ions produced from one formula unit).

### Step 3: Use the Van't Hoff factor formula
The relationship between the Van't Hoff factor \( i \), the number of ions \( n \), and the degree of dissociation \( \alpha \) is given by:
\[ i = 1 + (n - 1) \alpha \]
Substituting the known values into the equation:
\[ 2.74 = 1 + (3 - 1) \alpha \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 2.74 = 1 + 2\alpha \]
Subtract 1 from both sides:
\[ 2.74 - 1 = 2\alpha \]
\[ 1.74 = 2\alpha \]

### Step 5: Solve for \( \alpha \)
Now, divide both sides by 2 to find \( \alpha \):
\[ \alpha = \frac{1.74}{2} = 0.87 \]

### Step 6: Calculate the percentage dissociation
To find the percentage dissociation, multiply \( \alpha \) by 100:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.87 \times 100 = 87\% \]

### Conclusion
The percentage dissociation of the salt \( Ba(NO_3)_2 \) in the solution is approximately \( 87\% \).