The van't Hoff factor for `0.1 M Ba(NO_(3))_(2)` solution is `2.74`. The degree of dissociation is

Options

A. `91.3%`
B. `87%`
C. `100%`
D. `74%`

Explanation

To find the degree of dissociation (\( \alpha \)) of the \( 0.1 \, M \, Ba(NO_3)_2 \) solution given that the van't Hoff factor (\( i \)) is \( 2.74 \), we can follow these steps:

### Step 1: Understand the van't Hoff factor
The van't Hoff factor (\( i \)) is given by the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
where:
- \( n \) is the number of ions produced when one formula unit of the solute dissociates.
- \( \alpha \) is the degree of dissociation.

### Step 2: Determine the number of ions (\( n \))
For \( Ba(NO_3)_2 \):
- When it dissociates, it produces:
- 1 \( Ba^{2+} \) ion
- 2 \( NO_3^{-} \) ions

Thus, the total number of ions (\( n \)) is:
\[ n = 1 + 2 = 3 \]

### Step 3: Substitute values into the van't Hoff factor equation
Now, we can substitute \( i = 2.74 \) and \( n = 3 \) into the van't Hoff factor formula:
\[ 2.74 = 1 + (3 - 1) \cdot \alpha \]

### Step 4: Simplify the equation
This simplifies to:
\[ 2.74 = 1 + 2 \cdot \alpha \]
Subtracting 1 from both sides gives:
\[ 1.74 = 2 \cdot \alpha \]

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

### Step 6: Convert to percentage
To express the degree of dissociation as a percentage:
\[ \alpha = 0.87 \times 100 = 87\% \]

### Final Answer
The degree of dissociation of the \( 0.1 \, M \, Ba(NO_3)_2 \) solution is \( 87\% \).

---