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 `SrCl₂`, we can use the van't Hoff factor (i) and the formula relating it to the degree of dissociation (α). Here’s a step-by-step solution:

### Step 1: Understand the dissociation of `SrCl₂`
Strontium chloride (`SrCl₂`) dissociates in water as follows:

\[ SrCl_2 \rightarrow Sr^{2+} + 2Cl^{-} \]

This means that one formula unit of `SrCl₂` produces a total of 3 ions (1 `Sr^{2+}` ion and 2 `Cl^{-}` ions).

### Step 2: Identify the van't Hoff factor (i)
The van't Hoff factor (i) is given as 1.6 at a concentration of 0.01 M. The formula for the van't Hoff factor in terms of degree of dissociation (α) is:

\[ i = 1 + (n - 1) \alpha \]

Where:
- \( n \) is the number of ions produced upon dissociation (which is 3 for `SrCl₂`).
- \( \alpha \) is the degree of dissociation.

### Step 3: Substitute values into the formula
Substituting \( n = 3 \) into the equation gives:

\[ i = 1 + (3 - 1) \alpha \]
\[ 1.6 = 1 + 2\alpha \]

### Step 4: Solve for α
Rearranging the equation to solve for α:

\[ 1.6 - 1 = 2\alpha \]
\[ 0.6 = 2\alpha \]
\[ \alpha = \frac{0.6}{2} = 0.3 \]

### Step 5: Calculate the percent dissociation
To find the percent dissociation, we multiply α by 100:

\[ \text{Percent dissociation} = \alpha \times 100 = 0.3 \times 100 = 30\% \]

### Final Answer
The percent dissociation of `SrCl₂` at 0.01 M is **30%**.

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