If `Hg_(2)Cl_(2)` is 100% ionised, then find out its van't Hoff factor.

Explanation

To find the van't Hoff factor (i) for the compound \( \text{Hg}_2\text{Cl}_2 \) when it is 100% ionized, we can follow these steps:

### Step 1: Understand the dissociation of \( \text{Hg}_2\text{Cl}_2 \)

When \( \text{Hg}_2\text{Cl}_2 \) dissociates in solution, it breaks down into ions. The dissociation can be represented as:
\[ \text{Hg}_2\text{Cl}_2 \rightarrow \text{Hg}^{2+} + 2 \text{Cl}^- \]
From this equation, we can see that one formula unit of \( \text{Hg}_2\text{Cl}_2 \) produces a total of 3 ions (1 \( \text{Hg}^{2+} \) ion and 2 \( \text{Cl}^- \) ions).

### Step 2: Determine the total number of ions (N)

From the dissociation reaction, we find:
\[ N = 3 \]
where \( N \) is the total number of ions produced from one formula unit of the solute.

### Step 3: Calculate the degree of ionization (α)

Given that \( \text{Hg}_2\text{Cl}_2 \) is 100% ionized, we can express the degree of ionization (α) as:
\[ \alpha = \frac{100}{100} = 1 \]

### Step 4: Use the formula relating α, i, and N

The relationship between the degree of ionization (α), van't Hoff factor (i), and the number of ions (N) is given by the formula:
\[ \alpha = \frac{i - 1}{N - 1} \]
Substituting the known values into this equation:
\[ 1 = \frac{i - 1}{3 - 1} \]
This simplifies to:
\[ 1 = \frac{i - 1}{2} \]

### Step 5: Solve for the van't Hoff factor (i)

To find \( i \), we can rearrange the equation:
\[ 2 = i - 1 \]
Adding 1 to both sides gives:
\[ i = 3 \]

### Conclusion

Thus, the van't Hoff factor \( i \) for \( \text{Hg}_2\text{Cl}_2 \) when it is 100% ionized is:
\[ \boxed{3} \]