The van't Hoff factor of `BaCl_2` at 0.01 M concentration is 1.98. The percentage of dissociation of `BaCl_2` at this concentration is:

Options

A. 49
B. 69
C. 89
D. 79

Answer

To find the percentage of dissociation of \( BaCl_2 \) at a concentration of 0.01 M, we can follow these steps:

### Step 1: Understand the dissociation of \( BaCl_2 \)
When \( BaCl_2 \) dissociates in solution, it breaks down into one barium ion (\( Ba^{2+} \)) and two chloride ions (\( 2Cl^- \)):
\[ BaCl_2 \rightarrow Ba^{2+} + 2Cl^- \]

### Step 2: Set up the initial concentrations
At the start (time \( t = 0 \)), the concentration of \( BaCl_2 \) is 0.01 M. We will denote the degree of dissociation as \( x \).

- Initial concentration of \( BaCl_2 \) = 0.01 M
- At equilibrium:
- Concentration of \( Ba^{2+} \) = \( x \)
- Concentration of \( Cl^- \) = \( 2x \)
- Remaining concentration of \( BaCl_2 \) = \( 0.01 - x \)

### Step 3: Calculate the van't Hoff factor
The van't Hoff factor \( i \) is defined as the number of particles in solution after dissociation divided by the number of particles before dissociation.

Before dissociation, we have 1 particle of \( BaCl_2 \). After dissociation, we have:
\[ 1 \text{ (from } Ba^{2+}) + 2 \text{ (from } 2Cl^-) = 3 \text{ particles} \]
Thus, the van't Hoff factor \( i \) can be expressed as:
\[ i = \frac{(0.01 - x) + x + 2x}{0.01} = \frac{0.01 + 2x}{0.01} \]

### Step 4: Substitute the given van't Hoff factor
We know from the problem that \( i = 1.98 \). Therefore, we can set up the equation:
\[ 1.98 = \frac{0.01 + 2x}{0.01} \]

### Step 5: Solve for \( x \)
Now, we can rearrange the equation to find \( x \):
\[ 1.98 \times 0.01 = 0.01 + 2x \]
\[ 0.0198 = 0.01 + 2x \]
\[ 2x = 0.0198 - 0.01 \]
\[ 2x = 0.0098 \]
\[ x = \frac{0.0098}{2} = 0.0049 \]

### Step 6: Calculate the percentage of dissociation
The percentage of dissociation \( \alpha \) is given by:
\[ \alpha = \frac{x}{\text{initial concentration}} \times 100 \]
Substituting the values:
\[ \alpha = \frac{0.0049}{0.01} \times 100 = 49\% \]

### Final Answer
The percentage of dissociation of \( BaCl_2 \) at this concentration is **49%**.

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