Colligative properties of a solution depend upon the number of moles of a solute dissolved and do not depend upon the nature of the solute. However, they are applicable only to dilute solutions in which the solutes do not undergo any association or dissociation. For solutes undergoing such changes, van't Hoff introduced a factor, called van't Hoff factor (i). This has helped not only to explain the abnormal molecular masses of such solutes in the solution but has also helped to calculate the degree of association or dissociation.
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 (α) of the solute \( Ba(NO_3)_2 \) in a 0.1 M solution, we can use the van't Hoff factor (i) and the relationship between the degree of dissociation and the van't Hoff factor.

### Step-by-Step Solution:

1. **Understanding the Dissociation**:
The dissociation of \( Ba(NO_3)_2 \) in water can be represented as:
\[ Ba(NO_3)_2 \rightarrow Ba^{2+} + 2 NO_3^{-} \]
From this equation, we can see that 1 mole of \( Ba(NO_3)_2 \) produces 1 mole of \( Ba^{2+} \) and 2 moles of \( NO_3^{-} \), leading to a total of 3 moles of ions.

2. **Initial Concentration**:
The initial concentration of \( Ba(NO_3)_2 \) is given as 0.1 M. Therefore, initially, we have:
- Moles of \( Ba(NO_3)_2 = 0.1 \)
- Moles of \( Ba^{2+} = 0 \)
- Moles of \( NO_3^{-} = 0 \)

3. **Dissociation Representation**:
Let α be the degree of dissociation. After dissociation, the concentrations will be:
- Moles of \( Ba(NO_3)_2 = 0.1 - α \)
- Moles of \( Ba^{2+} = α \)
- Moles of \( NO_3^{-} = 2α \)

4. **Total Moles After Dissociation**:
The total number of moles after dissociation will be:
\[ \text{Total moles} = (0.1 - α) + α + 2α = 0.1 + 2α \]

5. **Van't Hoff Factor (i)**:
The van't Hoff factor (i) is given as 2.74. The relationship between the van't Hoff factor and the degree of dissociation is:
\[ i = 1 + 2α \]

6. **Setting Up the Equation**:
We can set up the equation using the given van't Hoff factor:
\[ 2.74 = 1 + 2α \]

7. **Solving for α**:
Rearranging the equation to solve for α:
\[ 2α = 2.74 - 1 \]
\[ 2α = 1.74 \]
\[ α = \frac{1.74}{2} = 0.87 \]

8. **Interpreting the Result**:
The degree of dissociation (α) is 0.87, which means that 87% of the \( Ba(NO_3)_2 \) has dissociated in the solution.

### Final Answer:
The degree of dissociation of \( Ba(NO_3)_2 \) in a 0.1 M solution is **87%**.