van't Hoff factor for `SrCl_(2)` at 0.01 M is 1.6. Percent dissociation of `SrCl_(2)` is

Options

A. 70
B. 55
C. 40
D. 30

Answer

To find the percent dissociation of \( \text{SrCl}_2 \) given its van't Hoff factor \( i \) at 0.01 M is 1.6, we can follow these steps:

### Step 1: Understand the dissociation of \( \text{SrCl}_2 \)
The dissociation of \( \text{SrCl}_2 \) can be represented as:
\[ \text{SrCl}_2 \rightarrow \text{Sr}^{2+} + 2 \text{Cl}^- \]
From this equation, we can see that one formula unit of \( \text{SrCl}_2 \) produces 3 ions in total (1 \( \text{Sr}^{2+} \) ion and 2 \( \text{Cl}^- \) ions).

### Step 2: Define the variables
Let:
- \( n \) = number of particles produced upon dissociation = 3 (from the dissociation equation)
- \( \alpha \) = percent dissociation (expressed as a fraction)

### Step 3: Use the van't Hoff factor formula
The van't Hoff factor \( i \) is given by the formula:
\[ i = 1 + (n - 1) \alpha \]
Substituting the known values:
\[ 1.6 = 1 + (3 - 1) \alpha \]

### Step 4: Simplify the equation
This simplifies to:
\[ 1.6 = 1 + 2\alpha \]
Subtract 1 from both sides:
\[ 0.6 = 2\alpha \]

### Step 5: Solve for \( \alpha \)
Now, divide both sides by 2:
\[ \alpha = \frac{0.6}{2} = 0.3 \]

### Step 6: Convert \( \alpha \) to percent dissociation
To find the percent dissociation, multiply \( \alpha \) by 100:
\[ \text{Percent dissociation} = 0.3 \times 100 = 30\% \]

### Final Answer
The percent dissociation of \( \text{SrCl}_2 \) is **30%**.

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