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) for the 0.1 M Ba(NO₃)₂ solution with a van't Hoff factor (i) of 2.74, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Dissociation of Ba(NO₃)₂**:
- The dissociation of barium nitrate can be represented as:
\[ \text{Ba(NO}_3\text{)}_2 \rightarrow \text{Ba}^{2+} + 2\text{NO}_3^{-} \]
- From this equation, we see that 1 mole of Ba(NO₃)₂ produces 1 mole of Ba²⁺ and 2 moles of NO₃⁻, leading to a total of 3 moles of ions from 1 mole of Ba(NO₃)₂.

2. **Setting Up the Initial and Final Moles**:
- Let’s assume we start with 1 mole of Ba(NO₃)₂.
- After dissociation, if α is the degree of dissociation, the moles of each species will be:
- Ba²⁺: α
- NO₃⁻: 2α
- Remaining Ba(NO₃)₂: \(1 - α\)

3. **Calculating Total Moles After Dissociation**:
- The total number of moles after dissociation will be:
\[ \text{Total moles} = (1 - α) + α + 2α = 1 + 2α \]

4. **Using the van't Hoff Factor**:
- The van't Hoff factor (i) is defined as the total number of particles in solution after dissociation divided by the number of formula units initially present.
- Therefore, we can express this as:
\[ i = \frac{\text{Total moles after dissociation}}{\text{Initial moles}} = \frac{1 + 2α}{1} \]
- Given that \(i = 2.74\), we can set up the equation:
\[ 2.74 = 1 + 2α \]

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

6. **Calculating the Degree of Dissociation in Percentage**:
- To express α as a percentage, we multiply by 100:
\[ \text{Degree of dissociation} = α \times 100 = 0.87 \times 100 = 87\% \]

### Final Answer:
The degree of dissociation of the 0.1 M Ba(NO₃)₂ solution is **87%**.

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