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. 0.87
C. 1
D. 0.74

Answer

To find the degree of dissociation (α) for the solution of barium nitrate, we can follow these steps:

### Step 1: Understand the van't Hoff factor (i)
The van't Hoff factor (i) is defined as the ratio of the number of particles in solution after dissociation to the number of formula units initially dissolved. For barium nitrate, Ba(NO₃)₂, when it dissociates in solution, it produces barium ions and nitrate ions.

### Step 2: Write the dissociation reaction
The dissociation of barium nitrate in water can be represented as:
\[ \text{Ba(NO}_3\text{)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{NO}_3^{-} \]

From this reaction, we can see that one formula unit of Ba(NO₃)₂ produces:
- 1 Ba²⁺ ion
- 2 NO₃⁻ ions

This results in a total of:
\[ n = 1 + 2 = 3 \text{ ions} \]

### Step 3: Use the formula for the van't Hoff factor
The formula for the van't Hoff factor is given by:
\[ i = 1 + \alpha (n - 1) \]
Where:
- \( i \) = van't Hoff factor
- \( \alpha \) = degree of dissociation
- \( n \) = number of ions produced from one formula unit

### Step 4: Substitute the known values
Given:
- \( i = 2.74 \)
- \( n = 3 \)

Substituting these values into the formula:
\[ 2.74 = 1 + \alpha (3 - 1) \]
\[ 2.74 = 1 + 2\alpha \]

### Step 5: Solve for α
Rearranging the equation:
\[ 2\alpha = 2.74 - 1 \]
\[ 2\alpha = 1.74 \]
\[ \alpha = \frac{1.74}{2} \]
\[ \alpha = 0.87 \]

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

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

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