`K_(2)HgI_(4)` is `55%` ionized in aqueous solution. The value of Van't Hoff factor is

Options

A. 2.1
B. 4.3
C. 1.9
D. 3.7

Answer

To find the Van't Hoff factor (i) for \( K_2HgI_4 \) which is 55% ionized in aqueous solution, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( K_2HgI_4 \) in aqueous solution can be represented as:
\[ K_2HgI_4 \rightleftharpoons 2K^+ + HgI_4^{2-} \]
From this equation, we can see that one mole of \( K_2HgI_4 \) produces 2 moles of \( K^+ \) ions and 1 mole of \( HgI_4^{2-} \) ions.

### Step 2: Determine the number of particles produced (n)
From the dissociation equation, we can calculate the total number of ions produced:
- 2 moles of \( K^+ \)
- 1 mole of \( HgI_4^{2-} \)

Thus, the total number of particles (n) produced from 1 mole of \( K_2HgI_4 \) is:
\[ n = 2 + 1 = 3 \]

### Step 3: Determine the degree of ionization (α)
The degree of ionization (α) is given as 55%, which can be expressed as a decimal:
\[ \alpha = \frac{55}{100} = 0.55 \]

### Step 4: Use the formula for Van't Hoff factor (i)
The Van't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
Substituting the values of n and α into the formula:
\[ i = 1 + (3 - 1) \cdot 0.55 \]
\[ i = 1 + 2 \cdot 0.55 \]
\[ i = 1 + 1.1 \]
\[ i = 2.1 \]

### Step 5: Conclusion
The value of the Van't Hoff factor (i) for \( K_2HgI_4 \) is:
\[ \boxed{2.1} \]